'''
Created on 16-Apr-2021

@author: rishi_priyansh
'''
from sympy import symbols, Eq, solve, Symbol, latex, sqrt, UnevaluatedExpr
from xv.km.ui import SvgHelper  
import matplotlib.pyplot as plt

import random
import inspect
import traceback

from ..helper import BaseHelper

from xv.km.managers import BaseManager
from xv.km.kelements import KProblemTemplate
import os

class MiddleschoolManager(BaseManager):
    '''
    MiddleschoolManager
    Example:
        ke = MiddleschoolManager()
        ke.getRandomProblem()
        ke.getRandomProblem(problem_type = 0)
        ...
        
        ke.printProblem()
        ke.printAnswer()
        ke.printSolution()
        
        ke.printProblemTypes()
        
    '''    
    _OPR_TYPE_COUNT = 10
    
    def __new__(cls, *args, **kwargs):
        instance = super(MiddleschoolManager, cls).__new__(cls, *args, **kwargs)
        for base_class in MiddleschoolManager.__bases__:
            base_class.__init__(instance, *args, **kwargs)
        return instance
    
    def __init__(self, *args, **kwargs):
        self.base_helper = BaseHelper(**kwargs)
        self.rah = self.base_helper.get_radio_activity_helper()
        pass

    def _init_problem_types(self, *args, **kwargs): 
        self._append_problem_types([
            '_problem_47_7',
            '_problem_47_6',
            '_problem_47_5',
            '_problem_47_4',
            '_problem_47_3',
            '_problem_47_2',
            '_problem_47_1',
            '_problem_46_5',
            '_problem_46_4',
            '_problem_46_3',
            '_problem_46_2',
            '_problem_46_1',
            '_problem_45_5',
            '_problem_45_4',
            '_problem_45_3',
            '_problem_45_2',
            '_problem_45_1',
            '_problem_44_7',
            '_problem_44_6',
            '_problem_44_5',
            '_problem_44_4',
            '_problem_44_3',
            '_problem_44_2',
            '_problem_44_1',
            '_problem_43_7',
            '_problem_43_6',
            '_problem_43_5',
            '_problem_43_4',
            '_problem_43_3',
            '_problem_43_2',
            '_problem_43_1',
            '_problem_42_2',
            '_problem_42_1',
            '_problem_41_10',
            '_problem_41_9',
            '_problem_41_8',
            '_problem_41_7',
            '_problem_41_6',
            '_problem_41_5',
            '_problem_41_4',
            '_problem_41_3',
            '_problem_41_2',
            '_problem_41_1',
            '_problem_40_3',
            '_problem_40_2',
            '_problem_40_1',
            '_problem_39_7',
            '_problem_39_6',
            '_problem_39_5',
            '_problem_39_4',
            '_problem_39_3',
            '_problem_39_2',
            '_problem_39_1',
            '_problem_38_6',
            '_problem_38_5',
            '_problem_38_4',
            '_problem_38_3',
            '_problem_38_2',
            '_problem_38_1',
            '_problem_37_10',
            '_problem_37_9',
            '_problem_37_8',
            '_problem_37_7',
            '_problem_37_6',
            '_problem_37_5',
            '_problem_37_4',
            '_problem_37_3',
            '_problem_37_2',
            '_problem_37_1',
            '_problem_36_6',
            '_problem_36_5',
            '_problem_36_4',
            '_problem_36_3',
            '_problem_36_2',
            '_problem_36_1',
            '_problem_35_7',
            '_problem_35_6',
            '_problem_35_5',
            '_problem_35_4',
            '_problem_35_3',
            '_problem_35_2',
            '_problem_35_1',
            '_problem_34_7',
            '_problem_34_6',
            '_problem_34_5',
            '_problem_34_4',
            '_problem_34_3',
            '_problem_34_2',
            '_problem_34_1',
            '_problem_33_11',
            '_problem_33_10',
            '_problem_33_9',
            '_problem_33_8',
            '_problem_33_7',
            '_problem_33_6',
            '_problem_33_5',
            '_problem_33_4',
            '_problem_33_3',
            '_problem_33_2',
            '_problem_33_1',
            '_problem_32_5',
            '_problem_32_4',
            '_problem_32_3',
            '_problem_32_2',
            '_problem_32_1',
            '_problem_31_3',
            '_problem_31_2',
            '_problem_31_1',
            '_problem_30_5',
            '_problem_30_4',
            '_problem_30_3',
            '_problem_30_2',
            '_problem_30_1',
            '_problem_29_12',
            '_problem_29_11',
            '_problem_29_10',
            '_problem_29_9',
            '_problem_29_8',
            '_problem_29_7',
            '_problem_29_6',
            '_problem_29_5',
            '_problem_29_4',
            '_problem_29_3',
            '_problem_29_2',
            '_problem_29_1',
            '_problem_28_10',
            '_problem_28_9',
            '_problem_28_8',
            '_problem_28_7',
            '_problem_28_6',
            '_problem_28_5',
            '_problem_28_4',
            '_problem_28_3',
            '_problem_28_2',
            '_problem_28_1',
            '_problem_27_14',
            '_problem_27_13',
            '_problem_27_12',
            '_problem_27_11',
            '_problem_27_10',
            '_problem_27_9',
            '_problem_27_8',
            '_problem_27_7',
            '_problem_27_6',
            '_problem_27_5',
            '_problem_27_4',
            '_problem_27_3',
            '_problem_27_2',
            '_problem_27_1',
            '_problem_26_6',
            '_problem_26_5',
            '_problem_26_4',
            '_problem_26_3',
            '_problem_26_2',
            '_problem_26_1',
            '_problem_25_8',
            '_problem_25_7',
            '_problem_25_6',
            '_problem_25_5',
            '_problem_25_4',
            '_problem_25_3',
            '_problem_25_2',
            '_problem_25_1',
            '_problem_24_5',
            '_problem_24_4',
            '_problem_24_3',
            '_problem_24_2',
            '_problem_24_1',
            '_problem_23_11',
            '_problem_23_10',
            '_problem_23_9',
            '_problem_23_8',
            '_problem_23_7',
            '_problem_23_6',
            '_problem_23_5',
            '_problem_23_4',
            '_problem_23_3',
            '_problem_23_2',
            '_problem_23_1',
            '_problem_22_7',
            '_problem_22_6',
            '_problem_22_5',
            '_problem_22_4',
            '_problem_22_3',
            '_problem_22_2',
            '_problem_21_7',
            '_problem_21_6',
            '_problem_21_5',
            '_problem_21_4',
            '_problem_21_3',
            '_problem_21_2',
            '_problem_21_1',
            '_problem_20_10',
            '_problem_20_9',
            '_problem_20_8',
            '_problem_20_7',
            '_problem_20_6',
            '_problem_20_5',
            '_problem_20_4',
            '_problem_20_3',
            '_problem_20_2',
            '_problem_20_1',
            '_problem_19_12',
            '_problem_19_11',
            '_problem_19_10',
            '_problem_19_9',
            '_problem_19_8',
            '_problem_19_7',
            '_problem_19_6',
            '_problem_19_5',
            '_problem_19_4',
            '_problem_19_3',
            '_problem_19_2',
            '_problem_19_1',
            '_problem_18_8',
            '_problem_18_7',
            '_problem_18_6',
            '_problem_18_5',
            '_problem_18_4',
            '_problem_18_3',
            '_problem_18_2',
            '_problem_18_1',
            '_problem_17_13',
            '_problem_17_12',
            '_problem_17_11',
            '_problem_17_10',
            '_problem_17_9',
            '_problem_17_8',
            '_problem_17_7',
            '_problem_17_6',
            '_problem_17_5',
            '_problem_17_4',
            '_problem_17_3',
            '_problem_17_2',
            '_problem_17_1',
            '_problem_16_9',
            '_problem_16_8',
            '_problem_16_7',
            '_problem_16_6',
            '_problem_16_5',
            '_problem_16_4',
            '_problem_16_3',
            '_problem_16_2',
            '_problem_16_1',
            '_problem_15_10',
            '_problem_15_9',
            '_problem_15_8',
            '_problem_15_7',
            '_problem_15_6',
            '_problem_15_5',
            '_problem_15_4',
            '_problem_15_3',
            '_problem_15_2',
            '_problem_15_1',
            '_problem_14_5',
            '_problem_14_4',
            '_problem_14_3',
            '_problem_14_2',
            '_problem_14_1',
            '_problem_13_8',
            '_problem_13_7',
            '_problem_13_6',
            '_problem_13_5',
            '_problem_13_4',
            '_problem_13_3',
            '_problem_13_2',
            '_problem_13_1',
            '_problem_12_10',
            '_problem_12_9',
            '_problem_12_8',
            '_problem_12_7',
            '_problem_12_6',
            '_problem_12_5',
            '_problem_12_4',
            '_problem_12_3',
            '_problem_12_2',
            '_problem_12_1',
            '_problem_11_11',
            '_problem_11_10',
            '_problem_11_9',
            '_problem_11_8',
            '_problem_11_7',
            '_problem_11_6',
            '_problem_11_5',
            '_problem_11_4',
            '_problem_11_3',
            '_problem_11_2',
            '_problem_11_1',
            '_problem_10_5',
            '_problem_10_4',
            '_problem_10_3',
            '_problem_10_2',
            '_problem_10_1',
            '_problem_9_11',
            '_problem_9_10',
            '_problem_9_9',
            '_problem_9_8',
            '_problem_9_7',
            '_problem_9_6',
            '_problem_9_5',
            '_problem_9_4',
            '_problem_9_3',
            '_problem_9_2',
            '_problem_9_1',
            '_problem_8_11',
            '_problem_8_10',
            '_problem_8_9',
            '_problem_8_8',
            '_problem_8_7',
            '_problem_8_6',
            '_problem_8_5',
            '_problem_8_4',
            '_problem_8_3',
            '_problem_8_2',
            '_problem_8_1',
            '_problem_7_5',
            '_problem_7_4',
            '_problem_7_3',
            '_problem_7_2',
            '_problem_7_1',
            '_problem_6_5',
            '_problem_6_4',
            '_problem_6_3',
            '_problem_6_2',
            '_problem_6_1',
            '_problem_5_18',
            '_problem_5_17',
            '_problem_5_16',
            '_problem_5_15',
            '_problem_5_14',
            '_problem_5_13',
            '_problem_5_12',
            '_problem_5_11',
            '_problem_5_10',
            '_problem_5_9',
            '_problem_5_8',
            '_problem_5_7',
            '_problem_5_6',
            '_problem_5_5',
            '_problem_5_4',
            '_problem_5_3',
            '_problem_5_2',
            '_problem_5_1',
            '_problem_4_5',
            '_problem_4_4',
            '_problem_4_3',
            '_problem_4_2',
            '_problem_4_1',
            '_problem_3_10',
            '_problem_3_9',
            '_problem_3_8',
            '_problem_3_7',
            '_problem_3_6',
            '_problem_3_5',
            '_problem_3_4',
            '_problem_3_3',
            '_problem_3_2',
            '_problem_3_1',
            '_problem_2_9',
            '_problem_2_8',
            '_problem_2_7',
            '_problem_2_6',
            '_problem_2_5',
            '_problem_2_4',
            '_problem_2_3',
            '_problem_2_2',
            '_problem_2_1',
            '_problem_1_8',
            '_problem_1_7',
            '_problem_1_6',
            '_problem_1_5',
            '_problem_1_4',
            '_problem_1_3',
            '_problem_1_2',
            '_problem_1_1',
            
            
            ])
        pass

    def _problem_47_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Intergalactic space travelers need to know the relative velocities and masses of their space ships. Each ship, therefore, has a $1.0 \\mathrm{{~m}}$ long bar painted on the side of the ship alongside their rest mass. As you pass by a ship you measure this $1.0 \\mathrm{{~m}}$ bar as $0.93 \\mathrm{{~m}}$. What is your relative velocity? You also observe their rest mass printed as $365,000 \\mathrm{{~kg}}$. What is their mass relative to you? What does an observer in the other ship measure for your $1.0 \\mathrm{{~m}}$ bar?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The length you observe is $\\displaystyle L=L_o \\sqrt{{1-v^2 / \\mathrm{{c}}^2}}$ where $L_v$ is the $1.0 \\mathrm{{~m}}$, the length an observer at rest with respect to the vehicle would measure, and $L$ is the length you measure so
        $\\displaystyle 0.93=1.0 \\sqrt{{1-v^2 / \\mathrm{{c}}^2}}$ or $v=0.37 \\mathrm{{c}}$
        The relativistic mass you observe is
        $\\displaystyle m=\\frac{{m_o}}{{\\sqrt{{1-v^2 / \\mathrm{{c}}^2}}}}=\\frac{{365,000 \\mathrm{{~kg}}}}{{0.93}}=392,000 \\mathrm{{~kg}}$
        Observers in the other space ship measure your bar as $0.93 \\mathrm{{~m}}$ and relative speed as $0.37 \\mathrm{{c}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_47_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the effective mass of an electron moving at $0.80 \\mathrm{{c}}$ ?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle m=\\frac{{m_o}}{{\\sqrt{{1-v^2 / \\mathrm{{c}}^2}}}}=\\frac{{9.1 \\times 10^{{-31}} \\mathrm{{~kg}}}}{{\\sqrt{{1-0.80^2}}}}=15.2 \\times 10^{{-31}} \\mathrm{{~kg}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_47_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        How fast must a meter stick be traveling relative to your reference frame for you to observe a $2 \\%$ contraction?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Use equation 47-3 with $L=0.98 L_{{\\circ}}$
        $\\displaystyle 0.98 L_o=L_i \\sqrt{{1-v^2 / \\mathrm{{c}}^2}}$ or $v=0.20 \\mathrm{{c}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_47_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        In a soaring space ship (a vehicle capable of both space flight and conventional aircraft flight) you return from a space journey to find a new landing runway. Passing over this runway at $0.92 \\mathrm{{c}}$ you measure its length as $1960 \\mathrm{{~m}}$. What is its length at your landing speed of $200 \\mathrm{{~m}} / \\mathrm{{s}}$ ?
 
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        At $200 \\mathrm{{~m}} / \\mathrm{{s}}$ there is no ohservable relativistic effect. Use equation 47-3 to find the length of the runway at your landing speed.
        $\\displaystyle L_0=\\frac{{L}}{{\\sqrt{{1-v^2 / \\mathrm{{c}}^2}}}}=\\frac{{1960 \\mathrm{{~m}}}}{{\\sqrt{{1-0.92^2}}}}=5000 \\mathrm{{~m}}$ 
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_47_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A "strange particle" is observed to move at $0.96 \\mathrm{{c}}$ and have a lifetime of $2.0 \\times 10^{{-8}} \\mathrm{{~s}}$. What is the lifetime of the particle in its own reference frame?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        In equation 47-1 $\\Delta t_0$ is the (proper) time an observer traveling with the "strange particle" would measure so find
        $\\displaystyle \\Delta t_o=\\Delta t \\sqrt{{1-v^2 / \\mathrm{{c}}^2}}=2.0 \\times 10^{{-8}} \\mathrm{{~s}} \\sqrt{{1-0.96^2}}=0.56 \\times 10^{{-8}} \\mathrm{{~s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_47_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation of the previous problem take the moving observer from the train to a $2500 \\mathrm{{~m}} / \\mathrm{{s}}$ jet plane and then to a rocket ship traveling at 0.90 c. (It is common, and very convenient in relativistic calculations, to express velocity as a fraction or decimal times $\\mathrm{{c}}$.)

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Again using the binomial expansion and looking at the first term
        $\\displaystyle \\frac{{1}}{{2}} v^2=\\frac{{1}}{{2}}\\left(\\frac{{2500}}{{3.0 \\times 10^8}}\\right)^2=3.5 \\times 10^{{-11}}$
        This small a time difference is observable but it is a difficult experiment.
        For the rocket ship traveling at $0.90 \\mathrm{{c}}$, use equation 47-1 directly.
        $\\displaystyle \\Delta t=\\frac{{\\Delta t_o}}{{\\sqrt{{1-0.90^2}}}}=\\frac{{\\Delta t_o}}{{0.43}}=\\frac{{0.45 \\mathrm{{~s}}}}{{0.43}}=1.0 \\mathrm{{~s}}$
        The observer in the rocket ship measures the time for the falling coin as 1.0s, over twice what the stationary observer sees. These two problems illustrate how relativistic effects are observable only as relative velocities approach c.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_47_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        You are in a railroad car moving at constant velocity of zero with respect to the surface of the earth. What do you measure as the time it takes for a coin to drop $1.0 \\mathrm{{~m}}$ to the floor of the railroad car? What does an observer in another railroad car traveling at a constant $25 \\mathrm{{~m}} / \\mathrm{{s}}$ measure for this time?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        You apply $s=\\frac{{1}}{{2}} g t^2$ and observe $\\displaystyle \\quad t=\\sqrt{{\\frac{{2 s}}{{g}}}}=\\sqrt{{\\frac{{2 \\cdot 1.0 \\mathrm{{~m}}}}{{9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2}}}}=$
        The moving observer measures a time given by equation $47-1$. For small $v / c$ the best way to make this calculation is to use a binomial expansion of the radical (see, the Introduction. Mathematical Background).
        $\\displaystyle \\Delta t=\\frac{{\\Delta t_o}}{{\\sqrt{{1-v^2 / \\mathrm{{c}}^2}}}}=\\Delta t_o\\left(1-v^2 / \\mathrm{{c}}^2\\right)^{{-1 / 2}}=\\Delta t_o\\left[1+\\frac{{1}}{{2}} \\frac{{v^2}}{{\\mathrm{{c}}^2}}+\\frac{{3}}{{8}} \\frac{{v^4}}{{\\mathrm{{c}}^4}}+\\ldots\\right] \\quad \\text {{ (47-2) }}$

        $\\displaystyle \\text {{ The first, and largest, relativistic correction term is }} \\frac{{1}}{{2}} \\frac{{v^2}}{{\\mathrm{{c}}^2}}=\\frac{{1}}{{2}}\\left(\\frac{{25}}{{3.0 \\times 10^8}}\\right)^2=3.5 \\times 10^{{-15}} \\text {{. }}$
        This means that the correction is $\\Delta t=\\Delta t_0\\left(1+3.5 \\times 10^{{-15}}\\right) s$. Time measurement this precise is beyond the capability of the cesium clocks. This is an unobservable effect.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_46_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A grating with 2000 lines per $\\mathrm{{cm}}$ is illuminated with a hydrogen gas discharge tube. Two of the hydrogen lines are at 410 and $434 \\mathrm{{~nm}}$. What is the first order spacing of these lines on a screen $1.0 \\mathrm{{~m}}$ from the grating. The experimental arrangement is shown in Fig. 46-4.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')

        file_name = 'how-to-solve-physics-problems-fig-46-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The phrase "first order" refers to the first maxima.
        First find the spacing $1.0 \\times 10^{{-2}} \\mathrm{{~m}} / 2000$ lines $=5.0 \\times 10^{{-6}} \\mathrm{{~m}}$.
        For the $410 \\mathrm{{~nm}}$ line
        $\\displaystyle y_1=\\frac{{1 \\cdot \\lambda R}}{{d}}=\\lambda \\frac{{R}}{{d}}=410 \\times 10^{{-9}} \\mathrm{{~m}} \\frac{{1.0 \\mathrm{{~m}}}}{{5.0 \\times 10^{{-6}} \\mathrm{{~m}}}}=0.082 \\mathrm{{~m}}$
        For the $434 \\mathrm{{~nm}}$ line
        $\\displaystyle y_1=\\frac{{1 \\cdot \\lambda R}}{{d}}=\\lambda \\frac{{R}}{{d}}=434 \\times 10^{{-9}} \\mathrm{{~m}} \\frac{{1.0 \\mathrm{{~m}}}}{{5.0 \\times 10^{{-6}} \\mathrm{{~m}}}}=0.087 \\mathrm{{~m}} .$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_46_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Light from a He-Ne laser of $633 \\mathrm{{~nm}}$ is incident on a $2.0 \\times 10^{{-4}} \\mathrm{{~m}}$ wide slit. What is the width of the central maxima (the distance between the dark fringes on either side of the central maxima) on a screen $2.0 \\mathrm{{~m}}$ away.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Rearranging equation $46-4$, the distance to the first dark fringe is
        $\\displaystyle y_1=\\frac{{1 \\cdot \\lambda R}}{{a}}=\\frac{{1.633 \\times 10^{{-9}} \\mathrm{{~m}}(2.0 \\mathrm{{~m}})}}{{2.0 \\times 10^{{-4}} \\mathrm{{~m}}}}=6.3 \\times 10^{{-3}} \\mathrm{{~m}}$
        The width of the central maxima is two times this value or $1.3 \\times 10^{{-2}} \\mathrm{{~m}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_46_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Monochromatic light is incident on a $4.0 \\times 10^{{-4}} \\mathrm{{~m}}$ wide slit producing a Fraunhofer diffraction pattern on a screen $1.5 \\mathrm{{~m}}$ away. The distance from the central maxima to the second dark fringe is $3.4 \\times 10^{{-3}} \\mathrm{{~m}}$. What is the wavelength of the light?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Reworking equation 46-4 yields
        $\\displaystyle \\lambda=\\frac{{a y_m}}{{m R}}=\\frac{{4.0 \\times 10^{{-4}} \\mathrm{{~m}} \\cdot 3.4 \\times 10^{{-3}} \\mathrm{{~m}}}}{{2 \\cdot 1.5 \\mathrm{{~m}}}}=453 \\mathrm{{~nm}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_46_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        In a different Young's experiment with $488 \\mathrm{{~nm}}$ light the second dark fringe is $1.2 \\times 10^{{-3}} \\mathrm{{~m}}$ away from the center of the central bright fringe. The screen is $1.2 \\mathrm{{~m}}$ from the slits. What is the separation of the slits?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This second dark fringe corresponds to $3 / 2$ of a wavelength in path difference. One wavelength path difference is $2 / 3$ of the $1.2 \\times 10^{{-3}} \\mathrm{{~m}}$ or $8.0 \\times 10^{{-4}} \\mathrm{{~m}}$. This number now fits with equation $46-2$ (one wavelength path difference), and solving for $d$
        $\\displaystyle d=\\frac{{m \\lambda R}}{{y_m}}=\\frac{{1.488 \\times 10^{{-9}} \\mathrm{{~m}} \\cdot 1.2 \\mathrm{{~m}}}}{{8.0 \\times 10^{{-4}} \\mathrm{{~m}}}}=7.3 \\times 10^{{-4}} \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_46_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A Young's experiment is set up with slit separation $2.0 \\times 10^{{-4}} \\mathrm{{~m}}$, screen distance $0.80 \\mathrm{{~m}}$, and distance from central maxima to second bright fringe $3.9 \\times 10^{{-3}} \\mathrm{{~m}}$. What is the wavelength of the source?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle \\lambda=\\frac{{d y_m}}{{m R}}=\\frac{{2.0 \\times 10^{{-4}} \\mathrm{{~m}} \\cdot 3.9 \\times 10^{{-3}} \\mathrm{{~m}}}}{{2 \\cdot 0.80 \\mathrm{{~m}}}}=488 \\mathrm{{~nm}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_45_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A diverging lens has a $14 \\mathrm{{~cm}}$ focal length. Describe the image of a $4.0 \\mathrm{{~cm}}$ object placed $40 \\mathrm{{~cm}}$ from the lens.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')

        file_name = 'how-to-solve-physics-problems-fig-45-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Draw a ray from the top of the object to the lens parallel to the principal axis and refract it back to the focus. Next draw a line from the top of the object through the center of the lens. The intersection of these rays locates the top of the object. The image distance is from equation 45-1.
        $\\displaystyle \\frac{{1}}{{40 \\mathrm{{~cm}}}}+\\frac{{1}}{{i}}=-\\frac{{1}}{{14 \\mathrm{{~cm}}}} \\text {{ or }} i=-10 \\mathrm{{~cm}}$
        The negative sign for the focal length is because this is a diverging lens. The image distance is negative because it is on the same side of the lens as the object (opposite to the converging lens). The magnification is $10 / 40=$ 0.25 .

        The image is $0.25 \\times 4.0 \\mathrm{{~cm}}=1.0 \\mathrm{{~cm}}$ high, erect, virtual, and appears to come from a point $10 \\mathrm{{~cm}}$ from the lens on the same side as the object.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_45_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A converging lens of focal length $8.0 \\mathrm{{~cm}}$ forms an image of an object placed $20 \\mathrm{{~cm}}$ in front of the lens. Describe the image.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        file_name = 'how-to-solve-physics-problems-fig-45-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Draw a ray from the top of the object parallel to the axis then through the focus. Next draw a ray the top of the object through the center of the lens to intersect the first ray. This locates the top of the object. Use equation $45-1$ to find the image distance.
        $\\displaystyle \\frac{{1}}{{20 \\mathrm{{~cm}}}}+\\frac{{1}}{{i}}=\\frac{{1}}{{8.0 \\mathrm{{~cm}}}} \\text {{ or }} i=13 \\mathrm{{~cm}}$

        The magnification is (-) image distance over object distance or
        $\\displaystyle M=-\\frac{{i}}{{o}}=-\\frac{{13}}{{20}}=-0.67$
        The image is located $13 \\mathrm{{~cm}}$ on the side of the lens opposite the object with magnification 0.67 . It is inverted (minus sign) and real (rays pass through image).
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_45_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For a spherical convex mirror of $14 \\mathrm{{~cm}}$ radius of curvature, describe the image of a $2.5 \\mathrm{{~cm}}$ object placed $30 \\mathrm{{~cm}}$ out on the principal axis of the mirror.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Here is where we get into the signs. The focus and the image are on one side of the mirror and the object is on the other side. Therefore we take the focus as negative and expect the image distance to be negative. The image distance comes from equation 45-1
        $\\displaystyle \\frac{{1}}{{30 \\mathrm{{~cm}}}}+\\frac{{1}}{{i}}=-\\frac{{1}}{{7 \\mathrm{{~cm}}}} \\text {{ or }} \\quad i=-5.7 \\mathrm{{~cm}}$
        The magnitication comes from equation $45-2$
        $\\displaystyle M=-\\frac{{-5.7}}{{30}}=\\frac{{-14+5.7}}{{30+14}}=0.19$
        The height of the image is from $\\displaystyle \\frac{{h^{{\\prime}}}}{{2.5 \\mathrm{{~cm}}}}=0.19$, or $h^{{\\prime}}=0.48 \\mathrm{{~cm}}$
        The minus sign for the image distance indicates that the image is behind the mirror, or on the same side as all the other minus signs. The plus sign for the magnification indicates that the image is erect (not inverted). The image is virtual.


        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_45_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For a spherical concave mirror of $12 \\mathrm{{~cm}}$ radius of curvature describe the image of a $2.0 \\mathrm{{~cm}}$ height object placed $20 \\mathrm{{~cm}}$ on the center line of the mirror according to Fig. 45-2.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-45-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The image is inverted. The ray diagrams show this. Using the radius of curvature and the object distance find the image distance from
        $\\displaystyle \\frac{{1}}{{20 \\mathrm{{~cm}}}}+\\frac{{1}}{{i}}=\\frac{{2}}{{12 \\mathrm{{~cm}}}} \\text {{ or }} i=8.6 \\mathrm{{~cm}}$
        The magnification is from
        $\\displaystyle M=-\\frac{{8.6}}{{20}}=-\\frac{{12-8.6}}{{20-12}}=-0.43$
        The height of the image is from $\\displaystyle \\frac{{h^{{\\prime}}}}{{2.0 \\mathrm{{~cm}}}}=-0.43$ or $h^{{\\prime}}=-0.86 \\mathrm{{~cm}}$.
        The image is $8.6 \\mathrm{{~cm}}$ from the mirror, inverted (minus sign) and real (rays pass through image)
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_45_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A light source is $4.0 \\mathrm{{~cm}}$ in front of a plane mirror. Where does an observer looking into the mirror see the image and is it real or virtual?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The image is $4.0 \\mathrm{{~cm}}$ behind the mirror. It is a virtual image because light does not pass through this image point.
        A concave (converging) spherical mirror as shown in Fig. 45-2 can be analyzed by rays.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-45-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        An object of height $h$ placed at $o$, the object distance from the concave mirror, will produce a smaller image, $h^{{\\prime}}$ at $i$, the image distance, according to the formula
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_44_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A beam of unpolarized light in water is incident on what is reported to be diamond. By means of polaroids the angle (of incidence) for complete polarization is determined to be $61^{{\\circ}}$. Is this diamond?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        For light in water reflected from diamond the Brewster angle should be
        $\\displaystyle \\tan \\theta_p=\\frac{{n_{{r e f r}}}}{{n_{{\\text {{tnc }}}}}}=\\frac{{2.42}}{{1.33}}=1.81 \\quad \\text {{ or }} \\quad \\theta_p=61^{{\\circ}}$
        From the indexes of refraction given in Fig. 44-1 we conclude that this is diamond.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-44-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_44_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the polarization angle for light reflected from water?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle \\tan \\theta_p=\\frac{{n_{{refr}}}}{{n_{{inc}}}}=\\frac{{1.33}}{{1}} \\text {{ or }} \\theta_p=53^{{\\circ}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_44_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Light is incident at an angle of $35^{{\\circ}}$ on a slab of glass $2.0 \\mathrm{{~cm}}$ thick. Part of the light travels through the glass and part along its original direction beside the glass. Find the path of the two rays, the velocity of the ray in the glass, and the time difference between the two when they emerge from the opposite side of the glass. The geometry is shown in Fig. 44-6.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-44-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Apply Snell's law to find the path through the slab. $1.00 \\sin 35^{{\\circ}}=1.62 \\sin \\theta_r$, or $\\theta_t=21^{{\\circ}}$. Because of symmetry, the ray through the slab exits parallel to the original ray. The velocity of the ray in the glass is from $n=c / v$, or $v=c / n=3.0 \\times 10^8 \\mathrm{{~m}} / \\mathrm{{s}} / 1.62=1.85 \\times 10^8 \\mathrm{{~m}} / \\mathrm{{s}}$. Now that the velocities are known the time difference requires finding the distance the rays travel.

        The distance $a$ (the path through the slab) comes from $\\cos 21^{{\\circ}}=2.0 \\mathrm{{~cm}} / a$, or $a=2.14 \\mathrm{{~cm}}$. Construct a line perpendicular to the two rays at the point where the refracted one leaves the slab.
        The distance $b$ in the original ray direction comes from $\\cos 14^{{\\circ}}=b / 2.14 \\mathrm{{~cm}}$, or $b=2.08 \\mathrm{{~cm}}$.
        The time for the refracted beam is $\\tau_{{\\mathrm{{r}}}}=0.0211 \\mathrm{{~m}} / 1.85 \\times 10^8 \\mathrm{{~m}} / \\mathrm{{s}}=11.6 \\times 10^{{-11}} \\mathrm{{~S}}$.
        The time for the unaltered beam is $\\tau=0.0208 \\mathrm{{~m}} / 3.0 \\times 10^8 \\mathrm{{~m}} / \\mathrm{{s}}=6.9 \\times 10^{{-11}} \\mathrm{{~S}}$.
        The time delay is $4.7 \\times 10^{{-11}} \\mathrm{{~S}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_44_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Find the critical angle for a fiber optic material with index 1.50 . 
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-44-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The critical angle is the one where the angle of refraction is $90^{{\\circ}}$. Applying Snell's law 
        $1.50 \\sin \\theta_c=1.00 \\sin 90^{{\\circ}}$ or $\\theta_c=42^n$
        Any beam that strikes the fiber-air interface at an angle greater than $42^{{\\circ}}$ will be internally reflected down the fiber.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_44_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A light ray in water is incident on the water-air interface at an angle of $45^{{\\circ}}$. What is the angle of reflection in the water and refraction into the air?        
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-44-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The problem is diagrammed in Fig. 44-2. The incident angle equals the reflected angle, so $\\theta_1=45^{{\\circ}}$ The refracted angle is from Snell's law
        $\\displaystyle 1.33 \\sin 45^{{\\circ}}=1.00 \\sin \\theta_r \\text {{ or }} \\theta_r=70^{{\\circ}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_44_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What are the angles of reflection and refraction for a light ray in air incident on glass at an angle of $35^{{\\circ}}$ ?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The angle of incidence is measured from the normal so the angle of reflection is $35^{{\\circ}}$ The angle of refraction is from Snell's law $n_1 \\sin \\theta_1=n_2 \\sin \\theta_2$
        $1.00 \\sin 35^{{\\circ}}=1.62 \\sin \\theta_2$ or $\\theta_2=21^{{\\circ}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_44_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A ray of light is incident on a reflecting surface at $70^{{\\circ}}$ with respect to the surface. What is the angle of the reflected ray?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The $70^{{\\circ}}$ with respect to the surface is $20^{{\\circ}}$ with respect to the normal so the reflected beam is $20^{{\\circ}}$ from the normal.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_43_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A light beam with energy flux $20 \\mathrm{{~W}} / \\mathrm{{cm}}^2$ falls normal onto a completely reflecting surface of $2.0 \\mathrm{{~cm}}^2$. What is the force on this reflector?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The average force is $F=\\Delta p / \\Delta t$ and $\\Delta p=2 \\Delta U / c$, but $\\Delta U=S A \\Delta t$, so
        $\\displaystyle F=\\frac{{2 S}}{{c}} A=\\frac{{2\\left(20 \\mathrm{{~W}} / \\mathrm{{cm}}^2\\right)}}{{3.0 \\times 10^8 \\mathrm{{~m}} / \\mathrm{{s}}}} 2.0 \\mathrm{{~cm}}^2=2.7 \\times 10^{{-7}} \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_43_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the maximum clectric field strength and intensity of a $500 \\mathrm{{~W}}$ spherically symmetric light source at $10 \\mathrm{{~m}}$ radius?
        
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The area of a sphere $10 \\mathrm{{~m}}$ in radius is $A=4 \\pi(10 \\mathrm{{~m}})^2=1.2 \\times 10^3 \\mathrm{{~m}}^2$. The entire $500 \\mathrm{{~W}}$ is delivered to this area. The power, intensity, and area are related by
        $\\displaystyle S=\\frac{{P}}{{A}}=\\frac{{500 \\mathrm{{~W}}}}{{1.2 \\times 10^3 \\mathrm{{~m}}^2}}=0.40 \\mathrm{{~W}} / \\mathrm{{m}}^2$
        The electric field intensity comes from
        $\\displaystyle E_{{\\max }}-\\left[2 \\mu_o c S\\right]^{{1 / 2}}=\\left[2\\left(4 \\pi \\times 10^{{-7}} \\mathrm{{~T}} \\cdot \\mathrm{{m}} / \\mathrm{{A}}\\right)\\left(3.0 \\times 10^8 \\mathrm{{~m}} / \\mathrm{{s}}\\right)\\left(0.40 \\mathrm{{~W}} / \\mathrm{{m}}^2\\right)\\right]^{{1 / 2}}=17 \\mathrm{{~V}} / \\mathrm{{m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_43_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The maximum value of electric field $2.0 \\mathrm{{~m}}$ away from a spherically symmetric source is $1.8 \\mathrm{{~V}} / \\mathrm{{m}}$. Find the maximum value of magnetic field, average intensity, and power output of the source.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The magnetic field is from
        $\\displaystyle B=\\frac{{E}}{{c}}=\\frac{{1.8 \\mathrm{{~V}} / \\mathrm{{m}}}}{{3.0 \\times 10^8 \\mathrm{{~m}} / \\mathrm{{s}}}}=6.0 \\times 10^{{-9}} \\mathrm{{~T}}$
        The average intensity is from
        $\\displaystyle S=\\frac{{1}}{{2}} \\varepsilon_o c E_{{\\max }}^2=\\frac{{1}}{{2}} 8.8 \\times 10^{{-12}} \\frac{{\\mathrm{{C}}^2}}{{\\mathrm{{~N}} \\cdot \\mathrm{{m}}^2}} 3.0 \\times 10^8 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}\\left(\\frac{{1.8 \\mathrm{{~V}}}}{{\\mathrm{{~m}}}}\\right)^2=4.3 \\times 10^{{-3}} \\frac{{\\mathrm{{W}}}}{{\\mathrm{{m}}^2}}$
        As an exercise work through the units in this problem.
        The total power is the power per unit area at $2.0 \\mathrm{{~m}}$ radius times the total area.
        $\\displaystyle P=4.3 \\times 10^{{-3}} \\mathrm{{~W}} / \\mathrm{{m}}^2 \\cdot 4 \\pi(2.0 \\mathrm{{~m}})^2=0.21 \\mathrm{{~W}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-43-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_43_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Take the intensity of sunlight at the earth's surface as $600 \\mathrm{{~W}} / \\mathrm{{m}}^2$. Assuming $100 \\%$ collection efficiency, how much energy is collected on a $1.0 \\mathrm{{~m}}^2$ panel exposed to this amount of sunlight for 10 hours?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-44-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The key to getting problems involving energy, power, and intensity correct is to watch the units closely. In this case notice how the units dictate how to make the calculation.
        $\\displaystyle E=\\frac{{600 \\mathrm{{~W}} \\mathrm{{~J}} / \\mathrm{{s}}}}{{\\mathrm{{m}}^2}} \\cdot 1.0 \\mathrm{{~m}}^2 \\cdot 10 \\mathrm{{hr}} \\frac{{60 \\mathrm{{~min}}}}{{\\mathrm{{hr}}}} \\frac{{60 \\mathrm{{~s}}}}{{\\min }}=2.2 \\times 10^7 \\mathrm{{~J}}$
        Another way to do this problem is to find the energy in W.hr rather than Joules.
        $E=\\frac{{600 \\mathrm{{~W}}}}{{\\mathrm{{~m}}^2}} 1.0 \\mathrm{{~m}}^2 \\cdot 10 \\mathrm{{hr}}=6.0 \\mathrm{{~kW}} \\cdot \\mathrm{{hr}}$
        This is a strange looking unit, but it gives us a feel for the amount of energy that can be collected this way. This $6000 \\mathrm{{~W}} \\cdot \\mathrm{{hr}}$ would light sixty $100 \\mathrm{{~W}}$ light bulbs for an hour or run a $1.0 \\mathrm{{~kW}}$ microwave oven for 6 hours.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_43_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        How far does light travel in $1.0 \\mathrm{{~ns}}\\left(1.0 \\times 10^{{-9}} \\mathrm{{~s}}\\right)$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-44-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        $\\displaystyle d=c t=\\left(3.0 \\times 10^8 \\mathrm{{~m}} / \\mathrm{{s}}\\right)\\left(1.0 \\times 10^{{-9}} \\mathrm{{~s}}\\right)=0.30 \\mathrm{{~m}} \\approx 1 \\mathrm{{ft}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_43_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        How long does it take for an electromagnetic wave to travel to the moon, $3.8 \\times 10^8 \\mathrm{{~m}}$ away?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle t=\\frac{{d}}{{c}}=\\frac{{3.8 \\times 10^8 \\mathrm{{~m}}}}{{3.0 \\times 10^8 \\mathrm{{~m}} / \\mathrm{{s}}}}=1.3 \\mathrm{{~s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_43_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A radio signal at a certain point has a measured maximum electric field of $5.0 \\times 10^{{-3}} \\mathrm{{~V}} / \\mathrm{{m}}$. What is the maximum magnetic field?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The magnetic field maximum is at right angles to the electric field and has magnitude
        $\\displaystyle B=\\frac{{E}}{{c}}=\\frac{{5.0 \\times 10^{{-3}} \\mathrm{{~V}} / \\mathrm{{m}}}}{{3.0 \\times 10^8 \\mathrm{{~m}} / \\mathrm{{s}}}}=1.7 \\times 10^{{-11}} \\mathrm{{~T}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_42_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Find the maximum magnetic field at the edge of a parallel plate capacitor with radius $6.0 \\mathrm{{~cm}}$ and separation 2.0cm connected to a sinusoidal voltage source of $1000 \\mathrm{{~Hz}}$ and $200 \\mathrm{{~V}}$ maximum.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The basic equation is $\\displaystyle B=\\mu_o \\varepsilon_o \\frac{{r}}{{2}} \\frac{{d E}}{{d t}}$ but for this case $\\displaystyle \\frac{{d E}}{{d t}}=\\frac{{1}}{{d}} \\frac{{d V}}{{d t}}$ so
        $\\displaystyle \\left.B\\right|_{{m \\sigma x}}=\\left.\\mu_o E_o \\frac{{r}}{{2}} \\frac{{1}}{{d}} \\frac{{d V}}{{d t}}\\right|_{{\\max}}$
        The voltage is $\\displaystyle v=\\mathrm{{V}}_{{\\mathrm{{o}}}} \\sin \\omega t$ and $\\displaystyle \\frac{{d V}}{{d t}}=V_o \\omega \\cos \\omega t$ so $\\displaystyle\\left.\\frac{{d V}}{{d t}}\\right|_{{\\max }}=V_o \\omega$ and
        $\\displaystyle B_{{\\max }}=\\mu_o \\varepsilon_o \\frac{{r}}{{2 d}} V_o \\omega=\\left(4 \\pi \\times 10^{{-7}}\\right)\\left(8.8 \\times 10^{{-12}}\\right) \\frac{{6.0}}{{2 \\cdot 2.0}}(200)(2 \\pi \\cdot 1000) \\mathrm{{T}}=2.1 \\times 10^{{-11}} \\mathrm{{~T}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_42_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Find the displacement current and the magnetic field in a parallel plate capacitor with circular plates of radius $5.0 \\mathrm{{~cm}}$ at an instant when $E$ is changing at $1.0 \\times 10^{{12}} \\mathrm{{~V}} / \\mathrm{{m}} . \\mathrm{{s}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The displacement current is
        $\\displaystyle i_d=\\varepsilon_o A \\frac{{d E}}{{d t}}=8.8 \\times 10^{{-12}} \\mathrm{{C}}^2 / \\mathrm{{N}} \\cdot \\mathrm{{m}}^2\\left(\\pi \\cdot 25 \\times 10^{{-4}} \\mathrm{{~m}}^2\\right) 1.0 \\times 10^{{12}} \\mathrm{{~V}} / \\mathrm{{m}} \\cdot \\mathrm{{s}}=0.069 \\mathrm{{~A}}$
        The magnetic field is
        $\\displaystyle B=\\mu_0 \\varepsilon_0 \\frac{{r}}{{2}} \\frac{{d E}}{{d t}}=\\left(4 \\pi \\times 10^{{-7}} \\mathrm{{~T}} \\cdot \\mathrm{{m}} / \\mathrm{{A}}\\right)\\left(88 \\times 10^{{-12}} \\mathrm{{C}}^2 / \\mathrm{{N}} \\cdot \\mathrm{{m}}^2\\right)\\left(25 \\times 10^{{-2}} \\mathrm{{~m}}\\right)\\left(10 \\times 10^{{12}} \\mathrm{{~V}} / \\mathrm{{m}} \\cdot \\mathrm{{s}}\\right)=28 \\times 10^{{-7}} \\mathrm{{~T}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_41_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A power distribution transformer steps voltage down from $8.5 \\mathrm{{kV}}$ to $120 \\mathrm{{~V}}$ (both rms values). If the $120 \\mathrm{{~V}}$ side of the transformer supplies $500 \\mathrm{{~A}}$ to a resistive load, what current is taken from the primary (high) side of the transformer and what is the turns ratio?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First find the turns ratio
        $\\displaystyle \\frac{{8.5 \\mathrm{{kV}}}}{{N_p}}=\\frac{{120 \\mathrm{{~V}}}}{{N_s}} \\text {{ or }} \\frac{{N_p}}{{N_s}}=\\frac{{8500}}{{120}}=71 .$
        The current is
        $\\displaystyle I_p=\\frac{{N_s}}{{N_p}} I_s=\\frac{{1}}{{71}} 500 \\mathrm{{~A}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_41_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the circuit described in problem 41-6 calculate the power loss in each of the components.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        There is no energy loss in an inductor. Energy goes in to build up the magnetic field and is released in the collapse of the magnetic field. Look at the curves of Fig. 11-7. The product vi averaged over one cycle adds to zero. A similar argument can be made for a capacitor.
        For the resistor the voltage is $v=(11.6 \\mathrm{{~V}}) \\cos \\omega t$ and the current is $i=(0.058 \\mathrm{{~A}}) \\cos \\omega t$, so the power is $p=$ 11.6.0.058W $\\cos ^2 \\omega t$. The average power is
        $\\displaystyle P=11.6 \\cdot 0.058(1 / 2) \\mathrm{{W}}=0.34 \\mathrm{{~W}}$
        Using the rms voltages and currents
        $\\displaystyle P=\\frac{{11.6 \\mathrm{{~V}}}}{{\\sqrt{{2}}}} \\frac{{0.058 \\mathrm{{~A}}}}{{\\sqrt{{2}}}}=0.34 \\mathrm{{~W}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_41_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An R-L-C circuit consists of a $300 \\Omega$ resistor, $0.15 \\mathrm{{H}}$ inductor, and $4.5 \\mu \\mathrm{{F}}$ capacitor driven by an ac source of $800 \\mathrm{{rad}} / \\mathrm{{s}}$. Find the phase angle of the source voltage with respect to the current.
        
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Notice that the maximum voltage is not given in the problem, yet a phase diagram is to be drawn. The problem is written this way to emphasize that it is not necessary to know the maximum source voltage to find the phase angle. The phase diagrams so far have used voltage.
        The voltage across each component is the resistance or reactance times a constant, the current. Thus the diagram conld as well he drawn with resistance and reactances as well as voltage.
        Calculate the reactances:
        $\\displaystyle X_L=\\omega L=800 \\mathrm{{rad}} / \\mathrm{{s}} \\cdot 0.15 \\mathrm{{H}}=120 \\Omega$
        $$\\displaystyle X_C=\\frac{{1}}{{\\omega C}}=\\frac{{1}}{{(800 \\mathrm{{rad}} / \\mathrm{{s}}) 4.5 \\times 10^{{-6}} \\mathrm{{~F}}}}=278 \\Omega$
        Now draw the phasor diagram as shown in Fig. 41-11. Draw $I$ at some arbitrary angle $\\omega t$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-41-11.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        Draw $R$ along $I$ proportional to $300 \\Omega$. Draw $X_L$ leading $R$ by $90^{{\\circ}}$ proportional to $120 \\Omega$. Draw $X_C$ lagging $R$ by $90^{{\\circ}}$ proportional to $278 \\Omega$. Draw $X_L-X_C=-158 \\Omega$. Complete the rectangle formed by $R$ and $X_L-X_C$ and draw the diagonal, which is proportional to $Z$. The phase angle is
        $\\displaystyle \\tan \\phi=\\frac{{X_L-X_C}}{{R}}=\\frac{{-158}}{{300}} \\text {{ or }} \\phi=-28^{{\\circ}}$
        The load is resistive and capacitive, and the phase angle is $-28^{{\\circ}}$ with the source voltage lagging the current.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_41_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the circuit described in problem $41-6$ construct the phasor diagram and find the angle of the source voltage with respect to the current.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The phasor diagram is shown in Fig. 41-10. Again, follow the logic in the construction
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-41-10.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        of the diagram. Draw the phasor representing $I=0.058 \\mathrm{{~A}}$. at an arbitrary angle $\\omega t$. Draw the phasor representing $V_R=11.6 \\mathrm{{~V}}$ in the same direction as $I$. Draw $V_L=14.6 \\mathrm{{~V}}$ leading $V_R$ by $90^{{\\circ}}$. $\\Lambda$ dd $V_C=30.7 \\mathrm{{~V}}$ lagging $V_R$ by $90^{{\\circ}}$. Complete the rectangle with side $V_L-V_C$ and $V_R$. Since $V_C>V_L$ this vector points in the same direction as $V_C$. The source voltage $V$ is the diagonal of this rectangle. The phase angle is determined from equation 41-4.
        $\\displaystyle \\tan \\phi=\\frac{{V_L-V_C}}{{V_R^{{\\prime}}}}=\\frac{{14.6-30.7}}{{11.6}} \\text {{ or }} \\phi=-54^{{\\circ}}$
        The load is resistive and capacitive and (from the diagram) the source voltage lags the current by $54^{{\\circ}}$.
        After you are clear on the calculations go back over these last two problems and concentrate on the logic. The biggest pitfall in R-L-C circuit problems is losing your way!
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_41_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An R-L-C circuit with $R=200 \\Omega, L=0.40 \\mathrm{{H}}$, and $\\mathrm{{C}}=3.0 \\mu \\mathrm{{F}}$ is driven by an ac source of $20 \\mathrm{{~V}}$ maximum and frequency $100 \\mathrm{{~Hz}}$. Find the reactances, impedance, and maximum current and voltages across each of the components.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        R-L-C circuit problems can be confusing. The key to successfully solving them is to follow a logical path through the problem. The current is everywhere the same and is determined by the source voltage and the impedance. The impedance is determined by the resistance and the reactances, and the reactances are frequency dependent. As you proceed through this problem be aware of the logic in the calculations. The schematic of the circuit is shown in Fig. 41-1.
        $\\displaystyle X_L=\\omega L=2 \\pi \\cdot 100 \\mathrm{{~Hz}} \\cdot 0.40 \\mathrm{{H}}-251 \\Omega$
        $\\displaystyle X_C=\\frac{{1}}{{\\omega C}}=\\frac{{10^6}}{{2 \\pi \\cdot 100 \\mathrm{{~Hz}} \\cdot 3.0 \\mathrm{{~F}}}}=530 \\Omega$
        $\\displaystyle \\left(X_L-X_C\\right)^2=(251 \\Omega-530 \\Omega)^2=77800 \\Omega^2$
        $\\displaystyle Z=\\sqrt{{R^2+\\left(X_L-X_C\\right)^2}}=\\sqrt{{(200)^2+77800 \\Omega}}=343 \\Omega$
        $\\displaystyle I=V / Z=20 \\mathrm{{~V}} / 343 \\Omega=0.058 \\mathrm{{~A}}$
        $\\displaystyle V_R=0.058 \\mathrm{{~A}} \\cdot 200 \\Omega=11.6 \\mathrm{{~V}}$
        $\\displaystyle V_L=0.058 \\mathrm{{~A}} \\cdot 251 \\Omega=14.6 \\mathrm{{~V}}$
        $\\displaystyle V_C=0.058 \\mathrm{{~A}} \\cdot 530 \\Omega=30.7 \\mathrm{{~V}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_41_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the circuit described in problem $41-4$, what is the current when the source voltage is 12 and increasing?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Look at the graphs of voltage and current, and voltage versus time as shown in Fig. 41-8 and note where on the graph the voltage is $12 \\mathrm{{~V}}$ and increasing. This point (12 V and increasing) is between $270^{{\\circ}}$ and $360^{{\\circ}}$.
        The curve between $270^{{\\circ}}$ and $360^{{\\circ}}$ is a sine curve so the angle $\\beta=\\sin ^{{-1}}(12 / 20)=37^{{\\circ}}$.
        Therefore the point where the voltage is $12 \\mathrm{{~V}}$ and increasing is $270^{{\\circ}}+37^{{\\circ}}=307^{{\\circ}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-41-8.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The current at this point is $\\left.i_L\\right|_{{12 \\mathrm{{~V}}+}}=I_L \\sin \\omega t=(0.45 \\mathrm{{~A}}) \\sin 307^o$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_41_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An ac source of $100 \\mathrm{{~Hz}}$ and maximum voltage of $20 \\mathrm{{~V}}$ is connected to a $70 \\mathrm{{mH}}$ inductor. What is the maximum current? When the current is maximum, what is the voltage of the source?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First calculate the inductive reactance
        $\\displaystyle X_L=\\omega L=2 \\pi \\cdot 100 \\mathrm{{~Hz}} \\cdot 0 \\cdot 070 \\mathrm{{H}}=44 \\Omega$
        The maximum current is $\\displaystyle \\left.I_L\\right|_{{\\max }}=V_L / X_L=20 \\mathrm{{~V}} / 44 \\Omega=0.45 \\mathrm{{~A}}$
        Look at the graphs in Fig. 41-7 and note that when the current is a maximum, the voltage is zero.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-41-7.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_41_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        At what frequency do a $65 \\mathrm{{mH}}$ inductor and a $20 \\mu \\mathrm{{F}}$ capacitor have the same reactance?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        When $X_L=X_O$ or $\\omega L=\\frac{{1}}{{\\omega C}}$, The frequency then is
        $\\displaystyle \\omega=\\frac{{1}}{{\\sqrt{{L C}}}} \\text {{ or }} f=\\frac{{1}}{{2 \\pi \\sqrt{{L C}}}}$
        This is the condition for resonance or oscillation in the L-C circuit!
        $\\displaystyle f=\\frac{{1}}{{2 \\pi \\sqrt{{65 \\times 10^{{-3}} \\mathrm{{H}} \\cdot 20 \\times 10^{{-6}} \\mathrm{{~F}}}}}}=140 \\mathrm{{~Hz}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_41_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $120 \\mathrm{{mH}}$ inductor is connected to a variable frequency ac source with maximum voltage $10 \\mathrm{{~V}}$. What is the inductive reactance at $100 \\mathrm{{~Hz}}$, and $1.0 \\mathrm{{MHz}}$ ?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle \\left.X_L\\right|_{{100}}=\\omega L=2 \\pi \\cdot 100 \\mathrm{{~Hz}} \\cdot 0.12 \\mathrm{{H}}=75 \\Omega$
        $\\displaystyle \\left.X_L\\right|_{{10 \\mathrm{{M}}}}=2 \\pi \\cdot 1.0 \\times 10^6 \\mathrm{{~Hz}} \\cdot 0.12 \\mathrm{{H}}=7.5 \\times 10^5 \\Omega$
        The basic relation $v=L(d i / d t)$ shows that $[\\mathrm{{H}}=\\mathrm{{V}} . \\mathrm{{s}} / \\mathrm{{A}}]$ making $X_L$, have the units of $\\Omega$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_41_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $20 \\mu \\mathrm{{F}}$ capacitor is connected to a variable frequency ac source with maximum voltage $30 \\mathrm{{~V}}$. What is the capacitive reactance at $60 \\mathrm{{~Hz}}, 600 \\mathrm{{~Hz}}$, and $60 \\mathrm{{kHz}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle \\left.X_C\\right|_{{60}}=\\frac{{1}}{{\\omega C}}=\\frac{{10^6 \\mathrm{{~s}}}}{{2 \\pi \\cdot 60 \\cdot 20 \\mathrm{{~F}}}}=\\left.130 \\Omega \\quad X_C\\right|_{{600}}=\\left.13 \\Omega \\quad X_C\\right|_{{60 \\mathrm{{k}}}}=0.13 \\Omega$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_40_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An R-L-C circuit consists of a $10 \\Omega$ resistor, $200 \\mathrm{{mH}}$ inductor, and a $10 \\mu \\mathrm{{F}}$ capacitor. The capacitor is initially charged to $10 \\mu \\mathrm{{C}}$. What is the maximum possible voltage on the capacitor (as dictated by the exponential envelope) at $0.10 \\mathrm{{~s}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The voltage on the capacitor is proportional to the charge. The value of the exponential function at any time is determined by taking the cosine function as 1 . Using equation $40-6$ the charge is
        $\\displaystyle q=q_0 e^{{-R t / 2 L}}=(10 \\mu \\mathrm{{C}}) \\exp \\left[-\\frac{{10 \\Omega \\cdot 0.10 \\mathrm{{~s}}}}{{2 \\cdot 0.20 \\mathrm{{H}}}}\\right]=10 \\exp [-2.5] \\mu \\mathrm{{C}}=0.82 \\mu \\mathrm{{C}}$
        The notation for $e$ has been switched to exp for convenience.
        The voltage at this time is
        $\\displaystyle V=\\frac{{q}}{{C}}=\\frac{{0.82 \\mu \\mathrm{{C}}}}{{10 \\mu \\mathrm{{F}}}}=0.082 \\mathrm{{~V}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_40_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the circuit described in problem 40-1, what is the maximum energy and the maximum current in the circuit?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Maximum energy is $\\displaystyle \\frac{{q_{{\\max }}^2}}{{2 \\mathrm{{C}}}}=\\frac{{\\left(1.0 \\times 10^{{-4}} \\mathrm{{C}}\\right)^2}}{{2.4 .0 \\times 10^{{-5}} \\mathrm{{~F}}}}=1.25 \\times 10^{{-4}} \\mathrm{{~J}}$
        The maximum current is from $\\displaystyle \\frac{{L i_{{\\text {{max }}}}^2}}{{2}}=1.25 \\times 10^{{-4}} \\mathrm{{~J}}$, so
        $\\displaystyle i_{{\\max }}=\\left[\\frac{{2 \\cdot 1.25 \\times 10^{{-4}} \\mathrm{{~J}}}}{{50 \\times 10^{{-3}} \\mathrm{{H}}}}\\right]^{{1 / 2}}-0.071 \\mathrm{{~A}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_40_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An L-C circuit consists of a $50 \\mathrm{{mH}}$ inductor and a $40 \\mu \\mathrm{{F}}$ capacitor. What is the frequency of oscillation of this circuit? If the maximum charge on the capacitor is $100 \\mu \\mathrm{{C}}$, what is the maximum voltage on the components?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The frequency
        $\\displaystyle f=\\frac{{1}}{{2 \\pi \\sqrt{{L C}}}}=\\frac{{1}}{{2 \\pi \\sqrt{{50 \\times 10^{{-3}} \\mathrm{{H}} \\cdot 4.0 \\times 10^{{-5}} \\mathrm{{C}}}}}}=112 \\mathrm{{~Hz}}$
        The maximum voltage on either component is the same as the maximum voltage on the capacitor, which comes from the basic definition of capacitance
        $\\displaystyle V_{{\\max }}=\\frac{{q_{{\\max }}}}{{\\mathrm{{C}}}}=\\frac{{100 \\mu \\mathrm{{C}}}}{{40 \\mu \\mathrm{{F}}}}=2.5 \\mathrm{{~V}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_39_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        In the circuit described in problem 39-5, what is the maximum energy stored in the (magnetic field of the) coil?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle U=\\frac{{L i^2}}{{2}}=\\frac{{10 \\mathrm{{H}}}}{{2}}\\left(\\frac{{10 \\mathrm{{~V}}}}{{12 \\Omega}}\\right)^2=3.5 \\mathrm{{~J}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
         
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_39_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the circuit described in problem 39-5 operating in decreasing current mode, calculate the time for the voltage to decay to one-half the steady state value. Then find the voltages across $R$ and $L$ at this time.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The time constant, calculated earlier, is $0.83 \\mathrm{{~s}}$. The time for the current to drop to onehalf the steady state value is found hy replacing $i$ in equation 39-3 with $0.50 i_o$.
        $\\displaystyle 0.50 i_o=i_o e^{{-t / 0.83 \\mathrm{{~s}}}}$
        A more convenient form for logarithms is
        $\\displaystyle 1 / 2=e^{{-t / 0.83 s}}$
        Taking logarithms and remembering that $1 n(1 / 2)=1 n 1-1 n 2=0-1 n 2=-1 n 2$,
        $\\displaystyle -\\ln 2=-1 / 0.83 \\mathrm{{~s}} \\quad \\text {{ or }} \\quad t=0.83 \\ln 2 \\mathrm{{~s}}=0.58 \\mathrm{{~s}}$
        The voltage across the inductor is
        $\\displaystyle v_L=L \\frac{{d i}}{{d t}}=-V e^{{-R t / L}}=-10 e^{{-0.58 / 0.83}} \\mathrm{{~V}}=-5.0 \\mathrm{{~V}}$
        The voltage across the resistor is
        $\\displaystyle v_R=V e^{{-R t / L}}=10 e^{{-0.58 / 0.83}} \\mathrm{{~V}}=5.0 \\mathrm{{~V}}$
        The voltages $v_R+v_L$ add to zero as they must since they are the entire circuit!
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_39_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Construct the circuit of Fig. 39-1 with an inductor of $10 \\mathrm{{H}}$, a resistor of $12 \\Omega$, and a $10 \\mathrm{{~V}}$ battery. Calculate the time constant, the rate of change of current, and the voltages across the resistor and inductor at one time constant.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-39-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The time constant is
        $\\displaystyle \\tau=\\frac{{L}}{{R}}=\\frac{{10 \\mathrm{{H}}}}{{12 \\Omega}}=0.83 \\mathrm{{~s}}$
        The rate of change of current is
        $\\displaystyle \\frac{{d i}}{{d t}}=\\frac{{V}}{{R}} e^{{-t / 0.83 \\mathrm{{~s}}}}=\\frac{{10 \\mathrm{{~V}}}}{{12 \\Omega}} e^{{-1}}=0.31 \\mathrm{{~A}} / \\mathrm{{s}}$
        The voltage across the resistor is $v_R=i R=V\\left(1-e^{{-/ / 0.83 \\mathrm{{~s}}}}\\right)=10\\left(1-e^{{-1}}\\right) \\mathrm{{V}}=6.3 \\mathrm{{~V}}$.
        The voltage across the inductor is
        $\\displaystyle v_L=L \\frac{{d i}}{{d t}}=V e^{{-t / 0.83 \\mathrm{{~s}}}}=10 e^{{-1}} \\mathrm{{~V}}=3.7 \\mathrm{{~V}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-41-8.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_39_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the same circuit described in problem $39-3$, what is the rate of current increase when $i$ has reached $90 \\%$ of its final value?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Translated into mathematics, the question reads, "Find di/dt when $i=0.90(V / R)$." First find $t$ when $i=$ $0.90(V / R)$. Set $i=0.90 V / R$ in equation $39-1$ and solve for $t$.

        $0.90(V / R)=(V / R)\\left(1-e^{{-t / 28 \\times 10^{{-4}} \\varepsilon}}\\right)$ or $e^{{t / 2.8 \\times 10^{{-4}} 3}}=0.10$ or $e^{{t / 2.8 \\times 10^{{-4}} s}}=10$
        Solve for $t$ by switching to logarithms: $t=2.8 \\times 10^{{-4}} 1 \\mathrm{{n}} 10=6.4 \\times 10^{{-4}} \\mathrm{{~s}}$

        This is not a common type of problem, so go back over these equations and be sure you know how to manipulate the exponents and logarithms.
        Now place this time in equation 39-2.
        $\\displaystyle \\frac{{d i}}{{d t}}=\\frac{{V}}{{L}} e^{{-t / 2.8 \\times 10^{{-4}} \\mathrm{{~s}}}}=\\frac{{60 \\mathrm{{~V}}}}{{50 \\times 10^{{-3}} \\mathrm{{H}}}} e^{{-6.4 \\times 10^{{-4}} / 2.8 \\times 10^{{-4}}}}=122 \\mathrm{{~A}} / \\mathrm{{s}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_39_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the same circuit, what is the rate of current increase at $t=1.0 \\times 10^{{-4}} \\mathrm{{~s}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Using $\\frac{{d i}}{{d t}}=\\frac{{V}}{{L}} e^{{-t / 2.8 \\times 10^{{-4}} \\mathrm{{~s}}}}$
        $\\displaystyle \\frac{{d i}}{{d i}}=\\frac{{60 \\mathrm{{~V}}}}{{50 \\times 10^{{-3}} \\mathrm{{H}}}} e^{{-1 / 2.8}}=840 \\mathrm{{~A}} / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_39_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the same circuit, what is the time constant?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The time constant
        $\\displaystyle \\tau=\\frac{{L}}{{R}}=\\frac{{50 \\times 10^{{-3}} \\mathrm{{H}}}}{{180 \\Omega}}=2.8 \\times 10^{{-4}} \\mathrm{{~s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_39_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Place a $60 \\mathrm{{~V}}$ battery in series with an inductor of $50 \\times 10^{{-3}} \\mathrm{{H}}$ and a resistor of $180 \\Omega$. What is the current and the rate of change of current at $t=0$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Using $\\displaystyle i=\\frac{{V}}{{R}}\\left(1-e^{{-R t / L}}\\right)$ and $\\displaystyle \\frac{{d i}}{{d t}}=\\frac{{V}}{{L}} e^{{-R t / L}}$ it is clear that at $t=0, i=0$, but $\\displaystyle \\frac{{d i}}{{d i}}=\\frac{{V}}{{L}}=\\frac{{60 \\mathrm{{~V}}}}{{50 \\times 10^{{-3}} \\mathrm{{H}}}}=1.2 \\times 10^3 \\mathrm{{~A}} / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_38_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the energy density in the large coil of problem $38-3$ for a current of $0.10 \\mathrm{{~A}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle u_B=\\frac{{\\mu_o n^2 i^2}}{{2}}=\\frac{{\\left(4 \\pi \\times 10^7 \\mathrm{{~T}} \\cdot \\mathrm{{m}} / \\mathrm{{A}}\\right)(1200 / 0.10 \\mathrm{{~m}})^2(0.10 \\mathrm{{~A}})^2}}{{2}}=0.90 \\mathrm{{~J}} / \\mathrm{{m}}^3$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_38_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Two coils wound together as a single package have an inductance of $6.0 \\times 10^{{-3}} \\mathrm{{H}}$. Find the voltage induced in the second coil when the current is changing in the first at the instantaneous rate of $20 \\times 10^{{-3}} \\mathrm{{~A}} / \\mathrm{{s}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-39-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        $e m f=L(d i / d i)=6.0 \\times 10^{{-3}} \\mathrm{{H}} \\cdot 20 \\times 10^{{-3}} \\mathrm{{~A}} / \\mathrm{{s}}=1.2 \\times 10^{{-4}} \\mathrm{{~V}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_38_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation of problem 38-3 find the voltage induced in the second coil as a result of a steady change in current of $0.50 \\mathrm{{~A}} / \\mathrm{{s}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle e m f=L \\frac{{\\Delta i}}{{\\Delta t}}=9.5 \\times 10^{{-4}} \\mathrm{{H}} \\frac{{0.50 \\mathrm{{~A}}}}{{\\mathrm{{~s}}}}=4.7 \\times 10^{{-4}} \\mathrm{{~V}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_38_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider a coil within a coil. The larger coil is $3.0 \\mathrm{{~cm}}$ in radius, $10 \\mathrm{{~cm}}$ in length with 1200 turns. The smaller coil is $2.0 \\mathrm{{~cm}}$ in radius, $4.0 \\mathrm{{~cm}}$ in length with 50 turns. Calculate the inductance.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-38-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The current in the large solenoid sets up a magnetic field within the coil of $B=\\mu_N N i /$. Assume the flux is uniform across the inside of the large coil so that the flux through the small coil is this $B$ times the area of the small coil $\\Phi_2=B A_2$. The 1 subscript refers to the large coil and 2 refers to the smaller coil. The mutual inductance, in accord with the definition of self inductance is the number of flux linkages divided by the current
        $M$ or $\\displaystyle L=\\frac{{N_2 \\Phi_2}}{{i_1}}=\\frac{{N_2}}{{i_1}} B A_2=\\frac{{N_2}}{{i_1}} \\frac{{\\mu_o N_1 i_1}}{{\\ell}} A_2=\\mu_o \\frac{{N_1}}{{\\ell}} N_2 A_2$
        The inductance is totally geometry dependent. For this particular combination of coils
        $\\displaystyle L=\\mu_o \\frac{{N_1}}{{\\ell}} N_2 A_2=4 \\pi \\times 10^{{-7}} \\mathrm{{~T}} \\cdot \\mathrm{{m}} / \\mathrm{{A}} \\frac{{1200}}{{0.10 \\mathrm{{~m}}}}(50) \\pi\\left(4.0 \\times 10^{{-4}} \\mathrm{{~m}}^2\\right)=9.5 \\times 10^{{-4}} \\mathrm{{H}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_38_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For a coil of 300 turns and inductance $12 \\times 10^{{-3}} \\mathrm{{H}}$, find the flux through the coil when the current is $5.0 \\times$ $10^{{-3}} \\mathrm{{~A}}$.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Use the relationship between flux linkages and inductance $N \\Phi=L i$.
        $\\displaystyle \\Phi-\\frac{{L i}}{{N}}-\\frac{{12 \\times 10^{{-3}} \\mathrm{{H}} \\cdot 5.0 \\times 10^{{-3}} \\mathrm{{~A}}}}{{300}}-2.0 \\times 10^{{-7}} \\mathrm{{~Wb}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_38_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the inductance of a solenoid of radius $0.0040 \\mathrm{{~m}}$, length $0.015 \\mathrm{{~m}}$, and $1.0 \\times 10^4$ turns per meter.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle L=\\mu_o n^2 \\ell A=\\left(4 \\pi \\times 10^{{-7}} \\mathrm{{~T}} \\cdot \\mathrm{{m}} / \\mathrm{{A}}\\right)\\left(1.0 \\times 10^8 1 / \\mathrm{{m}}^2\\right)(0.015 \\mathrm{{~m}}) \\pi(0.0040 \\mathrm{{~m}})^2=9.5 \\times 10^{{-5}} \\mathrm{{H}}$        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_37_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate how an electron placed at $0.30 \\mathrm{{~m}}$ radius would be accelerated in a changing magnetic field of 50 $\\times 10^{{-6}} \\mathrm{{~T}} / \\mathrm{{s}}$.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The force on the electron would be $e E=m a$ so
        $\\displaystyle a=\\frac{{e}}{{m}} \\frac{{r d B}}{{2}}=\\frac{{1.6 \\times 10^{{-19}} \\mathrm{{C}}}}{{9.1 \\times 10^{{-31}} \\mathrm{{~kg}}}} \\frac{{0.30 \\mathrm{{~m}}}}{{2}} \\frac{{50 \\times 10^{{-6}} \\mathrm{{~T}}}}{{\\mathrm{{~s}}}}=1.3 \\times 10^6 \\mathrm{{~m}} / \\mathrm{{s}}^2$
        Accelerating devices using this principle are called synchrotrons.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_37_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation described above the current in the wire increases from zero to $20 \\mathrm{{~A}}$ in 0.10 s. Find the induced $e m f$ over a rectangle with $a=0.20 \\mathrm{{~cm}}, b=0.60 \\mathrm{{~cm}}$, and $L=1.0 \\mathrm{{~cm}}$.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $e m f=\\Delta \\Phi / \\Delta i$ and $\\Delta \\Phi$ is the total flux, so
        $\\displaystyle e m f=\\frac{{\\mu_o I L}}{{2 \\pi}} \\ln \\frac{{b}}{{a}}=\\frac{{4 \\pi \\times 10^{{-7}} \\mathrm{{~T}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{A}} \\cdot 20 \\mathrm{{~A}} \\cdot 1.0 \\times 10^{{-2}} \\mathrm{{~m}}}}{{2 \\pi(0.10 \\mathrm{{~s}})}} \\ln 3=4.4 \\times 10^{{-7}} \\mathrm{{~V}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_37_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A variation of the problem done above is one where the loop is replaced by a "U"-shaped piece of wire with a sliding piece along the arms of the "U." Take a $0.60 \\mathrm{{~T}}$ field and a "U"shaped piece with width $0.30 \\mathrm{{~m}}$. The entire circuit has resistance of $20 \\Omega$ and the sliding bar is moving to the right at $6.0 \\mathrm{{~m}} / \\mathrm{{s}}$.
        
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The emf generated in the wire and moving rod is
        $e m f=-B L v=-0.60 \\mathrm{{~T}} \\cdot 0.30 \\mathrm{{~m}} \\cdot 6.0 \\mathrm{{~m}} / \\mathrm{{s}}=-1.1 \\mathrm{{~V}}$
        The current in the loop is $i=e m f / R=1.1 \\mathrm{{~V}} / 20 \\Omega=0.054 \\mathrm{{~A}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-37-8.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        When the bar is moving to the right, the number of lines of flux (amount) is increasing out of the page, so the current is clockwise so as to produce a field pointing into the page, i.e., a field opposing the increasing field, within the "loop" causing the current.
        The work performed in moving the bar is
        $W=F v=B i L v=0.60 \\mathrm{{~T}} \\cdot 0.054 \\mathrm{{~A}} \\cdot 0.30 \\mathrm{{~m}} \\cdot 6.0 \\mathrm{{~m}} / \\mathrm{{s}}=0.058 \\mathrm{{~J}}$
        Compare this with the Joule heat production $H=i^2 R=(0.054 \\mathrm{{~A}})^2(20 \\Omega)=0.058 \\mathrm{{~J}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_37_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Pull a loop of width $0.20 \\mathrm{{~m}}$, length $0.80 \\mathrm{{~m}}$, and resistance $200 \\Omega$ through a magnetic field of $0.40 \\mathrm{{~T}}$ at $0.20 \\mathrm{{~m}} / \\mathrm{{s}}$. First find the induced $e m f$ and current. Then find the force necessary to pull the loop, the work performed, and the power, the rate of doing the work.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The induced $e m f$ is $e m f=-B L v=-0.40 \\mathrm{{~T}} \\cdot 0 .20 \\mathrm{{~m}} \\cdot0 .20 \\mathrm{{~m}} / \\mathrm{{s}}=-1.6 \\times 10^{{-2}} \\mathrm{{~V}}$.
        The induced current (in the direction shown on Fig. 37-7) is
        $\\displaystyle i=\\frac{{e m f}}{{R}}=\\frac{{1.6 \\times 10^{{-3}} \\mathrm{{~V}}}}{{200 \\Omega}}=8.0 \\times 10^{{-5}} \\mathrm{{~A}}$
        The force necessary to pull the wire out of the field is
        $F=B 1 L=0.40 \\mathrm{{~T}} \\cdot 8.0 \\times 10^{{-5}} \\mathrm{{~A}} \\cdot 0.20 \\mathrm{{~m}}=6.4 \\times 10^{{-6}} \\mathrm{{~N}}$
        The work performed in completely removing the loop is this force times the length of the loop.
        $W=F \\cdot x=6.4 \\times 10^{{-6}} \\mathrm{{~N}} \\cdot 0.80 \\mathrm{{~m}}=5.1 \\times 10^{{-6}} \\mathrm{{~J}}$
        The power delivered to the loop is $P=F . v=6.4 \\times 10^{{-6}} \\mathrm{{~N}} .0 .20 \\mathrm{{~m}} / \\mathrm{{s}}=1.3 \\times 10^{{-6}} \\mathrm{{~W}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_37_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the coil and magnetic field situation of Fig. $37-5$ consider continuous rotation at some angular velocity $\\omega$. The maximum flux through the coil is $\\Phi_0$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The flux is $\\Phi=\\Phi_0  \\cos \\omega t$.
        The time rate of change of flux is $\\displaystyle \\frac{{d \\Phi}}{{d t}}=-\\Phi_o \\omega \\sin \\omega t$
        The instantanenus emf is $\\displaystyle e m f=-N \\frac{{d \\Phi}}{{d t}}=N \\Phi \\omega \\sin \\omega t$.
        These three quantities are shown in Fig. 37-6.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-37-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_37_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Now find the average $\\boldsymbol{{e m f}}$ induced over this time. This is the average value of the instantaneous $\\boldsymbol{{e m f}}$ function over this one-quarter cycle.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This requires the concept of the average value of a function from integral calculus. Review this procedure in the Introduction, Mathematical Background, if necessary. This integral is
        $\\displaystyle \\left.e m f\\right|_{{a v g}}=\\frac{{1}}{{\\pi / 2}} 7.85 \\mathrm{{~V}} \\int_0^{{\\pi / 2}} \\sin \\theta d \\theta=-\\left.5.0 \\mathrm{{~V}} \\cos \\theta\\right|_0 ^{{\\pi / 2}}=-5.0 \\mathrm{{~V}}$
        The integral is done over $\\theta$ for convenience rather than over $\\omega t$ and a time interval. The average value of the first quarter cycle of the sine function is the same as the average value for the first half cycle of the sine function and the same as the absolute value of the average value of the entire sine function. This is an excellent example of a problem where the average $e m f$ can be calculated using either $\\Delta \\Phi / \\Delta t$ (problem 37-3) or the average value of the function for $e m f$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_37_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Now instead of flipping the coil from perpendicular to parallel, rotate it with uniform angular velocity.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The expression for the flux, assuming the rotation is clockwise is
        $\\Phi=B A \\cos \\omega t \\quad$ where $\\displaystyle \\quad \\omega=\\frac{{\\Delta \\theta}}{{\\Delta t}}=\\frac{{\\pi / 2}}{{0.10 \\mathrm{{~s}}}}$.
        The coil rotates through an angle of $\\Delta / 2$ in $0.10 \\mathrm{{~s}}$ so $\\phi=0.50 \\mathrm{{~T}} .2 .0 \\times 10^{{-2}} \\mathrm{{~m}}^2 \\cos (\\Delta t / 0.20 \\mathrm{{~s}})$
        The instantaneous $e m f$ is
        $\\displaystyle e m f=-N \\frac{{d \\Phi}}{{d t}}=50 \\cdot 1.0 \\times 10^{{-2}} \\mathrm{{~T}} \\cdot \\mathrm{{m}}^2 \\frac{{\\pi}}{{0.20 \\mathrm{{~s}}}} \\sin \\frac{{\\pi t}}{{0.20 \\mathrm{{~s}}}}=7.85 \\mathrm{{~V}} \\sin \\frac{{\\pi t}}{{0.20 \\mathrm{{~s}}}}$
        The units in this calculation are helpful in understanding the $e m f$ generated in a loop. A T.m² is a Weber, a measure of the number of lines of flux. Webers per second is the number of flux lines per second that are changed within a loop. One Weber's worth of flux lines per second generates an $e m f$ of one Volt in the loop.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_37_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Take a 50 turn rectangular coil of dimensions $0.10 \\mathrm{{~m}}$ by $0.20 \\mathrm{{~m}}$ and rotate it from a position perpendicular to a field of $0.50 \\mathrm{{~T}}$ to parallel to the field in $0.10 \\mathrm{{~s}}$ and calculate the induced $e m f$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-37-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The flux $\\Phi=B A$. In addition, $\\Delta \\Phi=B A$ since the flux goes from maximum to zero.
        $\\Delta \\Phi=B A=0.50 \\mathrm{{~T}} \\cdot 2.0 \\times 10^{{-2}} \\mathrm{{~m}}^2=1.0 \\times 10^{{-2}} \\mathrm{{~T}} \\cdot \\mathrm{{m}}^2$
        The coil is 50 turns, so the induced $\\boldsymbol{{e m f}}$ is
        $\\displaystyle e m f=-N \\frac{{\\Delta \\Phi}}{{\\Delta t}}=-50 \\frac{{1.0 \\times 10^{{-2}} \\mathrm{{~T}} \\cdot \\mathrm{{m}}^2}}{{0.10 \\mathrm{{~s}}}}=-5.0 \\mathrm{{~V}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_37_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A circular loop $0.040 \\mathrm{{~m}}$ in radius is perpendicular to a magnetic field of $0.50 \\mathrm{{~T}}$. The loop is stretched, across a diameter, so that its area goes to zero in $0.20 \\mathrm{{~s}}$. What is the induced voltage and direction of the current through the resistor?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The flux through the loop is $\\Phi-B A-0.50 \\mathrm{{~T}} . \\Delta(0.04 \\mathrm{{~m}})^2-2.5 \\times 10^{{-3}} \\mathrm{{~Wb}}$.
        The $e m f$ is
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-37-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        $\\displaystyle e m f=-\\frac{{\\Delta \\Phi}}{{\\Delta t}}=-\\frac{{2.5 \\times 10^{{-3}} \\mathrm{{~Wb}}}}{{0.20 \\mathrm{{~s}}}}=-0.012 \\mathrm{{~V}}$
        The current is such as to produce a magnetic field out of the page (or in opposition to the collapsing magnetic field). The current is counterclockwise in the stretched loop and up on the resistor.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_37_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Place a 20 turn coil of radius $0.050 \\mathrm{{~m}}$ inside a solenoid where the magnetic field is changed from $40 \\times 10^{{-3}}$ Tin one direction to $40 \\times 10^{{-3}} \\mathrm{{~T}}$ in the other direction in $60 \\times 10^{{-3}} \\mathrm{{~s}}$. Find the induced $e m f$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        
        file_name = 'how-to-solve-physics-problems-fig-37-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Inside the coil the flux is $\\Psi=B A=40 \\times 10^{{-3}} \\mathrm{{~T}} \\pi(0.05 \\mathrm{{~m}})^2=3.1 \\times 10^{{-4}} \\mathrm{{~Wb}}$.
        The change in flux is $\\Delta \\Phi=6.2 \\times 10^{{-4}} \\mathrm{{~Wb}}$, and this occurs over $60 \\times 10^{{-3}} \\mathrm{{~s}}$ in the coil of 20 turns so the induced emf is
        $\\displaystyle e m f=-N \\frac{{\\Delta \\Phi}}{{\\Delta t}}=-20 \\frac{{6.2 \\times 10^{{-4}} \\mathrm{{~Wb}}}}{{60 \\times 10^{{-3}} \\mathrm{{~s}}}}=-0.21 \\mathrm{{~V}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_36_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the magnetic field at the center of a semicircular piece of wire with radius $0.20 \\mathrm{{~m}}$ and carrying 150A of current?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Start with equation 36-1 in differential form, and referring to Fig. 36-4, form dlx $\\hat{{r}}$ that shows $\\boldsymbol{{B}}$ into the page at the center of the semicircle. Integrating equation $36-1$ produces
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-36-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        $\\displaystyle B=\\frac{{\\mu_o}}{{4 \\pi}} I \\int \\frac{{d l \\times \\hat{{r}}}}{{r^2}}=\\frac{{\\mu_o}}{{4 \\pi}} \\frac{{I}}{{r^2}} \\int d l=\\frac{{\\mu_o}}{{4 \\pi}} \\frac{{I}}{{r^2}} \\pi r=\\frac{{\\mu_o I}}{{4 r}}=\\frac{{4 \\pi \\times 10^{{-7}} \\mathrm{{~T}} \\cdot \\mathrm{{m}} / \\mathrm{{A}}(150 \\mathrm{{~A}})}}{{4(0.20 \\mathrm{{~m}})}}=2.4 \\times 10^{{-4}} \\mathrm{{~T}}$
        The connecting wires are on radii out from the center of the semicircle, so $d l \\times^{{\\hat{{r}}}}=0$ and they contribute no magnetic field at the center of the semicircle
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_36_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $3.0 \\mu \\mathrm{{C}}$ charge at $x=0, y=1.0 \\mathrm{{~m}}$ has a velocity $v_x=5.0 \\times 10^7 \\mathrm{{~m}} / \\mathrm{{s}}$. A $-4.0 \\mu \\mathrm{{C}}$ charge at $x=2.0 \\mathrm{{~m}}, y=0$ has a velocity $v_y=8.0 \\times 10^7 \\mathrm{{~m}} / \\mathrm{{s}}$. What is the magnetic field at the origin due to the motion of these charges? 
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        For the first charge the $\\hat{{r_1}}$ vector points down (from the charge to the origin) and $v \\times \\hat{{r_1}}$ shows the magnetic field at the origin as into the page. The magnitude of this field is magnetic field at the origin as into the page. The magnitude of this field is
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        file_name = 'how-to-solve-physics-problems-fig-36-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        $\\displaystyle B=\\frac{{\\mu_o}}{{4 \\pi}} q \\frac{{v}}{{\\r_1^2}}=\\frac{{4 \\pi \\times 10^{{-7}} \\mathrm{{~T}} \\cdot \\mathrm{{m}} / \\mathrm{{A}}\\left(3.0 \\times 10^{{-6}} \\mathrm{{C}}\\right) 5.0 \\times 10^7 \\mathrm{{~m}} / \\mathrm{{s}}}}{{4 \\pi\\left(1.0 \\mathrm{{~m}}^2\\right)}}=1.5 \\times 10^{{-5}} \\mathrm{{~T}}$
        For the second charge the ${{ }}^{{r_2}}$ vector points to the left (from the charge toward the origin) and $v \\times \\hat{{r}}_2$ is a vector pointing out of the page, but the negative charge makes the field point into the page. This magnetic field has magnitude
        $\\displaystyle B=\\frac{{\\mu_o}}{{4 \\pi}} q \\frac{{y}}{{r_2^2}}=\\frac{{4 \\pi \\times 10^{{-7}} \\mathrm{{~T}} \\cdot \\mathrm{{m}} / \\mathrm{{A}}\\left(3.0 \\times 10^{{-6}} \\mathrm{{C}}\\right) 8.0 \\times 10^7 \\mathrm{{~m}} / \\mathrm{{s}}}}{{4 \\pi\\left(4.0 \\mathrm{{~m}}^2\\right)}}=6.0 \\times 10^{{-6}} \\mathrm{{~T}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_36_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For problem 36-3, what is the field $5.0 \\times 10^{{-1}} \\mathrm{{~m}}$ distance along the axis?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle B_x=\\frac{{\\mu_o \\mathrm{{~N}} I \\mathrm{{a}}^2}}{{2\\left(x^2+\\mathrm{{a}}^2\\right)^{{3 / 2}}}}=\\frac{{4 \\pi \\times 10^{{-7}} \\mathrm{{~T}} \\cdot \\mathrm{{m}} / \\mathrm{{A}}(40) 0.25 \\mathrm{{~A}} \\cdot 25 \\times 10^{{-4}} \\mathrm{{~m}}^2}}{{2\\left(25 \\times 10^{{-2}}+25 \\times 10^{{-4}}\\right)^{{3 / 2}} \\mathrm{{~m}}^2}}=1.2 \\times 10^{{-7}} \\mathrm{{~T}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_36_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the magnetic field at the center of a circular coil of radius $5.0 \\times 10^{{-2}} \\mathrm{{~m}}$ carrying a current of $0.25 \\mathrm{{~A}}$ and having 40 turns?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The field is perpendicular to the plane of the loop at the center and has value
        $\\displaystyle B=\\frac{{\\mu_o \\mathrm{{~N}} l}}{{2 \\mathrm{{a}}}}=\\frac{{4 \\pi \\times 10^{{-7}} \\mathrm{{~T}} \\cdot \\mathrm{{m}} / \\mathrm{{A}}(40) 0.25 \\mathrm{{~A}}}}{{2 \\cdot 5.0 \\times 10^{{-2}} \\mathrm{{~m}}}}=1.2 \\times 10^{{-4}} \\mathrm{{~T}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_36_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the magnetic field on the axis of a circular loop of current.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-37-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The Biot-Savart law is most convenient for solving this problem. The cross product of $d l$ and $\\hat{{r}}$ gives the direction of the field along the axis as being at right angles to $r$ and in the plane defined by $x$ and $r$. The loop of current and $d l$ are in the $y-z$ plane. Start with the law in differential form (equation 36-1) and write the $x$ and $y$ components of the field.
        $\\displaystyle d B_x=d B \\sin \\alpha=\\frac{{\\mu_0 l}}{{4 \\pi}} \\frac{{d \\ell}}{{x^2+y^2}} \\frac{{\\mathrm{{a}}}}{{\\left(x^2+y^2\\right)^{{1 / 2}}}}$
        $\\displaystyle d B_y=d B \\cos \\alpha=\\frac{{\\mu_o I}}{{4 \\pi}} \\frac{{d \\ell}}{{x^2+y^2}} \\frac{{x}}{{\\left(x^2+y^2\\right)^{{1 / 2}}}}$

        For every differential increment of length on the loop there is another differential increment of length across a diameter also producing a contribution to the field. Looking at the components of these fields, the $x$ components add while the $y$ components add to zero leaving only the $x$ component that contributes to the resultant field. The field then reduces to the integral of the $x$ component.
        $\\displaystyle B_x=\\frac{{\\mu_o I}}{{4 \\pi}} \\int \\frac{{\\mathrm{{a}} d \\ell}}{{\\left(x^2+y^2\\right)^{{3 / 2}}}}=\\frac{{\\mu_o I}}{{4 \\pi}} \\frac{{\\mathrm{{a}} \\cdot 2 \\pi \\mathrm{{a}}}}{{\\left(x^2+y^2\\right)^{{3 / 2}}}}=\\frac{{\\mu_o I \\mathrm{{a}}^2}}{{2\\left(x^2+\\mathrm{{a}}^2\\right)^{{3 / 2}}}}$
        At the center of the loop, $x=0$ and the expression for the field reduces to

        $\\displaystyle B_x=\\frac{{\\mu_o I}}{{2 \\mathrm{{a}}}}$
        For a coil of $\\mathrm{{N}}$ loops the current is multiplied by $\\mathrm{{N}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_36_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the magnetic field a distance $x$ radially out from a wire carrying a current $I$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        
        soln_s = f"""
        Figure 36-1 shows the geometry for calculating the magnetic field due to an element of current in the wire. Use the differential form of the Biot-Savart law
        $\\displaystyle d \\boldsymbol{{B}}=\\frac{{\\mu_o I}}{{4 \\pi}} \\frac{{d l \\times \\hat{{\\boldsymbol{{r}}}}}}{{r^2}}=\\frac{{\\mu_o I}}{{4 \\pi}} \\frac{{d l \\sin \\theta}}{{r^2}}$
        and integrate. The vector $d l$ is in the direction of $l$, and the angle $\\theta$ is the angle between the vectors $d l$ and $r$. Cross $d \\boldsymbol{{l}}$ (in the direction of $I$ ) with $\\boldsymbol{{r}}$ to see that $\\boldsymbol{{B}}$, at the point $P$, is into the paper.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-36-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        $\\displaystyle B=\\frac{{\\mu_o}}{{4 \\pi}} \\int l \\frac{{x d y}}{{\\left(x^2+y^2\\right)^{{3 / 2}}}}$ $\\quad x$ is a constant
        Integrate from -a to a and then let a go to infinity to find the expression for the magnetic field due to a long straight wire. This specific integral was done in problem 25-5 in Chapter 25, Electric Field. Follow along the steps in problem 25-5 and work through this problem to confirm that the magnetic field outside a long currentcarrying wire is $\\mu_0 I / 2 \\pi r$, in conformity to Ampere's law.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_35_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the magnetic field in a toroid of 600 turns, current of $0.76 \\mathrm{{~A}}$ and center line radius $0.50 \\mathrm{{~m}}$.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displayatyle B=\\mu_v \\mathrm{{~N}} I_v=4 \\pi \\times 10^{{-7}} \\mathrm{{~T}} \\cdot \\mathrm{{m}} / \\mathrm{{A}}\\left(\\frac{{600}}{{2 \\pi \\cdot 0.50 \\mathrm{{~m}}}}\\right) 0.76 \\mathrm{{~A}}=18 \\times 10^{{-4}} \\mathrm{{~T}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_35_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the magnetic field inside a solenoid of 300 turns per centimeter carrying a current of $0.050 \\mathrm{{~A}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle B=\\mu_o I_o \\mathrm{{~N}}=\\left(4 \\pi \\times 10^{{-7}} \\mathrm{{~T}} \\cdot \\mathrm{{m}} / \\mathrm{{A}}\\right)(0.050 \\mathrm{{~A}})\\left(300 / 10^{{-2}} \\mathrm{{~m}}\\right)=1.9 \\times 10^{{-3}} \\mathrm{{~T}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-37-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_35_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the average value of the magnetic field from 0.50 to $0.10 \\mathrm{{~m}}$ radially out from a long wire carrying $100 \\mathrm{{~A}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The magnetic field falls off radially according to $B=\\mu_0 I / 2 \\Pi$. The average value of the function over this interval is the integral of the function (the area under the curve of $B$ vs $r$ ) divided by the radial distance. See the Introduction, Mathematical Background, for the definition of the average value of a function.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        file_name = 'how-to-solve-physics-problems-fig-35-8.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        $\\displaystyle B_{{a v g}}=\\frac{{1}}{{0.10-0.050}} \\frac{{\\mu_o I}}{{2 \\pi}} \\int_{{0.050}}^{{0.10}} \\frac{{1}}{{r}} d r=\\left.\\frac{{\\mu_o I}}{{0.050 \\cdot 2 \\pi}} \\ln r\\right|_{{0.050}} ^{{0.10}}=\\frac{{\\mu_o I}}{{0.10 \\cdot \\pi}} \\ln 2$
        $\\displaystyle B_{{a v g}}=\\frac{{4 \\pi \\times 10^{{-7}} \\mathrm{{~T}} \\cdot \\mathrm{{m}} / \\mathrm{{A}}(100 \\mathrm{{~A}})}}{{0.10 \\pi \\mathrm{{m}}}} \\ln 2=2.8 \\times 10^{{-4}} \\mathrm{{~T}}$
        HTML(s)        
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_35_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the magnetic field $10 \\mathrm{{~cm}}$ perpendicular to the bisector of the line connecting two wires $2.0 \\mathrm{{~cm}}$ apart and carrying antiparallel currents of $100 \\mathrm{{~A}}$ each as shown in Fig. 35-5.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-35-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        
        soln_s = f"""
        The magnitude of the magnetic field from each wire is $\\mu_0 I / 2 \\pi d$. Because of the geometry the vertical components of $B_1$ and $B_2$ add to zero. The horizontal components add to a vector pointing toward the centerline of the wires with magnitude $B=2 B_1 \\sin \\theta$. The dimension $d=\\sqrt{{10^2+1^2}}=\\sqrt{{101}} \\mathrm{{~cm}}$. The $\\sin \\theta=1.0 / \\sqrt{{101}}$. The magnitude of the resultant field is
        $\\displaystyle B=\\frac{{2 \\mu_o I}}{{2 \\pi d}} \\sin \\theta=\\frac{{2.4 \\pi \\times 10^{{-7}} \\mathrm{{~T}} \\cdot \\mathrm{{m}} / \\mathrm{{A}}(100 \\mathrm{{~A}})}}{{2 \\pi \\sqrt{{101}} \\times 10^{{-2}} \\mathrm{{~m}}}} \\frac{{1.0}}{{\\sqrt{{101}}}}=4.0 \\times 10^{{-5}} \\mathrm{{~T}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_35_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider two wires coming out of the page carrying equal, and oppositely directed, currents as shown in Fig. 35-4. The wires are suspended from a common point with $0.40 \\mathrm{{~m}}$ long cords. The wires have a mass per unit length of $30 \\times 10^{{-3}} \\mathrm{{~kg}} / \\mathrm{{m}}$. Calculate the current to produce an angle of $6^{{\\circ}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This is a force-balance problem. The horizontal component of the tension (force) in the cord must equal the force due to the currents. Look at $1.0 \\mathrm{{~m}}$ of the wire. The mass of $1.0 \\mathrm{{~m}}$ of wire is $30 \\times 10^{{-3}} \\mathrm{{~kg}}$, and the force due to gravity $m g=0.29 \\mathrm{{~N}}$. This must equal the vertical component of the tension in the cord. From the geometry
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-35-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        In Fig. 35-4 the separation of the wires is $2 s$ where $s$ is defined in the figure through $\\sin 6^{{\\circ}}=s / 0.40 \\mathrm{{~m}}$ or $s=$ $0.40 \\mathrm{{~m}} . \\sin 6^{{\\circ}}$. The electric force produced by the current in one wire producing a magnetic field that acts on the current in the other wire is
        $\\displaystyle F=B l=\\frac{{\\mu_o I^2}}{{2 \\pi(2 s)}}$
        and this force must equal the mechanical force, or tension, in the cord so
        $\\displaystyle 0.031 \\mathrm{{~N}}=\\frac{{4 \\pi \\times 10^{{-7}} \\mathrm{{~T}} \\cdot \\mathrm{{m}} / \\mathrm{{A}} \\cdot I^2}}{{2 \\pi\\left(2 \\cdot 0.40 \\mathrm{{~m}} \\cdot \\sin 6^{{\\circ}}\\right)}}$
        and solving for $I$
        $\\displaystyle I=\\left[\\frac{{0.40 \\cdot \\sin 6^{{\\circ}}(0.031)}}{{10^{{-7}}}} \\mathrm{{~A}}^2\\right]^{{1 / 2}}=114 \\mathrm{{~A}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_35_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Two long parallel wires separated by $0.60 \\mathrm{{~m}}$ carry antiparallel currents (the currents are physically parallel but in opposite directions) of $7.0 \\mathrm{{~A}}$ each. What is the resulting field along the line between the wires and $0.40 \\mathrm{{~m}}$ from the lower wire?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-35-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        At the specified position the magnetic field due to wire 1 is
        $\\displaystyle B_1=\\frac{{\\mu_o I}}{{2 \\pi r}}=\\frac{{4 \\pi \\times 10^{{-7}} \\mathrm{{~Wb}} / \\mathrm{{A}} \\cdot \\mathrm{{m}}(7.0 \\mathrm{{~A}})}}{{2 \\pi \\cdot 0.20 \\mathrm{{~m}}}}=7.0 \\times 10^{{-6}} \\mathrm{{~T}}$
        and directed as shown (right thumb in direction of current with fingers curled). The magnetic field due to wire 2 is
        $\\displaystyle B_2=\\frac{{\\mu_0 I}}{{2 \\pi r}}=\\frac{{4 \\pi \\times 10^{{-7}} \\mathrm{{~Wb}} / \\mathrm{{A}} \\cdot \\mathrm{{m}}(7.0 \\mathrm{{~A}})}}{{2 \\pi \\cdot 0.40 \\mathrm{{~m}}}}=3.5 \\times 10^{{-6}} \\mathrm{{~T}}$
        and directed as shown. The total magnetic field is $10.5 \\times 10^{{-6}} \\mathrm{{~T}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_35_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the magnetic field $7.0 \\mathrm{{~m}}$ below a power line carrying $12000 \\mathrm{{~A}}$ of current.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle B=\\frac{{\\mu_0 i}}{{2 \\pi r}}=\\frac{{4 \\pi \\times 10^{{-7}} \\mathrm{{~Wb}} / \\mathrm{{A}} \\cdot \\mathrm{{m}}(12000 \\mathrm{{~A}})}}{{2 \\pi \\cdot 7.0 \\mathrm{{~m}}}}=3.4 \\times 10^{{-4}} \\mathrm{{~T}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_34_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A coil of $5.0 \\times 8.0 \\mathrm{{~cm}}$ dimensions and carrying a current of $0.0020 \\mathrm{{~A}}$ is rotated $45^{{\\circ}}$ by a $20 \\mathrm{{~T}}$ magnetic field from parallel to the plane of the field to an angle of $45^{{\\circ}}$ with respect to the field (see Fig. 34-5). Calculate the energy of the system.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The energy of the system is
        $\\displaystyle U=-\\left.m B \\cos \\alpha\\right|_{{90^{{\\circ}}}} ^{{45^{{\\circ}}}}=-m B \\cos 45^{{\\circ}}=-40 \\times 10^{{-4}} \\mathrm{{~m}}^2 \\cdot 0.0020 \\mathrm{{~A}} \\cdot 20 \\mathrm{{~T}} \\cdot \\cos 45^{{\\circ}}=-1.1 \\times 10^{{-4}} \\mathrm{{~J}}$
        This is the energy stored in the system. An exlernal agent would have to perform this amount of work to move the dipole (coil) back to parallel with the external magnetic field.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_34_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An interesting problem is to ask the question, "What dimensions of rectangle will produce maximum torque for the same length of wire?"  
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Take a rectangle with length $l$, width $w_{{22}}$ and perimeter equal to $p=2 l+2 w$. The perimeter relating $l$ to $w$ is the constraint in the problem. The torque is $T=m B=I A B=I l w B$. Before looking for a maximum we need to write $l$ or $w$ in terms of the constraint $p$.
        $\\displaystyle \\tau=I w \\frac{{p-2 w}}{{2}} B=\\frac{{I B}}{{2}}\\left(p w-2 w^2\\right)$
        Now take two derivatives of $T$ with respect to $w$
        $\\displaystyle \\frac{{d \\tau}}{{d w}}=\\frac{{I B}}{{2}}(p-4 w) \\quad \\frac{{d^2 T}}{{d w^2}}=-2 I B$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-37-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_34_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The plane of a 10 turn rectangular loop of wire $3.0 \\mathrm{{~cm}}$ by $4.0 \\mathrm{{~cm}}$ carrying $2.0 \\mathrm{{~A}}$ is at an angle of $60^{{\\circ}}$ with respect to a magnetic field of $20 \\mathrm{{~T}}$. What is the torque on the loop in this position?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First calculate the magnetic moment noting that the current is not due to one loop but ten loops. $m=I$ $A=10.2 .0 \\mathrm{{~A}} .12 \\times 10^{{-4}} \\mathrm{{~m}}^2=2.4 \\times 10^{{-2}} \\mathrm{{~A}} . \\mathrm{{m}}^2$. Refer to Fig. $34-5$ and note that if the angle between the plane of the coil and the magnetic field is $60^{{\\circ}}$, then the angle between $\\boldsymbol{{m}}$ and $\\boldsymbol{{B}}$ is $30^{{\\circ}}$, so the torque is
        $\\displaystyle \\tau=m B \\sin \\alpha=2.4 \\times 10^{{-2}} \\mathrm{{~A}} \\cdot \\mathrm{{m}}^2 \\cdot 20 \\mathrm{{~T}} \\cdot \\sin 30^{{\\circ}}=0.24 \\mathrm{{~N}} \\cdot \\mathrm{{m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_34_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Find the maximum torque on a circular coil of $0.050 \\mathrm{{~m}}$ radius with current of $0.50 \\mathrm{{~A}}$ and in a magnetic field of $0.0050 \\mathrm{{~T}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The maximum torque is
        $\\displaystyle \\tau_{{\\max }}=\\pi a^2 I B=\\pi(0.050 \\mathrm{{~m}})^2(0.50 \\mathrm{{~A}})(0.0050 \\mathrm{{~T}})=2.0 \\times 10^{{-5}} \\mathrm{{~N}} \\cdot \\mathrm{{m}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_34_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Find an expression for the torque on a circular coil of radius $\\alpha$ carrying a current $I$ placed at an angle to a magnetic field.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-34-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The definition of the magnetic moment and the expression for the torque based on the magnetic moment is most convenient. The magnetic moment is $m=I A=\\pi a^2 I$. The magnitude of the torque is $T=m B \\sin$ $\\alpha=\\pi a^2 I B \\sin \\alpha$ and the direction is the direction determined by $\\boldsymbol{{m}} \\times \\boldsymbol{{B}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_34_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Another problem illustrating this force is a rod or bar resting on current-carrying rails with a perpendicular magnetic field. Take the field as 0.50 gauss (approximately the earth's field) and the separation as $3.0 \\mathrm{{~m}}$ and calculate the current needed through the bar of mass $0.20 \\mathrm{{~kg}}$ with coefficient of friction 0.070 to make it slide along the rails.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-34-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The orientation of current (either direction in the bar) and magnetic field is such that the force produced by the $\\boldsymbol{{il}} \\times \\boldsymbol{{B}}$ will slide the bar along the rails. Verify this with the right-hand rule. The force opposing this sliding is $\\mu \\mathrm{{mg}}$ so this is a force-balance problem.
        $i l B=\\mu m g \\quad$ or $\\displaystyle\\quad i=\\frac{{\\mu m g}}{{l B}}=\\frac{{0.070 \\cdot 20 \\mathrm{{~kg}} \\cdot 9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2}}{{3.0 \\mathrm{{~m}} \\cdot 0.50 \\times 10^{{-4}} \\mathrm{{~T}}}}=9.1 \\times 10^4 \\mathrm{{~A}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_34_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A typical problem illustrating the force on a current-carrying wire is the problem of a wire of $0.010 \\mathrm{{~kg}}$ and 1.0m in length suspended by springs in a $\\boldsymbol{{B}}$ field of strength $30 \\mathrm{{~T}}$ directed out of the page. What current will cause zero tension in the springs?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-34-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        First consider the direction of the current. The force due to the magnetic field has to be up so the current has to be from right to left. With the right hrand, form $\\boldsymbol{{I}} \\times \\boldsymbol{{B}}$ with $i$ in buth directions to verify his conclusion. The value of the current that makes for zero tension in the springs occurs when the force due to the magnetic field balances gravity. This is a force-balance problem.
        $\\displaystyle m g-i l B \\text {{ or }} i=\\frac{{m g}}{{I B}}=\\frac{{0.010 \\mathrm{{~kg}} \\cdot 9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2}}{{1.0 \\mathrm{{~m}} \\cdot 30 \\mathrm{{~T}}}}=3.3 \\times 10^{{-3}} \\mathrm{{~A}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_33_11(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        In a Hall experiment the current is $2.0 \\times 10^{{-3}} \\mathrm{{~A}}$ when $2.0 \\mathrm{{~V}}$ is applied to a semiconductor of length $6.0 \\mathrm{{~cm}}$, width $3.0 \\mathrm{{~cm}}$, and depth $0.20 \\mathrm{{~cm}}$. A magnetic field of $2.0 \\mathrm{{~T}}$ produces a Hall voltage of $2.0 \\times 10^{{-4}} \\mathrm{{~V}}$. Find the sign of the charge carriers, their number density, and the resistivity of the material.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-33-9.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        
        soln_s = f"""
        The sign of the Hall voltage indicates negative charge carriers. Now calculate the current density.
        $\\displaystyle j=\\frac{{i}}{{A}}=\\frac{{2.0 \\times 10^{{-3}} \\mathrm{{~A}}}}{{3.0 \\times 10^{{-2}} \\mathrm{{~m}} \\cdot 2.0 \\times 10^{{-3}} \\mathrm{{~m}}}}=33 \\mathrm{{~A}} / \\mathrm{{m}}^2$
        The Hall field is $E_H=V_H / d=2.0 \\times 10^{{-4}} \\mathrm{{~V}} / 3.0 \\times 10^{{-2}} \\mathrm{{~m}}=6.7 \\times 10^{{-3}} \\mathrm{{~V}} / \\mathrm{{m}}$.
        The number density of charge carriers is
        $\\displaystyle n=\\frac{{j B}}{{E_{{H^e}}}}=\\frac{{33 \\mathrm{{~A}} / \\mathrm{{m}}^2 \\cdot 2.0 \\mathrm{{~T}}}}{{6.7 \\times 10^{{-3}} \\mathrm{{~V}} / \\mathrm{{m}} \\cdot 1.6 \\times 10^{{-19}} \\mathrm{{C}}}}=6.2 \\times 10^{{22}} \\frac{{1}}{{\\mathrm{{~m}}^3}}$
        The resistivity of the sample is easily calculated from the defining statement $E=\\rho j$ and remember that this $E$ is not the Hall field but the field driving the current through the sample.
        $2.0 \\times 10^{{-4}} \\mathrm{{~V}} / 6.0 \\times 10^{{-2}} \\mathrm{{~m}}=\\rho\\left(33 \\mathrm{{~A}} / \\mathrm{{m}}^2\\right) \\text {{ or }} \\rho=1.0 \\times 10^{{-4}} \\Omega \\cdot \\mathrm{{m}}$
        These measurements of the sign of the charge carriers, their number density, and the resistivity are very important in the characterization of semiconductor material.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_33_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the charge to mass ratio for charged particles in a J. J. Thomson tube with accelerating voltage of $3000 \\mathrm{{~V}}$ and crossed fields of $E=6.5 \\times 10^4 \\mathrm{{~V}} / \\mathrm{{m}}$ and $B=2.0 \\times 10^3 \\mathrm{{~T}}$.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle \\frac{{e}}{{m}}-\\frac{{E^2}}{{2 V B^2}}=\\frac{{\\left(6.5 \\times 10^4 \\mathrm{{~V}} / \\mathrm{{m}}\\right)^2}}{{2(3000 \\mathrm{{~V}})\\left(2.0 \\times 10^{{-3}} \\mathrm{{~T}}\\right)^2}}=1.76 \\times 10^{{11}} \\mathrm{{C}} / \\mathrm{{kg}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_33_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An electric field of $1.0 \\times 10^5 \\mathrm{{~V}} / \\mathrm{{m}}$ and a magnetic field of $0.35 \\mathrm{{~T}}$ act on a stream of electrons so as to produce zero net force. What is the speed of the electrons?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle v=\\frac{{E}}{{B}}=\\frac{{1.0 \\times 10^3 \\mathrm{{~V}} / \\mathrm{{m}}}}{{0.35 \\mathrm{{~T}}}}=2.8 \\times 10^3 \\mathrm{{~m}} / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_33_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the resolution required of the photographic plate to differentiate between carbon 12 and carbon 14.        
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Use the diameter from the previous problem, $7.0 \\times 10^{{-2}} \\mathrm{{~m}}$, and $m$ as 12 amu with $\\Delta m$ as 2 amu.
        $\\displaystyle \\Delta x=\\frac{{x \\Delta m}}{{2 m}}=\\frac{{7.0 \\times 10^{{-2}} \\mathrm{{~m}} \\cdot 2 \\mathrm{{amu}}}}{{2 \\cdot 12 \\mathrm{{amu}}}}=5.8 \\times 10^{{-3}} \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_33_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the mass of a singly charged ion accelerated through $4120 \\mathrm{{~V}}$ and striking the collection area at $7.0 \\times 10^{{-2}} \\mathrm{{~m}}$ from the entrance point in a $1.0 \\mathrm{{~T}}$ field.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle m=\\frac{{q B^2 x^2}}{{8 V}}=\\frac{{1.6 \\times 10^{{-19}} \\mathrm{{C}}(1.0 \\mathrm{{~T}})^2 49 \\times 10^{{-4}} \\mathrm{{~m}}^2}}{{8.4120}}=2.4 \\times 10^{{-26}} \\mathrm{{~kg}}$
        The ion is carbon 14 , an isotope of carbon 12 .

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_33_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the maximum energy of a deuteron in this cyclotron with maximum available radius of $1.5 \\mathrm{{~m}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The kinetic energy is $(1 / 2) m v^2$, but $v$ can be written in terms of the $r, q, B$, and $m$ according to the basic equation defining the cyclotron orbits.
        $\\displaystyle K E=\\frac{{1}}{{2}} m v^2=\\frac{{1}}{{2}} m\\left(\\frac{{q B r}}{{m}}\\right)^2=\\frac{{(q B r)^2}}{{2 m}}=\\frac{{\\left(1.6 \\times 10^{{-19}} \\mathrm{{C}} \\cdot 2.0 \\mathrm{{~T}} \\cdot 1.5 \\mathrm{{~m}}\\right)^2}}{{2 \\cdot 2 \\cdot 1.7 \\times 10^{{-27}} \\mathrm{{~kg}}}}=3.4 \\times 10^{{-11}} \\mathrm{{~J}}=2.1 \\times 10^8 \\mathrm{{eV}}$
        The deuteron gained this large amount of energy by making many traverses of the distance between the Dees as the voltage was alternating at the cyclotron frequency.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-37-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_33_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the frequency of oscillation of the Dee voltage for a deuteron in a 2.0T magnetic field.        
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle f=\\frac{{q B}}{{2 \\pi m}}=\\frac{{1.6 \\times 10^{{-19}} \\mathrm{{C}} \\cdot 2.0 \\mathrm{{~T}}}}{{2 \\pi \\cdot 2 \\cdot 1.7 \\times 10^{{-27}} \\mathrm{{~kg}}}}-1.5 \\times 10^7 1 / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_33_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An $\\alpha$-particle moves in a circle of radius $5.0 \\times 10^{{-2}} \\mathrm{{~m}}$ in a magnetic tield of $2.0 \\mathrm{{~T}}$. Find the speed of the $\\alpha$ particle and the frequency and period of the motion.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        An $\\alpha$-particle is a helium nucleus, which has two protons and two neutrons. The speed of the $\\alpha$ particle is from the defining equation for the cyclotron.
        $\\displaystyle v=\\frac{{q r B}}{{m}}=\\frac{{2 \\cdot 1.6 \\times 10^{{-19}} \\mathrm{{C}}\\left(5.0 \\times 10^{{-2}} \\mathrm{{~m}}\\right) 2.0 \\mathrm{{~T}}}}{{4 \\cdot 1.7 \\times 10^{{-27}} \\mathrm{{~kg}}}}=4.7 \\times 10^6 \\mathrm{{~m}} / \\mathrm{{s}}$
        The angular velocity is (analogous to mechanics)
        $\\displaystyle \\omega=\\frac{{v}}{{r}}=\\frac{{q B}}{{m}}=\\frac{{2 \\cdot 1.6 \\times 10^{{-19}} \\mathrm{{C}} \\cdot 2.0 \\mathrm{{~T}}}}{{4 \\cdot 1.7 \\times 10^{{-27}} \\mathrm{{~kg}}}}=9.4 \\times 10^7 \\mathrm{{l}} / \\mathrm{{s}}$
        The frequency, or what is called the cyclotron frequency, is $f=w / 2 \\pi=1.5 \\times 10^7 1 / \\mathrm{{s}}$.
        The period, or time for one transit of the circle is $T=1 / f=6.7 \\times 10^{{-8}} \\mathrm{{~s}}$.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_33_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An electron of energy $4.0 \\mathrm{{KeV}}$ is circulating in a plane at right angles to a magnetic field of $3.0 \\mathrm{{~T}}$. What is the radius of the orhit?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The geometry is the same as in Fig. 33-4. First, calculate the velocity from the energy.
        $4.0 \\times 10^3 \\mathrm{{eV}}\\left(1.6 \\times 10^{{-19}} \\mathrm{{~J}} / \\mathrm{{eV}}\\right)=1 / 2\\left(9.1 \\times 10^{{-31}} \\mathrm{{~kg}}\\right) v^2$ or $\\mathrm{{v}}=3.8 \\times 10^7 \\mathrm{{~m}} / \\mathrm{{s}}$
        Now calculate $r$
        $\\displaystyle r=\\frac{{m v}}{{q B}}=\\frac{{9.1 \\times 10^{{-31}} \\mathrm{{~kg}} \\cdot 3.8 \\times 10^7 \\mathrm{{~m}} / \\mathrm{{s}}}}{{1.6 \\times 10^{{-19}} \\mathrm{{C}} \\cdot 3.0 \\mathrm{{~T}}}}=7.1 \\times 10^{{-5}} \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_33_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Take an electron with velocity (components) $v=2.0 \\times 10^6 \\boldsymbol{{i}}-3.0 \\times 10^6 \\boldsymbol{{j}} \\mathrm{{m}} / \\mathrm{{s}}$ and a $B$ field of $\\boldsymbol{{B}}=4.0 \\times 10^{{-3}}$ $K$ T.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The force vector in component form is
        $\\displaystyle F=1.6\\left|\\begin{{matrix}}i & j & k\\\\2.0 & -3.0 & 0\\\\0 & 0 & 4.0\\end{{matrix}}\\right| \\times 10^{{-16}} \\mathrm{{~N}}=1.6\\left[ i\\left|\\begin{{matrix}}-3.0 & 0\\\\0 & 4.0\\end{{matrix}}\\right| - j \\left|\\begin{{matrix}}2.0 & 0\\\\0 & 4.0\\end{{matrix}}\\right|\\right] \\times 10^{{-16}} \\mathrm{{~N}}$
        $F=1.6[-12 i-8.0 j] \\times 10^{{-16}} \\mathrm{{~N}}=[-1.9 i-1.3 j] \\times 10^{{-15}} \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_33_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the deflection of the electrons in a typical video display tube due to the earth's magnetic field. Take the electrons with 10KeV energy. Orient the tube north to south and with the electron beam at right angles to the earth's $B$ field of 0.40 gauss.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First calculate the velocity of the electrons using an energy relation $e V=(1 / 2) m v^2$.
        $10 \\times 10^3 \\mathrm{{eV}}\\left(1.6 \\times 10^{{-19}} \\mathrm{{~J}} / \\mathrm{{eV}}\\right)=(1 / 2) m v^2=(1 / 2) 9.1 \\times 10^{{-31}} \\mathrm{{~kg}} \\cdot v^2$
        $\\displaystyle v=\\left[\\frac{{2 \\cdot 1.6 \\times 10^{{-15}} \\mathrm{{~J}}}}{{9.1 \\times 10^{{-31}} \\mathrm{{~kg}}}}\\right]^{{1 / 2}}=5.9 \\times 10^7 \\mathrm{{~m}} / \\mathrm{{s}}$
        Convert gauss to Tesla: 0.40 gauss $\\left(1 \\mathrm{{~T}} / 10^4\\right.$ gauss $)=4.0 \\times 10^5 \\mathrm{{~T}}$.
        The force on the electron is
        $F=q v B=1.6 \\times 10^{{-19}} \\mathrm{{C}}\\left(5.9 \\times 10^7 \\mathrm{{~m}} / \\mathrm{{s}}\\right) 4.0 \\times 10^{{-5}} \\mathrm{{~T}}=3.8 \\times 10^{{-16}} \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_32_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An RC circuit is observed during discharge to have an initial capacitor potential of $100 \\mathrm{{~V}}$ and atter $3.0 \\mathrm{{~s}}$ to have a potential of $20 \\mathrm{{~V}}$. How long will it take for the capacitor to discharge to $1.0 \\mathrm{{~V}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The voltage across the capacitor at any time is determined by equation $32-5$ rewritten as $q / C$, or $V_c=V_o e^{{-t R C}}$, where $V_o$ is the voltage at $t=0$.
        Take $V_o=100, V_c=20 \\mathrm{{~V}}$, and $t=3.0 \\mathrm{{~s}}$ and write
        $\\displaystyle 20 \\mathrm{{~V}}=100 \\mathrm{{~V}} e^{{-30 / R C}} \\text {{ or }} 2 / 10=e^{{-3.0 / R C}} \\text {{ or }} e^{{3.0 / R C}}=5$
        Switching to logarithms
        $\\displaystyle \\frac{{3.0}}{{R C}}=\\ln 5 \\text {{ or }} \\frac{{3.0}}{{\\ln 5}}=R C \\text {{ or }} R C=1.9 \\mathrm{{~s}}$
        Knowing this number, the specific decay law for this circuit can be written $V=100 \\mathrm{{~V}} e^{{-t / 1.9}}$.
        Now calculate the time for the voltage to drop to $1.0 \\mathrm{{~V}}$.
        $\\displaystyle 1.0 \\mathrm{{~V}}=100 \\mathrm{{~V}} e^{{-t / 1.9}} \\text {{ or }}(1 / 100)=e^{{-t / 1.9}}$
        Switching to logarithms
        $\\displaystyle -\\ln 100=-(t / 1.9) \\text {{ or }} t=1.9 \\cdot \\ln 100=8.6 \\mathrm{{~s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_32_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the circuit described in prohlem 32-1 placed in the discharge mode, how long does it take for the circuit to discharge to $50 \\%$ of its original (total) charge?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Solve equation $32-5$ for $t$ when $q=0.50 \\mathrm{{CV}}$
        $\\displaystyle 0.50 \\mathrm{{CV}}=\\mathrm{{CV}} e^{{-t / R C}}$ or $0.50=e^{{-t / R C}}$
        It is more convenient to write 0.50 as $1 / 2$, so when the statement is converted to logarithms,
        $\\displaystyle \\ln 0.50=-\\frac{{t}}{{R C}}$ is $\\displaystyle \\ln 1-\\ln 2=-\\frac{{t}}{{R C}}$ and since $\\ln 1=0$
        $t=R C \\ln 2=1.0 \\times 10^4 \\Omega \\cdot 20 \\times 10^{{-6}} \\mathrm{{~F}} \\cdot \\ln 2=(0.20)(0.69 \\mathrm{{~s}})=0.14 \\mathrm{{~s}}$
        In terms of time constants this would be $1 \\mathrm{{n}} 2$ time constants, or 0.69 of a time constant.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_32_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the circuit of problem $32-1$, how long does it take for the capacitor to reach $80 \\%$ of its final charge?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This problem is similar to radioactive decay problems where we want to know the time for half the material to decay. There is a fair amount of algebraic manipulation that is easier to follow without numbers, so the problem will be worked as far as possible with symbols.
        Start with $q=C V\\left(1-e^{{-t / R C}}\\right)$ and note that the final (fully charged) $q$ is equal to $C V$.
        Mathematically, when $t \\rightarrow \\infty, e^{{-t / R C}} \\rightarrow 1$ so $q \\rightarrow C V$.
        To find the time for $80 \\%$ charge set $q$ equal to $80 \\%$ of the final charge, or $q=0.80 \\mathrm{{CV}}$, and solve for $t$.
        $\\displaystyle 0.80 C V=C V\\left(1-e^{{-t / R C}}\\right) \\text {{ or }} 0.80=1-e^{{-t / R C}} \\text {{ or }} e^{{-t / R C}}=0.20$
        For convenience switch to positive exponents so $1 / e^{{-t / R C}}=0.20$ or $e^{{-t / R C}}=1 / 0.20=5$
        In order to solve for $t$, switch the exponential equation to a logarithmic equation. One of the functions of logarithms is to solve for variables in exponents.
        $t / R C=\\ln 5$ or $t=R C \\ln 5$
        Now put in the values for $R$ and $C$.
        $\\displaystyle t=R C \\ln 5=1.0 \\times 10^4 \\Omega \\cdot 20 \\times 10^{{-6}} \\mathrm{{~F}} \\cdot \\ln 5=0.32 \\mathrm{{~s}}$
        As a check note that $1-e^{{-U / R C}}=1-e^{{-0.3210 .20}}=0.80$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_32_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the circuit of problem 32-1, find the time constant, and the charge, current, and $V_R$, and $V_C$ at a time equal to one time constant.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The time constant is $R C=10 \\times 10^3 \\Omega .20 \\times 10^{{-6}} \\mathrm{{~F}}=0.20 \\mathrm{{~s}}$.
        The charge on the capacitor at $t=0.20 \\mathrm{{~s}}$ is
        $\\left.q\\right|_{{t=R C}}=C V\\left(1-e^{{-1}}\\right)=12 \\mathrm{{~V}}\\left(20 \\times 10^{{-6}} \\mathrm{{~F}}\\right) 0.63=1.5 \\times 10^{{-4}} \\mathrm{{C}}$
        The current at $t=0.20 \\mathrm{{~s}}$ is
        $\\displaystyle \\left.i\\right|_{{t=R C}}=\\frac{{V}}{{R}} e^{{-1}}=\\frac{{12 \\mathrm{{~V}}}}{{1.0 \\times 10^4 \\Omega}} 0.37=4.4 \\times 10^{{-4}} \\mathrm{{~A}}$
        The voltage across the capacitor is $V_c=V\\left(1-\\mathrm{{e}}^{{-1}}\\right)=12 \\mathrm{{~V}} .0 .63=7.6 \\mathrm{{~V}}$.
        The voltage across the resistor is $12 \\mathrm{{~V}}-7.6 \\mathrm{{~V}}=4.4 \\mathrm{{~V}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_32_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $10 \\mathrm{{k}} \\Omega$ resistor and a $20 \\mu \\mathrm{{F}}$ capacitor are placed in series with a $12 \\mathrm{{~V}}$ battery. Find the charge on the capacitor, the current, and the voltages on the capacitor and resistor at the instant the switch is closed, $t=0$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        At $t=0$ the charge on the capacitur is zero.
        At $t=0$, the current is $i=V / R=12 \\mathrm{{~V}} / 10 \\times 10^3 \\Omega=1.2 \\times 10^{{-3}} \\mathrm{{~A}}$.
        At $t=0$, the voltage on the capacitor is zero (it has no charge) and the entire battery voltage of $12 \\mathrm{{~V}}$ is across the resistor.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_31_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the currents in the resistors of the circuit shown in Fig. 31-5.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-31-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Draw the battery and resistor circuit and $V$ - $I R$ equation to remind you of the algebraic sign convention. Apply the "fat wire law" to the line across the top of the circuit.
        $l_1-I_2-I_3+I_4=0$
        The loop around the left branch is $4-6 I_1-3 I_2+2-4 I_2=0$.
        The middle loop is $4 I_2-2+3 I_2-8 I_3=0$.
        The loop around the right branch is $-3+5 I_4+8 I_3=0$.
        $I_1-I_2-I_3+I_4 =0$
        $6 I_1+7 I_2 =6$
        $7 I_2-8 I_3 =2$
        $8 I_3+5 I_4 =3$
        Work the problem and verify that the currents are $I_1=0.41, I_2=0.51, I_3=0.19$, and $I_4=0.29$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_31_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider a more complicated problem as shown in Fig. 31-4 and calculate the currents in the resistors.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-31-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Draw the battery and resistor circuit and $V-I R$ equation to remind you of the algebraic sign convention.
        Note the current arrows. Remember that if one of the currents comes out negative it means only that the current is opposite to the direction of your arrow.
        Apply the "fat wire law" to the line across the top of the three resistors.
        $I_1+I_2+I_4=I_3$
        The loop around the top branch of the circuit is $6-8 I_1=0$.
        A loop around the lower left branch is $10-10 I_4-5 I_4-3=0$.
        A lonp around the lower right branch is $3+5 I_3+6 I_2-6=0$.
        Rewrite the equations in convenient form for solving.
        $I_1+I_2-I_3+I_4 =0$
        $8 I_1 =6$
        $6 I_2+5 I_3 =3$
        $5 I_3+10 I_4 =7$
        Immediately $I_1=3 / 4=0.75$. Work the prohlem and verify that the other currents are $I_2=-0.20, I_3=0.84$, and $I_4=0.28$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_31_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Solve for the current in each of the resistors in the circuit shown in Fig. 31-3.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        
        file_name = 'how-to-solve-physics-problems-fig-31-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Follow the procedure for Kirchhoff problems. This is a subject that is conceptually easy but because of the large number of manipulations often hard to actually do. Work this problem as you are following along in the book. Draw the circuit and begin.
        Draw the single battery and single resistor with current arrow and $V-I R=0$ equation.
        Place and label the current arrows.
        Write the statement of the "fat wire law" for the top center of the circuit.
        $l_1+l_3=I_2$
        Write a loop statement for the left side of the circuit.
        $6-8 I_1-10 I_2+2=0$
        Write a loop statement for the right side of the circuit.
        $-2+10 l_2+6 I_3-4=0$
        Now write these three equations in a convenient form for solution.
        $I_1-I_2+I_3 =0$
        $8 I_1+10 I_2 =8$
        $10 I_2+6 I_3 =6$
        As an exercise solve these equations and verify that the solutions are $I_1=17 / 47=0.36, I_2=24 / 47=0.51$, and $I_3=7 / 47=0.15$. Check these answers in the original equations.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_30_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        In the circuit of Fig. 30-6 calculate the voltage across, current through, and power requirements of the $8.0 \\Omega$ resistor when $15 \\mathrm{{~V}}$ is applied to the network.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-30-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The first step in the analysis is to find the equivalent of $R_7$ and $R_3$.
        $\\displaystyle \\frac{{1}}{{R_6}}=\\frac{{1}}{{R_2}}+\\frac{{1}}{{R_3}}=\\frac{{1}}{{8.0 \\Omega}}+\\frac{{1}}{{12 \\Omega}}=\\frac{{5}}{{24 \\Omega}} \\text {{ or }} \\quad R_6=\\frac{{24}}{{5}}=4.8 \\mathrm{{~S}} 2$
        A similar analysis on $R_4$ and $R_5$ yields
        $\\displaystyle \\frac{{1}}{{R_7}}=\\frac{{1}}{{R_4}}+\\frac{{1}}{{R_5}}=\\frac{{1}}{{6.0 \\Omega}}+\\frac{{1}}{{10 \\Omega}}=\\frac{{8}}{{30 \\Omega}} \\quad \\text {{ or }} \\quad R_6=\\frac{{30}}{{8}}=3.8 \\Omega$
        Now the equivalent can be drawn.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_30_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Apply $20 \\mathrm{{~V}}$ across this network and find the voltage across and current through each resistor.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        In order to find the current in a resistor in the original network, it is necessary to work back through the equivalent circuits. Start with the last circuit. Apply $20 \\mathrm{{~V}}$ to a $14.5 \\Omega$ resistor for a current of $I=V / R=20$ $\\mathrm{{V}} / 14.5 \\Omega=1.4 \\mathrm{{~A}}$. Working backwards, this current through $R_{{\\mathrm{{l}}}}$ produces a voltage across this resistor of $V_1=(1.4 \\mathrm{{~A}}) R_1=(1.4 \\mathrm{{~A}})(8.0 \\Omega)=11 \\mathrm{{~V}}$. This current passes through the equivalent resistor $R_4$ producing a voltage across it of $V_4=(1.4 \\mathrm{{~A}}) R_4=(1.4 \\mathrm{{~A}})(6.5 \\Omega)=9.1 \\mathrm{{~V}}$. $V_1$ and $V_4$ add to $20 \\mathrm{{~V}}$, the total applied, to within round off error range. The 1.4A divides between $R_2$ and $R_3$. The voltage across $R_2$ and $R_3$ is $V_4=9.1 \\mathrm{{~V}}$ so
        $I_2=9.1 \\mathrm{{~V}} / 12 \\Omega=0.76 \\mathrm{{~A}} \\quad I_3=9.1 \\mathrm{{~V}} / 14 \\Omega=0.65 \\mathrm{{~A}}$
        Again this total adds to $1.4 \\mathrm{{~A}}$, within round off error. This analysis produces the voltage across and current through each resistor in the network. The numbers are internally consistent (the currents through $R_2$ and $R_3$ add up to the current through their equivalence) giving confidence in the calculation.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_30_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the equivalent resistance of the series-parallel combination shown in Fig. 30-4.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-30-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        
        soln_s = f"""
        The first step in this circuit analysis is to find the equivalent of the parallel combination.
        $\\displaystyle \\frac{{1}}{{R_4}}=\\frac{{1}}{{R_2}}+\\frac{{1}}{{R_3}}=\\frac{{1}}{{12 \\Omega}}+\\frac{{1}}{{14 \\Omega}}=\\frac{{13}}{{84 \\Omega}} \\text {{ or }} R_4=\\frac{{84 \\Omega}}{{13}}=6.5 \\Omega$
        The circuit then becomes two series resistors and they add to a single resistor.
        NOTE: The equivalent resistance of any two resistors is lower than either of the two alone.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-30-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_30_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the current in each resistor for a parallel combination of $3.0 \\Omega, 3.5 \\Omega$, and $4.0 \\Omega$ resistors with $8.0 \\mathrm{{~V}}$ ?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle I_1=\\frac{{8.0 \\mathrm{{~V}}}}{{3.0 \\Omega}}=2.7 \\mathrm{{~A}} \\quad I_2=\\frac{{8.0 \\mathrm{{~V}}}}{{3.5 \\Omega}}=2.3 \\mathrm{{~A}} \\quad I_3=\\frac{{8.0 \\mathrm{{~V}}}}{{4.0 \\Omega}}=2.0 \\mathrm{{~A}}$
        The total current delivered by the power source is $I=2.7 \\mathrm{{~A}}+2.3 \\mathrm{{~A}}+2.0 \\mathrm{{~A}}=7.0 \\mathrm{{~A}}$. Now check the total by using the equivalence
        $\\displaystyle \\frac{{1}}{{R_{{e q}}}}=\\frac{{1}}{{3.0 \\Omega}}+\\frac{{1}}{{3.5 \\Omega}}+\\frac{{1}}{{4.0 \\Omega}}=\\frac{{36.5}}{{42 \\Omega}} \\quad \\text {{ or }} \\quad R_{{e q}}=1.15 \\Omega$
        Using the equivalent resistance $(1.15 \\Omega)$ the total current is $I=V / R_{{e q}}=8.0 \\mathrm{{~V}} / 1.15 \\Omega=7.0 \\mathrm{{~A}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_30_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the current in each resistor for a parallel combination of $3.0 \\Omega, 3.5 \\Omega$, and $4.0 \\Omega$ resistors with $8.0 \\mathrm{{~V}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The equivalent resistance is $R_{{\\mathrm{{eq}}}}=3.0 \\Omega+4.0 \\Omega=7.0 \\Omega$. The current is the same in each resistor and has value $I=V / \\mathrm{{R}}_{{\\mathrm{{eq}}}}=10 \\mathrm{{~V}} / 7.0 \\Omega=1.4 \\mathrm{{~A}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_29_12(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A battery of $12 \\mathrm{{~V}} e \\mathrm{{emf}}$ has internal resistance of $0.50 \\Omega$ and is connected to a $6.0 \\Omega$ load. What is the terminal voltage of the battery under these conditions and the power dissipated in the external resistor?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The total current comes from $e m f=\\mathrm{{i}}\\left(\\mathrm{{r}}_{{\\mathrm{{i}}}}+R\\right)$ or $\\displaystyle i=\\frac{{12 \\mathrm{{~V}}}}{{6.50 \\Omega}}=1.85 \\mathrm{{~A}}$
        At this current the voltage drop due to the internal resistor is 1.85A.0.50S $2=0.92 \\mathrm{{~V}}$, so the output voltage or voltage applied to the $6.0 \\Omega$ resistor is $11.08 \\mathrm{{~V}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-29-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The power dissipated in the external resistor is $P=V i=11.08 \\mathrm{{~V}} .1 .85 \\mathrm{{~A}}=20.5 \\mathrm{{~W}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_29_11(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A battery with an $e m f$ of $10 \\mathrm{{~V}}$ has internal resistance of $0.70 \\Omega$. What is the voltage at the terminals of the battery when $1.0 \\mathrm{{~A}}$ is being drawn by an external load?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-29-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        
        soln_s = f"""
        Solution: The voltage drop inside the battery is $\\mathrm{{r}}_{{\\mathrm{{i}}}}=1.0 \\mathrm{{~A}} \\cdot 0 .70 \\Omega=0.70 \\mathrm{{~V}}$, so the output voltage is reduced to $9.30 \\mathrm{{~V}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_29_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the resistance of an iron bar when the temperature is raised from $0^{{\\circ}} \\mathrm{{C}}$ to $100^{{\\circ}} \\mathrm{{C}}$.
        The initial resistance is $1.43 \\Omega$, and the temperature coefficient of resistivity is $0.00501 / \\mathrm{{C}}^{{\\circ}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $R=R_{{\\circ}}\\left[1+\\alpha\\left(T-T_a\\right)\\right]=1.43 \\Omega\\left[1+0.00501 / \\mathrm{{C}}^{{\\circ}}\\left(100 \\mathrm{{C}}^{{\\circ}}\\right)\\right]=2.16 \\Omega$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_29_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation of problem $29-8$, if the voltage is reduced to $110 \\mathrm{{~V}}$ (assume no change in resistance) how does the heat output change?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This is best handled with a variational equation. In this case start with $P=V^2 / R$ and ask how does $P$ change with $V$, while $R$ remains constant. The left side becomes $\\Delta P$, and the right side becomes $(1 / R)$ times the differential of $V^2$. Writing in delta format
        $\\displaystyle \\Delta P=\\frac{{1}}{{R}} \\Delta\\left(V^2\\right)=\\frac{{1}}{{R}} 2 V \\Delta V$
        The $2 V \\Delta V$ is the difference between $(V+\\Delta V)^2$ and $V^2$.
        $(V+\\Delta V)^2-V^2=V^2+2 V \\Delta V+\\Delta V^2-V^2=2 V \\Delta V$
        The $\\Delta V^2$ term is dropped because it is small compared with $2 V \\Delta V$. For this case them
        $\\displaystyle \\Delta P=\\frac{{1}}{{R}} 2 V \\Delta V-\\frac{{1}}{{13.2 \\Omega}}2 \\cdot 115 \\mathrm{{~V}} \\cdot 5 \\mathrm{{~V}}-87 \\mathrm{{~W}}$
        What is the fractional change in power output, $\\Delta P / P$ ?
        NOTE: The fractional change is the change divided by the original amount.
        $\\displaystyle \\frac{{\\Delta P}}{{P}}=\\frac{{2 V \\Delta V}}{{R}} \\frac{{R}}{{V^2}}-2 \\frac{{\\Delta V}}{{V}}-2 \\frac{{5}}{{115}}-0.087$
        There is an $8.7 \\%$ decline in power for a $4.35 \\%$ decline in voltage.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_29_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $1000 \\mathrm{{~W}}$ electric heater operates at $115 \\mathrm{{~V}}$. Calculate the current, resistance, and energy generated in one hour.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The current in the heater is
        $\\displaystyle i=\\frac{{P}}{{V}}=\\frac{{1000 \\mathrm{{~W}}}}{{115 \\mathrm{{~V}}}}=8.7 \\mathrm{{~A}}$
        The resistance is
        $\\displaystyle R=\\frac{{V^2}}{{P}}=\\frac{{(115 \\mathrm{{~V}})^2}}{{1000 \\mathrm{{~W}}}}=13.2 \\Omega$
        The energy generated in one hour is $U=P t=(1000 \\mathrm{{~J}} / \\mathrm{{s}})(3600 \\mathrm{{~s}})=3.6 \\times 10^6 \\mathrm{{~J}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_29_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Take a rectangular block of metal $1.0 \\times 10^{{-4}} \\mathrm{{~m}}^2$ in cross section, $50 \\mathrm{{~cm}}$ in length with a relaxation time of $3.0 \\times 10^{{-14}} \\mathrm{{~s}}$. Apply $4.0 \\mathrm{{~V}}$ for a current of $3.0 \\mathrm{{~A}}$. Calculate resistivity, electrons/sec. passing a cross section, current density, the acceleration of the electrons, and the drift velocity.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The resistivity is
        $\\displaystyle \\rho=\\frac{{R A}}{{\\ell}}=\\frac{{4.0 \\mathrm{{~V}}}}{{3.0 \\mathrm{{~A}}}} \\frac{{1.0 \\times 10^{{-4}} \\mathrm{{~m}}^2}}{{50 \\times 10^{{-2}} \\mathrm{{~m}}}}=2.7 \\times 10^{{-4}} \\Omega \\cdot \\mathrm{{m}} .$
        The number of $e^{{-/ s}}$ passing a cross section is
        $\\displaystyle N=\\frac{{3.0 \\mathrm{{C}}}}{{\\mathrm{{s}}}} \\frac{{1 e^{{-}}}}{{1.6 \\times 10^{{-19}} \\mathrm{{C}}}}=\\frac{{1.9 \\times 10^{{19}} e^{{-}}}}{{\\mathrm{{s}}}} .$
        Calculate the current density.
        $\\displaystyle j=\\frac{{i}}{{A}}=\\frac{{3.0 \\mathrm{{~A}}}}{{1.0 \\times 10^{{-4}} \\mathrm{{~m}}^2}}=3.0 \\times 10^4 \\frac{{\\mathrm{{A}}}}{{\\mathrm{{m}}^2}}$
        Calculate the acceleration of the electrons.
        $\\displaystyle a=\\frac{{e E}}{{m}}=\\frac{{1.6 \\times 10^{{-19}} \\mathrm{{C}}}}{{9.1 \\times 10^{{-31}} \\mathrm{{~kg}}}} \\frac{{4.0 \\mathrm{{~V}}}}{{50 \\times 10^{{-2}} \\mathrm{{~m}}}}-1.4 \\times 10^{{12}} \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}^2}}$
        Calculate the drift velocity. $v_d=a \\tau=\\left(1.4 \\times 10^{{12}} \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)\\left(3.0 \\times 10^{{-14}} \\mathrm{{~s}}\\right)=4.2 \\times 10^{{-2}} \\mathrm{{~m}} / \\mathrm{{s}}$
        NOTE: The mean free speed, the speed of the electrons before the application of the field, is the order of $10^6$ $\\mathrm{{m}} / \\mathrm{{s}}$.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_29_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The resistivity of copper is $1.7 \\times 10^{{-8}} \\Omega$.m. What current flows through a $2.0 \\mathrm{{~m}}$ long copper conductor of $1.0 \\times 10^{{-4}} \\mathrm{{~m}}^2$ cross section when $20 \\mathrm{{~V}}$ is applied?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First calculate the resistance:
        $\\displaystyle R=\\rho \\frac{{\\ell}}{{A}}=\\frac{{1.7 \\times 10^{{-8}} \\Omega \\cdot \\mathrm{{m}}(2.0 \\mathrm{{~m}})}}{{1.0 \\times 10^{{-4}} \\mathrm{{~m}}^2}}=3.4 \\times 10^{{-4}} \\Omega$
        Now the current:
        $\\displaystyle i=\\frac{{V}}{{R}}=\\frac{{20 \\mathrm{{~V}}}}{{3.4 \\times 10^{{-4}} \\Omega}}=5.9 \\times 10^4 \\mathrm{{~A}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_29_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A metal conductor with $2.0 \\times 10^{{25}}$ charge carriers $/ \\mathrm{{m}}^3$ and cross-sectional area $1.0 \\times 10^{{-4}} \\mathrm{{~m}}^2$ has a current of $4.0 \\times 10^{{-2}} \\mathrm{{~A}}$. What is the drift velocity of the charge carriers?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle v_d=\\frac{{j}}{{n e}}=\\frac{{4.0 \\times 10^{{-2}} \\mathrm{{~A}} / 1.0 \\times 10^{{-4}} \\mathrm{{~m}}^2}}{{2.0 \\times 10^{{25}}\\left(1 / \\mathrm{{m}}^3\\right) \\cdot 1.6 \\times 10^{{-19}} \\mathrm{{C}}}}=1.25 \\times 10^{{-4}} \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_29_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the current and current density in a $1.2 \\mathrm{{~cm}}$ diameter gas discharge tube if $3.0 \\times 10^{{17}}$ electrons and $2.0 \\times 10^{{16}}$ positive ions pass a cross section of the tube each second?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The current is due to electrons going one way and positive ions going the other way.
        $i=3.2 \\times 10^{{17}}(1 / \\mathrm{{s}}) \\cdot 1.6 \\times 10^{{-19}} \\mathrm{{C}}=0.051 \\mathrm{{~A}}$
        The current density is $\\displaystyle j=\\frac{{3.2 \\times 10^{{17}}(1 / \\mathrm{{s}}) \\cdot 1.6 \\times 10^{{-19}} \\mathrm{{C}}}}{{\\pi\\left(0.36 \\times 10^{{-4}} \\mathrm{{~m}}^2\\right)}}=453 \\frac{{\\mathrm{{A}}}}{{\\mathrm{{m}}^2}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_29_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the number of electrons per second.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle \\frac{{5.0 \\mathrm{{C}}}}{{\\mathrm{{s}}}} \\frac{{1 e^{{-}}}}{{1.6 \\times 10^{{-19}} \\mathrm{{C}}}}=3.1 \\times 10^{{19}} e^{{-}} \\text {{per second }}$
        Notice that a current on the order of $10^{{-15}} \\mathrm{{~A}}$ would amount to $10^4$ electrons per second.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_29_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the amount of charge that passes a cross section each second.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $5.0 \\mathrm{{~A}}=5.0 \\mathrm{{C}} / \\mathrm{{s}}$, so $5.0 \\mathrm{{C}}$ passes a cross second each second.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_29_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A current of $5.0 \\mathrm{{~A}}$ exists in a wire of $0.50 \\mathrm{{~cm}}$ diameter. Calculate the current density.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle j=\\frac{{i}}{{A}}=\\frac{{5.0 \\mathrm{{~A}}}}{{\\pi\\left(0.25 \\times 10^{{-4}} \\mathrm{{~m}}^2\\right)}}=6.4 \\times 10^4 \\frac{{\\mathrm{{A}}}}{{\\mathrm{{m}}^2}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_28_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The plates (problem 28-8) are physically pulled apart to double the original separation. Calculate the work performed by the outside agent pulling the plates apart, the average force exerted on the plates, and the average power if the separation occurs over 10s.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First, calculate the original separation from
        $\\displaystyle C=\\frac{{\\varepsilon_o A}}{{d}} \\text {{ or }} d=\\frac{{\\varepsilon_o A}}{{C}}=\\frac{{8.9 \\times 10^{{-12}} \\mathrm{{~F}}}}{{\\mathrm{{~m}}}} \\frac{{0.50 \\mathrm{{~m}}^2}}{{2.0 \\times 10^{{-9}} \\mathrm{{~F}}}}=2.2 \\times 10^{{-3}} \\mathrm{{~m}}$
        Substituting the expression for the capacitance of a parallel plate capacitor the total energy is
        $\\displaystyle U=\\frac{{1}}{{2}} C V^2=\\frac{{1}}{{2}} \\frac{{\\varepsilon_0 A}}{{d}} V^2$
        If the separation is doubled the energy is halved. The energy density is reduced by one-quarter. (Refer to equation 28-9.) In this case the energy is reduced from $90 \\times 10^{{-4}} \\mathrm{{~J}}$ to $45 \\times 10^{{-4}} \\mathrm{{~J}}$. This $45 \\times 10^{{-4}} \\mathrm{{~J}}$ energy loss must be the amount of work done in separating the plates. The work is force times distance, so the average force would be
        $\\displaystyle F_{{\\text {{avg }}}}=\\frac{{\\text {{ work }}}}{{\\text {{ distance }}}}=\\frac{{45 \\times 10^{{-4}} \\mathrm{{~J}}}}{{2.2 \\times 10^{{-3}} \\mathrm{{~m}}}}=2.0 \\mathrm{{~N}}$
        If this work were done over 10 seconds the power would be
        $\\displaystyle P=\\frac{{\\text {{ work }}}}{{\\text {{ time }}}}=\\frac{{45 \\times 10^{{-4}} \\mathrm{{~J}}}}{{10 \\mathrm{{~s}}}}=4.5 \\times 10^{{-4}} \\mathrm{{~W}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_28_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A parallel plate capacitor of $2.0 \\times 10, \\mathrm{{~F}}$ and plate area $0.50 \\mathrm{{~m}}^2$ is connected to $3000 \\mathrm{{~V}}$. Calculate the stored energy.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle U=\\frac{{1}}{{2}} C V^2=\\frac{{1}}{{2}} 2.0 \\times 10^{{-9}} \\mathrm{{~F}} \\cdot 9.0 \\times 10^6 \\mathrm{{~V}}^2=90 \\times 10^{{-4}} \\mathrm{{~J}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_28_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Go back to the previous problem of the $12 \\mu \\mathrm{{F}}$ capacitor charged to $200 \\mathrm{{~V}}$ and then connected to the $7.0 \\mu \\mathrm{{F}}$ capacitor and look at the question of energy.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The $12 \\mu \\mathrm{{F}}$ capacitor at $200 \\mathrm{{~V}}$ has a stored energy of
        $U=(1 / 2) C V^2=(1 / 2) 12 \\mu \\mathrm{{F}} \\cdot 40000 \\mathrm{{~V}}^2=0.24 \\mathrm{{~J}}$
        NOTE: The units of energy are $\\left[\\mathrm{{FV}}_2\\right]-\\left[(\\mathrm{{C}} / \\mathrm{{V}})(\\mathrm{{J}} / \\mathrm{{C}})_2\\right]-[\\mathrm{{J}}]$.
        After the capacitors are connected, the voltage drops to $126 \\mathrm{{~V}}$ and the energy stored in the two capacitors is
        $\\displaystyle U=\\frac{{1}}{{2}} 12 \\mu \\mathrm{{F}} \\cdot(126 \\mathrm{{~V}})^2+\\frac{{1}}{{2}} 7.0 \\mu \\mathrm{{F}}(126 \\mathrm{{~V}})^2=9.5 \\mu \\mathrm{{F}}(126 \\mathrm{{~V}})^2=0.15 \\mathrm{{~J}}$
        Where did the energy go? The difference between the $0.24 \\mathrm{{~J}}$ and the $0.15 \\mathrm{{~J}}$ is the kinetic energy needed to transport the charge or the work to move the charge.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_28_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Charge a $12 \\mu \\mathrm{{F}}$ capacitor with a $200 \\mathrm{{~V}}$ source, then place this capacitor in parallel with an uncharged $7.0 \\mu \\mathrm{{F}}$ capacitor and calculate the "new" voltage.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-28-9.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The $12 \\mu \\mathrm{{F}}$ capacitor initially placed on the $200 \\mathrm{{~V}}$ source has a charge of
        $\\displaystyle q=C V=12 \\mu \\mathrm{{F}} \\cdot 200 \\mathrm{{~V}}=2.4 \\times 10^{{-3}} \\mathrm{{C}}$
        When connected to the $7.0 \\mu \\mathrm{{F}}$ capacitor the charge distributes so that the total charge remains the same and the "new" voltages are equal as they must be for capacitors in parallel. Take $\\mathrm{{Q}}$ as the amount of charge that moves from the $12 \\mu \\mathrm{{F}}$ capacitor to the $7.0 \\mu \\mathrm{{F}}$ capacitor. The voltages as expressed by the ratio of $q$ to $C$ must be equal, so write
        $\\displaystyle \\frac{{2.4 \\times 10^{{-3}} \\mathrm{{C}}-Q}}{{12 \\mu \\mathrm{{F}}}}=\\frac{{Q}}{{7.0 \\mu \\mathrm{{F}}}} \\text {{ so }} Q=8.84 \\times 10^{{-4}} \\mathrm{{C}}$
        The voltage on each of the capacitors is
        $\\displaystyle V_7=\\frac{{q}}{{C}}=\\frac{{8.84 \\times 10^{{-4}} \\mathrm{{C}}}}{{7.0 \\mu \\mathrm{{F}}}}=126 \\mathrm{{~V}} \\quad \\mathrm{{~V}}_{{12}}=\\frac{{q}}{{C}}=\\frac{{1.52 \\times 10^{{-3}} \\mathrm{{C}}}}{{12 \\mu \\mathrm{{F}}}}=126 \\mathrm{{~V}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_28_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The analysis of multiple capacitor combinations is done with (sometimes several) equivalent circuits. Consider the circuit shown that has parallel and series combinations.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-28-7.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        soln_s = f"""
        The first equivalent circuit comes from the combination of $C_1$ and $C_2$.
        $\\displaystyle \\frac{{1}}{{C_4}}=\\frac{{1}}{{C_1}}+\\frac{{1}}{{C_2}}=\\frac{{1}}{{10 \\mu \\mathrm{{F}}}}+\\frac{{1}}{{8.0 \\mu \\mathrm{{F}}}}=\\frac{{9.0}}{{40 \\mu \\mathrm{{F}}}} \\text {{ or }} C_4$
        Now combine $C_3$ and $C_4$ to $C_5=C_3+C_4=4.4 \\mu \\mathrm{{F}}+6.0 \\mu \\mathrm{{F}}=10.4 \\mu \\mathrm{{F}}$ allowing construction of the final (simplest) equivalent circuit
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-28-8.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        Now calculate the total charge $q_t=C_5 V=10.4 \\mu \\mathrm{{F}} \\cdot 300 \\mathrm{{~V}}=3.1 \\times 10_{{.3}} \\mathrm{{C}}$
        The charge on $C_3$ is $q_3=C_3 V=6.0 \\mu \\mathrm{{F}} \\cdot 300 \\mathrm{{~V}}=1.8 \\times 10^{{-3}} \\mathrm{{C}}$.
        The remaining charge is $1.3 \\times 10^{{-3}} \\mathrm{{C}}$, and this is the charge on each of the other capacitors, $C_1$ and $C_2$. Remember that the charge on capacitors in series is the same. The voltage on each of these capacitors is
        $\\displaystyle V_1=\\frac{{q}}{{C_1}}=\\frac{{1.3 \\times 10^{{-3}} \\mathrm{{C}}}}{{10 \\mu \\mathrm{{F}}}}=130 \\mathrm{{~V}} \\quad V_2=\\frac{{q}}{{C_2}}=\\frac{{1.3 \\times 10^{{-3}} \\mathrm{{C}}}}{{8.0 \\mu \\mathrm{{F}}}}=163 \\mathrm{{~V}}$
        The error here of these voltages not adding up to $300 \\mathrm{{~V}}$ is about $2 \\%$, which is expected from round off.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_28_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $12 \\mu \\mathrm{{F}}$ capacitor and a $10 \\mu \\mathrm{{F}}$ capacitor are connected in parallel across a $100 \\mathrm{{~V}}$ battery. Find the charge and potential on each capacitor.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-28-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        soln_s = f"""
        The circuit is as shown. The voltage across each capacitor is the same, $100 \\mathrm{{~V}}$. The charge, however, depends on the capacitance.
        $q_1=C_1 V=12 \\mu \\mathrm{{F}} \\cdot 100 \\mathrm{{~V}}=12 \\times 10^{{-4}} \\mathrm{{C}} \\quad q_2=C_2 V=10 \\mu \\mathrm{{F}} \\cdot 100 \\mathrm{{~V}}=10 \\times 10^{{-4}} \\mathrm{{C}}$
        Look at the equivalence. Take this total charge and the applied voltage and find an equivalent capacitance.
        $\\displaystyle C=\\frac{{q}}{{V}}=\\frac{{22 \\times 10^{{-4}} \\mathrm{{C}}}}{{100 \\mathrm{{~V}}}}=22 \\mu \\mathrm{{F}}$
        This is the sum of the capacitances as expected in a parallel arrangement.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_28_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $6.0 \\mu \\mathrm{{F}}$ capacitor and an $8.0 \\mu \\mathrm{{F}}$ capacitor are connected in series. A potential of $200 \\mathrm{{~V}}$ is placed across them. Find the charge and potential on each.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-28-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        soln_s = f"""
        The circuit is as shown. Since the capacitors are in series the charge on each capacitor is the same. The equivalent capacitance is found from
        $\\displaystyle \\frac{{1}}{{C}}=\\frac{{1}}{{C_1}}+\\frac{{1}}{{C_2}}=\\frac{{1}}{{6.0 \\mu \\mathrm{{F}}}}+\\frac{{1}}{{8.0 \\mu \\mathrm{{F}}}}=\\frac{{4}}{{24 \\mu \\mathrm{{F}}}}+\\frac{{3}}{{24 \\mu \\mathrm{{F}}}}=\\frac{{7}}{{24 \\mu \\mathrm{{F}}}}$
        so the equivalent capacitance is $(24 / 7) \\mu \\mathrm{{F}}$, and the equivalent circuit is as shown.
        The charge is easily calculated from $q=C V=(24 / 7) \\mu \\mathrm{{F}} .200 \\mathrm{{~V}}=686 \\mu \\mathrm{{C}}$. The voltage on each capacitor is
        $\\displaystyle V_1=\\frac{{q}}{{C_1}}=\\frac{{686 \\mu \\mathrm{{C}}}}{{6.0 \\mu \\mathrm{{F}}}}=114 \\mathrm{{~V}} \\quad V_2=\\frac{{q}}{{C_2}}=\\frac{{686 \\mu \\mathrm{{C}}}}{{8.0 \\mu \\mathrm{{F}}}}=86 \\mathrm{{~V}}$
        These voltages add to the $200 \\mathrm{{~V}}$ applied to the combination.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_28_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the capacitance of a spherical capacitor with inner radius $a$ and outer radius $b$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-28-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        
        soln_s = f"""
        Set up the Gaussian surface symmetric about the inner sphere. $E$ is a constant over this surface and $\\displaystyle \\varepsilon_o \\oint E \\cdot d s=q \\quad E=\\frac{{q}}{{4 \\pi \\varepsilon_o r^2}}$
        Now calculate $V$ from
        $\\displaystyle V=-\\int_a^b E \\cdot d r=-\\frac{{q}}{{4 \\pi \\varepsilon_o}} \\int_a^b \\frac{{d r}}{{r^2}}=\\left.\\frac{{q}}{{4 \\pi \\varepsilon_o}} \\frac{{1}}{{r}}\\right|_a ^b=\\frac{{q}}{{4 \\pi \\varepsilon_o}}\\left(\\frac{{1}}{{a}}-\\frac{{1}}{{b}}\\right)=\\frac{{q}}{{4 \\pi \\varepsilon_o}} \\frac{{b-a}}{{a b}}$
        and by definition
        $\\displaystyle C=\\frac{{q}}{{V}} \\quad C=\\frac{{4 \\pi \\varepsilon_o a b}}{{b-a}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_28_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the capacitance of a cylindrical capacitor of inner radius $a$, outer radius $b$, and length $l$
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Set up a cylindrical Gaussian surface symmetrical about the inner cylinder and of length $l$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        file_name = 'how-to-solve-physics-problems-fig-28-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
    
        soln_s = f"""
        $E$ is constant over this surface and $\\displaystyle \\varepsilon_o \\oint E \\cdot d s=q \\quad$ so that $\\displaystyle E=\\frac{{q}}{{2 \\pi \\varepsilon_o r}}$.
        Continuing with the procedure, calculate $V$.
        $\\displaystyle V=-\\int_a^b \\boldsymbol{{E}} \\cdot d \\boldsymbol{{r}}=\\frac{{q}}{{2 \\pi \\varepsilon_o \\ell}} \\int_a^b \\frac{{1}}{{r}} d \\boldsymbol{{r}}=\\left.\\frac{{q}}{{2 \\pi \\varepsilon_o \\ell}} \\ln r\\right|_a ^b=\\frac{{q}}{{2 \\pi \\varepsilon_o \\ell}} \\ln \\frac{{b}}{{a}}$
        $\\displaystyle C=\\frac{{q}}{{V}} \\quad C=\\frac{{2 \\pi \\varepsilon_o \\ell}}{{\\ln (b / a)}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_28_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A capacitor with circular plates of $0.20 \\mathrm{{~m}}$ radius separated by $0.0010 \\mathrm{{~m}}$ is connected to a $100 \\mathrm{{~V}}$ battery. After a long time the battery is disconnected. What is the charge stored on the plates?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The capacitance of the arrangement is
        $\\displaystyle C=\\frac{{\\varepsilon_o A}}{{d}}=\\frac{{8.9 \\times 10^{{-12}} \\mathrm{{~F}}}}{{\\mathrm{{~m}}}} \\frac{{\\pi 0.040 \\mathrm{{~m}}^2}}{{0.0010 \\mathrm{{~m}}^2}}=1.1 \\times 10^{{-9}} \\mathrm{{~F}}$
        The charge is obtained from the definition of capacitance, equation 28-1
        $\\displaystyle q=C V=1.1 \\times 10^{{-9}} \\mathrm{{~F}} \\cdot 100 \\mathrm{{~V}}=1.1 \\times 10^{{-7}} \\mathrm{{C}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_27_14(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Another classic problem has to do with the calculation of the electric potential due to charge on a surface along a line from the center of a disk and perpendicular to the disk.
        $\\displaystyle V=\\frac{{1}}{{4 \\pi \\varepsilon_o}} \\int \\frac{{d q}}{{r}}$

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-27-8.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The potential is
        $\\displaystyle d V=\\frac{{1}}{{4 \\pi c_o}} \\frac{{d q}}{{r}}$
        In this problem $d q$ will have to be written in terms of charge per unit area times (differential) area. Start by writing the differential voltage in terms of the differential element of charge, $d q$, or
        $\\displaystyle d V=\\frac{{\\sigma(2 \\pi y) d y}}{{4 \\pi \\varepsilon_o \\sqrt{{a^2+y^2}}}}$
        The charge on the surface is assumed uniform with surface charge density $\\sigma$ (charge per unit area). Take the charge contained in the circular strip, $\\sigma(2 \\pi y) d y$ ( $2 \\pi y$ is the length of the strip and $d y$ is the width). The $r$ is $\\sqrt{{a^2+y^2}}$
        The integration is to be performed in $y$ from zero to $b$, the radius of the disk
        $\\displaystyle V=\\frac{{\\sigma}}{{4 \\varepsilon_o}} \\int_0^b \\frac{{2 y d y}}{{\\sqrt{{a^2+y^2}}}}$
        Make a change of variable. Let $u=a^2+y^2$ and $d u=2 y d y$ so the integral becomes (and we look at the integral only without the limits)
        $\\displaystyle \\int \\frac{{2 y d y}}{{\\sqrt{{a^2+y^2}}}}=\\int u^{{-1 / 2}} d u=\\frac{{u^{{1 / 2}}}}{{1 / 2}}=2 \\sqrt{{a^2+y^2}}$
        This integral is in the Introduction, Mathematical Background.
        Therefore
        $\\displaystyle V=\\frac{{\\sigma}}{{4 \\varepsilon_o}}\\left[2 \\sqrt{{a^2+y^2}}\\right]_0^b=\\frac{{\\sigma}}{{2 \\varepsilon_o}}\\left[\\sqrt{{a^2+b^2}}-a\\right]$
        Now look at the limiting case where $b / a<<1$, which is often the case. This is most conveniently done by first rewriting the expression in brackets as
        $\\displaystyle V=\\frac{{\\sigma}}{{2 \\varepsilon_o}}\\left[a \\sqrt{{1+b^2 / a^2}}-a\\right]$
        and then writing the square root as a binomial expansion remembering that $b / a<1$. The binomial expansion (see the Introduction, Mathematical Background tor the binomial expansion) is
        $\\displaystyle a\\left(1+b^2 / a^2\\right)^{{1 / 2}}=a\\left(1+\\frac{{1}}{{2}} \\frac{{b^2}}{{a^2}}+\\frac{{1}}{{2}}\\left(-\\frac{{1}}{{2}}\\right) \\frac{{b^4}}{{8 a^3}}+\\ldots\\right)$
        $\\displaystyle V=\\frac{{\\sigma}}{{2 \\varepsilon_o}}\\left[\\frac{{b^2}}{{2 a}}-\\frac{{b^4}}{{8 a^3}}+\\ldots\\right]$ and finally for $b / a<<1$
        $\\displaystyle V=\\frac{{\\sigma \\pi b^2}}{{4 \\pi \\varepsilon_o a}}$
        This is expected since $\\sigma \\pi \\mathrm{{b}}^2$ is the total charge and $a$ is the distance from that charge. A long way away a disk of charge "looks like" a point charge. The use of the binomial expansion is helpful in two ways. First, it allows a convenient calculation for the situation where $b / a<<1$; and second, the limiting case reduces to the simple expression for the potential due to a point charge, in effect verifying the calculation.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_27_13(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the work required to arrange the charges as shown in Fig. 27-7. In this problem the subscripts on the $q$ 's represent the position of the charges

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-27-7.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        
        soln_s = f"""
        To make this calculation place $q_0$ in position (corresponding to the origin of a coordinate system) and calculate the work necessary to place $q_1$ in position. This is
        $\\displaystyle W_{{0-1}}=-\\frac{{q^2}}{{4 \\pi \\varepsilon_o}}\\left(\\frac{{1}}{{r}}\\right)_{{\\infty}}^a=-\\frac{{q^2}}{{4 \\pi \\varepsilon_o a}}$
        Since the charges are opposite, the electric force is attractive and the work to put this charge in position is negative (this charge has to be restrained as it approaches $q_0$ ). Now with these charges in place, add $-2 q_2$ with an amount of work
        $\\displaystyle W_{{0-2}}=\\frac{{2 q^2}}{{4 \\pi \\varepsilon_o}}\\left(\\frac{{1}}{{\\sqrt{{2}} a}}-\\frac{{1}}{{a}}\\right)$
        There are two components to this work calculation, the work to bring $-2 q_2$ to within $\\sqrt{{2}}$ of $-q_0$ and the work to bring $-2 q_2$ to within $a$ of $+q_1$. Adding the third charge means taking into account the three charges already in place
        $\\displaystyle W_{{0-3}}=\\frac{{-q^2}}{{4 \\pi \\varepsilon_0}}\\left(\\frac{{1}}{{a}}-\\frac{{1}}{{\\sqrt{{2}} a}}+\\frac{{2}}{{a}}\\right)$
        The total work is the sum of these, or
        $\\displaystyle W=\\frac{{q^2}}{{4 \\pi \\varepsilon_o}}\\left[-\\frac{{1}}{{a}}+\\frac{{1}}{{\\sqrt{{2}} a}}-\\frac{{1}}{{a}}-\\frac{{1}}{{a}}+\\frac{{1}}{{\\sqrt{{2}} a}}-\\frac{{2}}{{a}}\\right]=\\frac{{q^2}}{{4 \\pi \\varepsilon_o}}\\left[-\\frac{{5}}{{a}}+\\frac{{\\sqrt{{2}}}}{{a}}\\right]=\\frac{{q^2}}{{4 \\pi \\varepsilon_o}}[\\sqrt{{2}}-5] \\frac{{1}}{{a}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_27_12(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A positron (positively charged electron) traveling at $3.0 \\times 10^7 \\mathrm{{~m}} / \\mathrm{{s}}$ is projected directly at a stationary proton. What is the distance of closest approach?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First Solution: The positron will approach the proton until the energy of the positron is expended in doing work against the Coulomb repulsive force. The positron will stop, then exit the vicinity of the proton having a velocity profile on exit the same as, but in the opposite direction, to the entrance.
        $\\displaystyle r_c=\\frac{{2 e^2}}{{4 \\pi \\varepsilon_o}} \\frac{{1}}{{m v^2}}=2\\left(1.6 \\times 10^{{-19}} \\mathrm{{C}}\\right)^2 \\frac{{9.0 \\times 10^9 \\mathrm{{~N}} \\cdot \\mathrm{{m}}^2}}{{\\mathrm{{C}}^2}} \\frac{{1}}{{9.1 \\times 10^{{-31}} \\mathrm{{~kg}}\\left(3.0 \\times 10^7 \\mathrm{{~m}} / \\mathrm{{s}}\\right)^2}}=5.6 \\times 10^{{-13}} \\mathrm{{~m}}$ 
        This work is set equal to the kinectic energy $\\displaystyle \\frac{{e^2}}{{4 \\pi \\varepsilon_o}} \\frac{{1}}{{r_c}}=\\frac{{m^2}}{{2}}$ and solved for 
        $1 \\mathrm{{eV}}=(1 \\mathrm{{~J}} / \\mathrm{{C}}) 1.6 \\times 10^{{-19}} \\mathrm{{C}}=1.6 \\times 10^{{-19}} \\mathrm{{~J}}$
        This is a very close approach being two orders of magnitude smaller than the distance from the nucleus to the first electron orbit radius in the hydrogen atom.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        soln_s = f"""
        
        Second Solution: This problem can be done using the concept of the electron-volt $(\\mathrm{{eV}})$ unit of energy. An electron accelerated through one volt has an amount of work performed on it equal to one volt (the work per unit charge) times the charge on the electron.
        $\\displaystyle K E=\\frac{{m v^2}}{{2}}=\\frac{{9.1 \\times 10^{{-31}} \\mathrm{{~kg}}\\left(3.0 \\times 10^7 \\mathrm{{~m}} / \\mathrm{{s}}\\right)^2}}{{2}}=4.1 \\times 10^{{-16}} \\mathrm{{~J}} \\frac{{1 \\mathrm{{eV}}}}{{1.6 \\times 10^{{-19}} \\mathrm{{~J}}}}=2600 \\mathrm{{eV}}$
        This equation defines the Joule-eV equivalence. An electron accelerated through a potential difference of one volt is said to gain one electron volt in energy. An electron accelerated through 1000 volts would gain $1000 \\mathrm{{eV}}$ in energy. An alpha particle (two positive electronic charges) accelerated through 1000 volts would gain 2000 eV in energy.
        In this particular problem the positron with a velocity of $3.0 \\times 10^7 \\mathrm{{~m}} / \\mathrm{{s}}$ has a kinetic energy of
        This means that the positron has been accelerated through $2600 \\mathrm{{~V}}$.
        A positron with this energy could approach a proton to a radius equivalent to 2600 volt potential.
        Set $\\displaystyle V=\\frac{{q}}{{4 \\pi \\varepsilon_o r_c}}$ equal to $2600 \\mathrm{{~V}}$ and solve for $r_c$
        $\\displaystyle r_c=\\frac{{q}}{{4 \\pi \\varepsilon_o V}}=\\frac{{1.6 \\times 10^{{-19}} \\mathrm{{C}} \\cdot 9.0 \\times 10^9 \\mathrm{{~N}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{C}}^2}}{{2600 \\mathrm{{~V}}}}=5.6 \\times 10^{{-13}} \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_27_11(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The water molecule $\\mathrm{{H}}_2 \\mathrm{{O}}$ has an asymmetric charge distribution leading to a dipole moment $p$ of $6.2 \\times 10^{{-30}} \\mathrm{{C}} . \\mathrm{{m}}$. Calculate the electric potential $1.0 \\mathrm{{~nm}}$ a) at right angles to the direction of the dipole moment and b) at $45^{{\\circ}}$ to the direction of the dipole moment. ( $1.0 \\mathrm{{~nm}}$ is approximately 10 hydrogen atom diameters)

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-27-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        
        soln_s = f"""
        The approximate expression for the electric potential is $\\displaystyle V=\\frac{{1}}{{4 \\pi \\varepsilon_o}} \\frac{{p \\cos \\theta}}{{r^2}}$
        and the electric potentials at $P_a$ and $P_b$ are
        $\\displaystyle V_a=\\frac{{9.0 \\times 10^9 \\mathrm{{~N}} \\cdot \\mathrm{{m}}^2}}{{\\mathrm{{C}}^2}} \\frac{{6.2 \\times 10^{{-30}} \\mathrm{{C}} \\cdot \\mathrm{{m}}\\left(\\cos 90^{{\\circ}}\\right)}}{{\\left(1.0 \\times 10^{{-9}} \\mathrm{{~m}}\\right)^2}}=0 \\mathrm{{~V}} \\quad \\text {{ and }} \\quad V_b=0.039 \\mathrm{{~V}}$
        $\\displaystyle W=\\int_{{\\infty}}^{{r_c}} \\boldsymbol{{F}} \\cdot d r=-\\frac{{e^2}}{{4 \\pi \\varepsilon_o}} \\int_{{\\infty}}^{{r_c}} \\frac{{d r}}{{r^2}}=\\frac{{e^2}}{{4 \\pi \\varepsilon_o}} \\frac{{1}}{{r_c}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_27_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the electric potential along the $x$-axis for the triangular array of charges shown.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-27-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        
        soln_s = f"""
        $\\displaystyle V=\\frac{{1}}{{4 \\pi \\varepsilon_o}} \\sum_i \\frac{{q_i}}{{r_i}}=\\frac{{1}}{{4 \\pi \\varepsilon_o}}\\left[\\frac{{q}}{{x-a}}+\\frac{{q}}{{x+a}}-\\frac{{q}}{{\\sqrt{{x^2+a^2}}}}\\right]$
        $\\displaystyle V=\\frac{{1}}{{4 \\pi \\varepsilon_o}}\\left[\\frac{{q(x+a)+q(x-a)}}{{x^2-a^2}}-\\frac{{q}}{{\\sqrt{{x^2+a^2}}}}\\right]=\\frac{{1}}{{4 \\pi \\varepsilon_o}}\\left[\\frac{{2 q x}}{{x^2-a^2}}-\\frac{{q}}{{\\sqrt{{x^2+a^2}}}}\\right]$
        for $\\displaystyle x<a, V=\\frac{{1}}{{4 \\pi \\varepsilon_o}} \\frac{{q}}{{x}}$
        which is what we would expect.
        A long way away from the collection of charges the collection looks like one positive charge.
        A classic problem is the calculation of the potential at a point a distance $r$ and at an angle $\\theta$ from the center of a dipole. The potential is written by summing the potential from the two charges and using the approximate trigonometric relation between $r, r_1$, and $r_2$.
        $\\displaystyle V=\\frac{{1}}{{4 \\pi \\varepsilon_o}}\\left[\\frac{{q}}{{r_1}}-\\frac{{q}}{{r_2}}\\right]=\\frac{{q}}{{4 \\pi \\varepsilon_o}} \\frac{{r_2-r_1}}{{r_1 r_2}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-27-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        If $r \\gg a$ then $r_1 r_2 \\approx r^2$ and also $\\theta_1 \\approx \\theta_2 \\approx$ so $r_2-r_1 \\approx 2 a \\cos \\theta$ and using the dipole definition $p=2 a q$
        $\\displaystyle V \\approx \\frac{{1}}{{4 \\pi \\varepsilon_o}} \\frac{{p \\cos \\theta}}{{r^2}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_27_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the array shown in Fig. 27-3 calculate the potential mid-way between $q_1$ and $q_2$.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First calculate the distance from this midpoint to $q_3$
        $\\displaystyle 4 a^2-(a / 2)^2=r^2 \\quad a^2(4-1 / 4)=r^2 \\quad a \\sqrt{{15}} / 2=r$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-27-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The potential at the point $P$ is the sum of the contributions from each of the charges.
        $\\displaystyle V=V_1+V_2+V_3=\\frac{{1}}{{4 \\pi \\varepsilon_0}}\\left[\\frac{{1.0 \\times 10^{{-8}} \\mathrm{{C}}}}{{a / 2}}+\\frac{{1.5 \\times 10^{{-8}} \\mathrm{{C}}}}{{a / 2}}-\\frac{{1.5 \\times 10^{{-8}} \\mathrm{{C}}}}{{a \\sqrt{{15}} / 2}}\\right]$
        $\\displaystyle V=\\frac{{2.0 \\times 10^{{-8}} \\mathrm{{C}}}}{{4 \\pi \\varepsilon_0 a}}\\left[2.5-\\frac{{1.5}}{{\\sqrt{{15}}}}\\right]=\\frac{{9.0 \\times 10^9 \\mathrm{{~N}} \\cdot \\mathrm{{m}}^2}}{{\\mathrm{{C}}^2}} \\frac{{2.0 \\times 10^{{-8}} \\mathrm{{C}}}}{{a}}[2.1]=\\frac{{380}}{{a}} \\mathrm{{~V}}$
        if $a$ is measured in meters.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_27_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Now calculate the electric field just outside this surface.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle E=\\frac{{q}}{{4 \\pi \\varepsilon_o}} \\frac{{1}}{{r^2}}=\\frac{{V}}{{r}}=\\frac{{4.5 \\times 10^7 \\mathrm{{~V}}}}{{0.1 \\mathrm{{~m}}}}=4.5 \\times 10^8 \\mathrm{{~N}} / \\mathrm{{C}}$
        As an exercise show that the unit $\\mathrm{{V}} / \\mathrm{{m}}$ is equal to $\\mathrm{{N}} / \\mathrm{{C}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_27_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the electric potential of a spherical (conducting) shell of radius 0.10 min carrying a charge of $5.0 \\times 10^{{-4}} \\mathrm{{C}}$.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        In calculating the potential on the spherical shell the charge acts as if it were at the center of the sphere.
        $\\displaystyle V=\\frac{{q}}{{4 \\pi \\varepsilon_0}} \\frac{{1}}{{r}}=\\frac{{9.0 \\times 10^9 \\mathrm{{~N}} \\cdot \\mathrm{{m}}^2}}{{\\mathrm{{C}}^2}} \\frac{{5.0 \\times 10^{{-4}} \\mathrm{{C}}}}{{0.10 \\mathrm{{~m}}}}=4.5 \\times 10^7 \\mathrm{{~V}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_27_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A 5.0g conducting sphere with charge of $20 \\mu \\mathrm{{C}}$ hangs by a non-conducting thread in an electric field produced by two plates separated by $8.0 \\mathrm{{~cm}}$. What potential will cause the ball to hang at $25^{{\\circ}}$ to the vertical?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-27-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        This is a force-balance problem, the electric force is balanced by the horizontal component of the tension in the thread. Refer to Fig. 27-2 and write the force statements.
        Divide one by the other to arrive at
        $\\displaystyle \\tan 25^{{\\circ}}=\\frac{{E q}}{{m g}}$ or $\\displaystyle E=\\frac{{m g}}{{q}} \\tan 25^{{\\circ}}=\\frac{{5.0 \\times 10^{{-3}} \\mathrm{{~kg}} \\cdot 9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2}}{{20 \\times 10^{{-6}} \\mathrm{{C}}}} \\tan 25^{{\\circ}}=1140 \\mathrm{{~N}} / \\mathrm{{C}}$
        The potential is from $E=V / d$ or $V=E d$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_27_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Use the situation (charge and potential calculations) of problems $27-3$ and $27-4$. If a charge of $5.0 \\times 10^{{-6}} \\mathrm{{C}}$ were moved from a radius of $0.50 \\mathrm{{~m}}$ to a radius of $0.20 \\mathrm{{~m}}$, how much work would be performed?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The work performed is the potential difference in Joule/Coulomb (or volt) times the charge in Coulomb (observe the units carefully).
        $W=2.9 \\times 10^4 \\mathrm{{~J}} / \\mathrm{{C}} \\cdot 5.0 \\times 10^{{-6}} \\mathrm{{C}}=0.15 \\mathrm{{~J}}$
        The work is independent of the path between these radii. This is the great value of potential calculations.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_27_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the electric potential for the same charge but at $0.50 \\mathrm{{~m}}$ away.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle V=\\frac{{q}}{{4 \\pi \\varepsilon_o}} \\frac{{1}}{{r}}=\\frac{{9.0 \\times 10^9 \\mathrm{{~N}} \\cdot \\mathrm{{m}}^2}}{{\\mathrm{{C}}^2}} \\frac{{1.0 \\times 10^{{-6}} \\mathrm{{C}}}}{{0.50 \\mathrm{{~m}}}}=1.8 \\times 10^4 \\mathrm{{~V}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_27_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the electric potential $0.20 \\mathrm{{~m}}$ away from a charge of $1.0 \\times 10^{{-6}} \\mathrm{{C}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle V=\\frac{{q}}{{4 \\pi \\varepsilon_o}} \\frac{{1}}{{r}}=\\frac{{9.0 \\times 10^9 \\mathrm{{~N}} \\cdot \\mathrm{{m}}^2}}{{\\mathrm{{C}}^2}} \\frac{{1.0 \\times 10^{{-6}} \\mathrm{{C}}}}{{0.20 \\mathrm{{~m}}}}=4.5 \\times 10^4 \\mathrm{{~V}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_27_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An electron placed between two charged parallel plates separated by $2.0 \\times 10^{{-2}} \\mathrm{{~m}}$ is observed to accelerate at $5.0 \\times 10^{{14}} \\mathrm{{~m}} / \\mathrm{{s}}^2$. What is the voltage on the plates?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First calculate the force on the electron as
        $F=m a=9.1 \\times 10^{{-31}} \\mathrm{{~kg}} \\cdot 5.0 \\times 10^{{14}} \\mathrm{{~m}} / \\mathrm{{s}}^2=4.6 \\times 10^{{-16}} \\mathrm{{~N}}$
        The electric field required to move this electron is
        $E=F / c^{{-}}=4.6 \\times 10^{{-16}} \\mathrm{{~N}} / 1.6 \\times 10^{{-19}} \\mathrm{{C}}-2.8 \\times 10^3 \\mathrm{{~N}} / \\mathrm{{C}}$
        The voltage for this separation is
        $V=E d=2.8 \\times 10^3 \\mathrm{{~N}} / \\mathrm{{C}} \\cdot 2.0 \\times 10^{{-2}} \\mathrm{{~m}}=57 \\mathrm{{~V}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_27_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        In a certain experiment an electric field of $2.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{C}}$ is required for parallel plates separated by $0.0020 \\mathrm{{~m}}$. What voltage will produce this field?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        For a constant field, $V=E d$ so
        $V=2.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{C}} \\cdot 2.0 \\times 10^{{-3}} \\mathrm{{~m}}=4.0 \\times 10^2 \\mathrm{{~J}} / \\mathrm{{C}}=400 \\mathrm{{~V}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_26_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the electric field between cylinders carrying charges as shown. The $+q$ is on the inner conductor and $-q$ on the inside of the outer conductor with another $-q$ on the outside of the outer conductor.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        On a surface between the conductors, $\\displaystyle \\quad \\varepsilon_o \\oint E \\cdot d s=q$ applics to the charge contained
        and $\\displaystyle \\varepsilon_o \\oint E \\cdot d s=E \\cdot 2 \\pi r \\ell$ with $2 \\pi r l$ the surface area of the Gaussian cylinder, corresponding to the point where the field is required, so in this region $E$ points inward and has value $E=q / 2 \\pi \\varepsilon_o r l$.

        Note that the electric field at a point between the cylinders is only due to the charge enclosed by the Gaussian surface. Because of the symmetry of the problem, the charge on the outer cylinder produces no net electric field at any point inside the outer cylinder.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-26-7.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        Outside the outer cylinder the $E$ field points inward because the net charge contained is negative, and at any point $r$ (outside the outer cylinder) measured from the center line of the two cylinders $E=q / 2 \\pi \\varepsilon_0 r l$.

        With this particular arrangement of charge, the field has the same algebraic form between the cylinders and outside the outer cylinder but points outward between the cylinders and inward outside the outer cylinder.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_26_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A planetary probe is traveling radially inward toward the center of the planet. At $600 \\mathrm{{~m}}$ above the surface the field is directed up and has value $180 \\mathrm{{~N}} / \\mathrm{{C}}$. At $500 \\mathrm{{~m}}$ it is still directed up but has value $125 \\mathrm{{~N}} / \\mathrm{{C}}$. Find the sign and density of the charge (density) in this region.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The radius of the planet is sufficiently large so that we can take the region as a cube $100 \\mathrm{{~m}}$ on a side. The electric fields are as shown in Fig. 26-6. Gauss' law can be written as
        $\\displaystyle \\varepsilon_0 \\oint E \\cdot d \\boldsymbol{{s}}=\\varepsilon_o\\left[\\int_{{\\text {{sides }}}} E \\cdot d s+\\int_{{\\text {{top }}}} E \\cdot d s+\\int_{{\\text {{bvulum }}}} E \\cdot d s\\right]=q_{{\\text {{enclosed }}}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        
        file_name = 'how-to-solve-physics-problems-fig-26-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The electric field at the sides is zero. At the top, the electric field and the vector representing $d s$ point in the same direction so the dot product has a plus sign. On the bottom, the electric field vector points up while the vector representing $d s$ points down giving the dot product a minus sign. Putting in the values for the integrals.
        $\\displaystyle \\varepsilon_o\\left[\\text {{ zero }}+1.8 \\times 10^6 \\mathrm{{~N}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{C}}-1.25 \\times 10^6 \\mathrm{{~N}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{C}}\\right]=q$
        the total charge enclosed within the volume is
        $q=4.9 \\times 10^{{-6}} \\mathrm{{C}}$
        The charge density then is $4.9 \\times 10^{{-12}} \\mathrm{{C}} / \\mathrm{{m}}^3$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_26_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The total charge (charge density) contained in an electron stream can be calculated with Gauss' law. If the field at the edges of a stream of square cross section $1.0 \\mathrm{{~cm}}$ on a side is
        $1.0 \\times 10^3 \\mathrm{{~N}} / \\mathrm{{C}}$, calculate the charge density in the stream. Assume that the $E$-field in the direction of the stream is constant.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-26-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The field is directed in, so the charge is negative and $\\displaystyle \\epsilon_o \\oint E \\cdot d s=(-)_q$ reduces to calculating four identical integrals. $\\displaystyle \\int E \\cdot d s=1.0 \\times 10^3 \\mathrm{{~N}} / \\mathrm{{C}} \\cdot 1.0 \\times 10^{{-4}} \\mathrm{{~m}}^2=-0.10 \\mathrm{{~N}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{C}}$.
        The charge in a cube $1.0 \\mathrm{{~cm}}$ on a side is now
        $\\displaystyle q=\\varepsilon_o \\int E \\cdot d s=8.9 \\times 10^{{-12}} \\mathrm{{C}}^2 / \\mathrm{{N}} \\cdot \\mathrm{{m}}^2(4) 0.10 \\mathrm{{Nm}}^2 / \\mathrm{{C}}=35.6 \\times 10^{{-13}} \\mathrm{{C}}$
        The charge density is
        $\\displaystyle \\frac{{q}}{{V}}=\\frac{{35 \\times 10^{{-13}} \\mathrm{{C}}}}{{1.0 \\times 10^{{-6}} \\mathrm{{~m}}^3}}=3.56 \\times 10^{{-6}} \\mathrm{{C}} / \\mathrm{{m}}^3$
        For the electron charge density we need to introduce the charge of the electron.
        $\\displaystyle \\frac{{e^{{-}}}}{{V}}=\\frac{{3.56 \\times 10^{{-6}} \\mathrm{{C}}}}{{\\mathrm{{m}}^3}} \\cdot \\frac{{e^{{-}}}}{{1.6 \\times 10^{{-19}} \\mathrm{{C}}}}=2.2 \\times 10^{{13}} e^{{-}} / \\mathrm{{m}}^3$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_26_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the sign and magnitude of the charge contained in a cube $10 \\mathrm{{~cm}}$ on a side oriented as shown where the $E$-field is given by
        $E_x=\\left(800 \\mathrm{{~N}} / \\mathrm{{C}} \\cdot \\mathrm{{m}}^{{1 / 2}}\\right) x^{{1 / 2}} \\qquad E_y=0 \\qquad E_z=0$
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-26-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        First calculate the $E$-fields entering and exiting the cube. Watch the algebraic signs carefully.
        $\\left.E_x\\right|_{{10 \\mathrm{{~cm}}}}=800 \\mathrm{{~N}} / \\mathrm{{C}} \\cdot \\mathrm{{m}}^{{1 / 2}}(0.10 \\mathrm{{~m}})^{{1 / 2}}=253 \\mathrm{{~N}} / \\mathrm{{C}}$ 
        $\\left.E_x\\right|_{{20 \\mathrm{{~cm}}}}=800 \\mathrm{{~N}} / \\mathrm{{C}} \\cdot \\mathrm{{m}}^{{1 / 2}}(0.20 \\mathrm{{~m}})^{{1 / 2}}=358 \\mathrm{{~N}} / \\mathrm{{C}}$
        There is more flux leaving the cube than there is entering it, so the net charge inside the cube must be positive. On the left side $E$ is to the right and the vector representing $d s$ is to the left, so
        $\\int_l E \\cdot d s=-(253 \\mathrm{{~N}} / \\mathrm{{C}})\\left(0.010 \\mathrm{{~m}}^2\\right)=-2.53 \\mathrm{{~N}} / \\mathrm{{C}}$
        Remember that the direction of $d s$ is always outward from the enclosed volume. On the right side $E$ is to the right and the vector representing $d s$ is to the right so
        $\\int_r E \\cdot d s=358 \\mathrm{{~N}} / \\mathrm{{C}}\\left(0.010 \\mathrm{{~m}}^2\\right)-3.58 \\mathrm{{Nm}}^2 / \\mathrm{{C}}$
        The integrals over $y$ and $z$ are zero and the integrals over $x$ add to $1.05 \\mathrm{{Nm}}^2 / \\mathrm{{C}}$. Now $q$ can be calculated
        $q=\\varepsilon_o \\oint E \\cdot d s=8.9 \\times 10^{{-12}} \\mathrm{{C}}^2 / \\mathrm{{Nm}}^2 \\cdot 1.05 \\mathrm{{Nm}}^2 / \\mathrm{{C}}=9.3 \\times 10^{{-12}} \\mathrm{{C}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_26_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The electric field at the surface of the earth is approximately $150 \\mathrm{{~N}} / \\mathrm{{C}}$ directed down. Calculate the sign and magnitude of the charge on the earth.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Gauss' law states that $\\displaystyle \\varepsilon_o \\oint \\boldsymbol{{E}} \\cdot d s=\\mathrm{{q}}_{{\\text {{enclosed }}}}$
        Construct a spherical Gaussian surface just outside the physical surface of the earth where $E$ is everywhere normal and of value $150 \\mathrm{{~N}} / \\mathbf{{C}}$. The integral of $\\boldsymbol{{E}}$. $d s$ is the sum of this field over the Gaussian surface. $\\boldsymbol{{E}}$ and $d \\boldsymbol{{s}}$ are $180^{{\\circ}}$ so their dot product is negative.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        file_name = 'how-to-solve-physics-problems-fig-26-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
            
        soln_s = f"""
        $\\displaystyle q=-\\varepsilon_o E\\left(4 \\pi R_E^2\\right)=-8.9 \\times 10^{{-12}} \\mathrm{{C}}^2 / \\mathrm{{N}} \\cdot \\mathrm{{m}}^2(150 \\mathrm{{~N}} / \\mathrm{{C}}) 4 \\pi\\left(6.4 \\times 10^6 \\mathrm{{~m}}\\right)^2=-6.9 \\times 10^5 \\mathrm{{C}}$
        The field pointing down indicates that the charge is negative as does the dot product being negative. Again notice that this integral over the Gaussian surface is quite easy. These past two problems are very illustrative of the "integration by inspection" approach to problems involving Gauss' law.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_26_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the flux through a square plate $1.0 \\mathrm{{~m}}$ on a side inclined at $120^{{\\circ}}$ with respect to a constant electric field of $100 \\mathrm{{~N}} / \\mathrm{{C}}$. You may at this point want to review the definition of dot product in Chapter 1, Vectors, and the concept of the integral in the Introduction, Mathematical Background.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-26-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The surface area $d s$ is represented by a vector normal to the surface. The vector product $\\boldsymbol{{E}}$.ds is $E d s\\left(\\cos 30^{{\\circ}}\\right)$. Integrating this product
        $\\displaystyle \\int E \\cdot d s=E: A(\\cos \\theta)=100 \\mathrm{{~N}} / \\mathrm{{C}}\\left(1.0 \\mathrm{{~m}}^2\\right) \\cos 30^{{\\circ}}=87 \\mathrm{{~N}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{C}}$
        This represents the flux through the surface.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_25_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Take the original position $\\theta_0=70^{{\\circ}}$, and the final position $\\theta_1=20^{{\\circ}}$ and calculate the work performed on the dipole and the energy stored in the system.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle W=-\\left.p E \\cos \\theta\\right|_{{70^{{\\circ}}}} ^{{20^{{\\circ}}}}=0.144 \\mathrm{{~N}} \\cdot \\mathrm{{m}}\\left(\\cos 70^{{\\circ}}-\\cos 20^{{\\circ}}\\right)=-0.086 \\mathrm{{~J}}$
        The electric field has done 0.086.J of work on the dipole, and this is the amount of energy stored in the dipolefield system.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_25_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A dipole with charge $6.0 \\times 10^{{-6}} \\mathrm{{C}}$ and separation $4.0 \\times 10^{{-3}} \\mathrm{{~m}}$ is immersed in an $E$-field of $3.0 \\times 10^6 \\mathrm{{~N}} / \\mathrm{{C}}$. Calculate the maximum torque.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\tau_{{\\max }}=2 a q E=2 \\cdot 4.0 \\times 10^{{-3}} \\mathrm{{~m}} \\cdot 6.0 \\times 10^{{-6}} \\mathrm{{C}} \\cdot 3.0 \\times 10^6 \\mathrm{{~N}} / \\mathrm{{C}}=144 \\times 10^{{-3}} \\mathrm{{Nm}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_25_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider an electron ballistics problem where an electron is projected between two plates (as shown in Fig. 25-6) at an initial velocity of $6.0 \\times 10^6 \\mathrm{{~m}} / \\mathrm{{s}}$ and angle of $45^{{\\circ}}$. The electric field is $2.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{C}}$ and directed up. The length of the plates is $10 \\mathrm{{~cm}}$ and separation $2.0 \\mathrm{{~cm}}$. Will the electron strike the plates? And if so. where?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-25-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        This is a projectile problem. Look through Chapter 4, Projectile Motion, and refresh your memory on the general approach to projectile motion problems before continuing with this problem. The electronic charge and mass are given on the constants page.

        First calculate the acceleration. The force on the electron comes from the electric field, $F=e E$, and this force makes the electron accelerate leading to $a=e E / m$. Using the values for $e$ and $m$ for the electron
        $\\displaystyle a=\\frac{{e E}}{{m}}=\\frac{{1.6 \\times 10^{{-19}} \\mathrm{{C}} \\cdot 2.0 \\times 10^3 \\mathrm{{~N}} / \\mathrm{{C}}}}{{9.1 \\times 10^{{-31}} \\mathrm{{~kg}}}}=3.5 \\times 10^{{14}} \\mathrm{{~m}} / \\mathrm{{s}}^2$

        The electric field is directed up, which would be the direction of acceleration of a positive charge. For the negatively charged electron the acceleration is down. Using this logic argument to determine the direction of the acceleration is easier than carrying minus signs throughout the problem.

        The most convenient origin for the coordinate system is the edge of the plate where the initial velocity and direction are known with $x$ positive to the right and $y$ positive up. Now write the six equations describing the motion.

        $\\begin{{array}}{{ll}}
        a_x=0 & a_y=-3.5 \\times 10^{{14}} \\mathrm{{~m}} / \\mathrm{{s}}^2 \\\\
        v_x=v_o \\cos 45^o=4.2 \\times 10^6 \\mathrm{{~m}} / \\mathrm{{s}} & v_y=4.2 \\times 10^6 \\mathrm{{~m}} / \\mathrm{{s}}-\\left(3.5 \\times 10^{{14}} \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t \\\\
        x=\\left(4.2 \\times 10^6 \\mathrm{{~m}} / \\mathrm{{s}}\\right) t & y=\\left(4.2 \\times 10^6 \\mathrm{{~m}} / \\mathrm{{s}}\\right) t-\\left(1.8 \\times 10^{{14}} \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2
        \\end{{array}}$
        To determine if the electron hits the upper plate, set $y=2.0 \\times 10^{{-2}} \\mathrm{{~m}}$ and solve for $t$. Without units
        $1.8 \\times 10^{{14}} t^2-4.2 \\times 10^6 t+2.0 \\times 10^{{-2}}=0$
        and
        $\\displaystyle t=\\frac{{4.2 \\times 10^6 \\pm \\sqrt{{18 \\times 10^{{12}}-4\\left(1.8 \\times 10^{{14}}\\right)\\left(2.0 \\times 10^{{-2}}\\right)}}}}{{3.5 \\times 10^{{14}}}}=\\frac{{4.2 \\times 10^6 \\pm 1.9 \\times 10^6}}{{3.5 \\times 10^{{14}}}}=0.66 \\times 10^{{-8}} \\mathrm{{~s}}$

        The smallest time is taken as this corresponds to the electron striking the top plate. If the solution for $t$ had contained an imaginary number, this would have indicated that the electron would never reach the top plate.
        Now with this time calculate the position in $x$ where the electron strikes the plate.
        $x=4.2 \\times 10^6 \\mathrm{{~m}} / \\mathrm{{s}} \\cdot 0.66 \\times 10^{{-8}} \\mathrm{{~s}}=2.8 \\times 10^{{-2}} \\mathrm{{~m}}$

        Another way to solve the problem is to go back to the six equations describing the motion and solve for $y$ as a function of $x$. First solve the $x$ equation for $t$ and substitute this expression for $t$ into the $y$ equation to obtain
        $y=x-1.8 \\times 10^{{14}} \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}^2}} \\frac{{x^2}}{{\\left(4.2 \\times 10^6 \\mathrm{{~m}} / \\mathrm{{s}}\\right)^2}}=x-10 x^2$

        When $y-0.020$ this equation becomes $10 x^2-x+0.020-0$ with solution
        $\\displaystyle x=\\frac{{1 \\pm \\sqrt{{1-4(10)(0.020)}}}}{{2 \\cdot 10}}=\\frac{{1 \\pm 0.45}}{{20}}=2.8 \\times 10^{{-2}} \\mathrm{{~m}}$
        Here again be sure to take the value of $x$ corresponding to the smallest positive time.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_25_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the electric field vertically out from a uniformly charged disk of radius $\\mathrm{{R}}$ and charge density (charge/area) $\\sigma$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The calculus definition of the electric field is that each element of charge produces an element of electric field according to equation 25-4 where $r$ is the distance from the element to the point where $d E$ is calculated. Because of the symmetry of the problem the $d E$ is produced by a $d q$ in the form of a ring of charge of radius $r$ and width $d r$. Refer to Fig. 25-5 and note that the $d \\boldsymbol{{E}}$ produced by this ring of charge is a cone with horizontal components adding to zero. A piece of the ring producing a piece of $d \\boldsymbol{{E}}$ has a corresponding piece across a diameter. The horizontal components of these corresponding pieces add to zero. The vertical component is $d E \\cos \\theta$ with $\\cos \\theta$ determined from the geometry as $\\cos \\theta=x / \\sqrt{{x^2+r^2}}$ Summing the contributions over $r$ produces the electric field pointing away (and perpendicular to) the charged disk. The differential element of charge is $d q=(2 \\pi r d r) \\sigma(2 \\pi r d r$ is the differential element of area), thus
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        file_name = 'how-to-solve-physics-problems-fig-25-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        $\\displaystyle E=\\frac{{1}}{{4 \\pi \\varepsilon_0}} \\int_0^R \\frac{{\\sigma 2 \\pi r d r}}{{x^2+r^2}} \\frac{{x}}{{\\sqrt{{x^2+r^2}}}}=\\frac{{\\sigma x}}{{2 \\varepsilon_0}} \\int_0^R \\frac{{r d r}}{{\\left(x^2+r^2\\right)^{{3 / 2}}}}$
        Remember the integral is over $r$ and not $x$. Make a change of variable with $u=x^2+r^2$ and $d u=2 r d r$. The integral transforms according to
        $\\displaystyle \\int \\frac{{r d r}}{{\\left(x^2+r^2\\right)^{{3 / 2}}}}=\\frac{{1}}{{2}} \\int \\frac{{d u}}{{u^{{3 / 2}}}}=-\\frac{{1}}{{\\sqrt{{u}}}}=-\\frac{{1}}{{\\sqrt{{x^2+r^2}}}}$
        Changing the sign and switching the limits
        $\\displaystyle E=\\frac{{\\sigma x}}{{2 \\varepsilon_o}}\\left[\\frac{{1}}{{\\sqrt{{x^2+r^2}}}}\\right]_R^0-\\frac{{\\sigma x}}{{2 \\varepsilon_o}}\\left[\\frac{{1}}{{x \\sqrt{{1+r^2 / x^2}}}}\\right]_R^0=\\frac{{\\sigma}}{{2 \\varepsilon_o}}\\left[1-\\frac{{1}}{{\\sqrt{{1+R^2 / x^2}}}}\\right]$
        When $R$ is large compared to $x$, the bracket approaches one and the electric field is $\\displaystyle \\frac{{\\sigma}}{{2 \\varepsilon_o}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_25_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A problem that illustrates this is the calculation of the electric field on the axis of a ring of positive charge. Set up the problem as shown in Fig. 25-4.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-25-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        Each element of charge $d q$ produces an element of electric field $d E$ at a point on the biscetor of a diameter of the ring. This $d E$ has components $d E \\cos \\theta$ along the direction of the bisector and $d E \\sin \\theta$ perpendicular to the bisector. An element of $d q$ at one spot on the ring produces $d E \\sin \\theta$, and across the diameter of the ring a similar element (of $d q$ ) produces a $d E \\sin \\theta$ component that cancels this one. Because of this symmetry the $d E \\sin \\theta$ components add to zero.

        The magnitude of the electric field due to each element of charge can be written as
        $\\displaystyle d E=\\frac{{1}}{{4 \\pi \\varepsilon_o}} \\frac{{d q}}{{a^2+x^2}}$
        but it is $d E \\cos \\theta$ which ultimately produces the field pointing outward from the center of the ring.

        From the geometry $\\cos \\theta=x / \\sqrt{{a^2+x^2}}$, so the differential electric field component is
        $\\displaystyle d E \\cos \\theta=\\frac{{1}}{{4 \\pi \\varepsilon_0}} \\frac{{d q}}{{a^2+x^2}} \\frac{{x}}{{\\sqrt{{a^2+x^2}}}}$

        This expression can be summed (integrated) without the need for formal integration. The left side is just the total electric field, the sum of the $d E \\cos \\theta$ contributions. The right side is the sum of all the elements of $d q$, which is just $q$, so
        $\\displaystyle E=\\frac{{1}}{{4 \\pi \\varepsilon_0}} \\frac{{q x}}{{\\left(a^2+x^2\\right)^{{3 / 2}}}}$
        Notice that at $x=0, E=0$, as it should and that for $\\displaystyle x>>, E \\approx \\frac{{1}}{{4 \\pi \\varepsilon_0}} \\frac{{q}}{{x^2}}$, again as it should. Far away the ring of charge should look like a point.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_25_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Find an expression for the electric field due to a collection of charges of equal magnitude all placed in a line and separated by a distance $d$. The charges have alternate signs, and the field to be calculated is the field on a line normal to the line of charges and opposite the eentral positive charge. Additionally, look for approximate expressions for long distances away from the line of charges.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-25-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The electric fields are laheled to correspond with the charges. The fields at a point $r$ distance away from the line of charge are:
        $\\displaystyle E_1=\\frac{{1}}{{4 \\pi \\varepsilon_o}} \\frac{{q}}{{r^2}} \\qquad E_2=E_4=\\frac{{1}}{{4 \\pi \\varepsilon_0}} \\frac{{(-q)}}{{r^2+d^2}} \\qquad E_3=E_5=\\frac{{1}}{{4 \\pi \\varepsilon_o r^2+4 d^2}}$
        Look at the geometry associated with $E_2$ and notice that this vector can be written in terms of a horizontal and a vertical component. The vector $E_4$ also has vertical and horizontal components.

        The vertical components of $\\boldsymbol{{E}}_2$ and $\\boldsymbol{{E}}_4$ add to zero while the horizontal components add along $r$. The same is true of vectors $\\boldsymbol{{E}}_3$ and $\\boldsymbol{{E}}_5$. The horizontal component of $\\boldsymbol{{E}}_2$ is $E_2$ times the cosine of the angle $\\boldsymbol{{\\theta}}$, thus the contribution of $\\boldsymbol{{E}}_2$ and $\\boldsymbol{{E}}_4$ is two times the magnitude of $\\boldsymbol{{E}}_2$ times the cosine of the angle $\\theta$

        $\\boldsymbol{{E}}_2$ and $\\boldsymbol{{E}}_4$ contribute $\\displaystyle 2 E_2 \\cos \\theta=\\frac{{1}}{{4 \\pi \\varepsilon_a}} \\frac{{-2 q}}{{r^2+d^2}} \\frac{{r}}{{\\sqrt{{r^2+d^2}}}}$ pointed in along $r$.
        Similarly $\\boldsymbol{{E}}_3$ and $\\boldsymbol{{E}}_5$ contribute $\\displaystyle \\frac{{1}}{{4 \\pi \\varepsilon_0}} \\frac{{2 q r}}{{r^2+d^2}} \\frac{{r}}{{\\sqrt{{r^2+4 d^2}}}}$ also pointed out along $r$

        Combining the contributions from all the charges
        $\\displaystyle E=\\frac{{1}}{{4 \\pi \\varepsilon_o}} q\\left[\\frac{{1}}{{r^2}}-\\frac{{2 r}}{{\\left(r^2+d^2\\right)^{{3 / 2}}}}+\\frac{{2 r}}{{\\left(r^2+4 d^2\\right)^{{3 / 2}}}}\\right]$

        If $r$ is comparable to $d$ then this formula is appropriate. However if $r$ is large compared to $d$, then the $r$ terms predominate in the denominators and the expression is approximately
        $\\displaystyle E \\approx \\frac{{1}}{{4 \\pi \\varepsilon_o}} q\\left[\\frac{{1}}{{r^2}}-\\frac{{2}}{{r^2}}+\\frac{{2}}{{r^2}}\\right] \\approx \\frac{{1}}{{4 \\pi \\varepsilon_o}} \\frac{{q}}{{r^2}}$
        This formula shows that at long distances the array looks like one positive charge.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_25_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the electric field due to two charges, $q_1=4.0 \\times 10^{{-6}} \\mathrm{{C}}$ and $q_2=-3.0 \\times 10^{{-6}} \\mathrm{{C}}$, separated by $3.0 \\mathrm{{~m}}$ and at a point $3.0 \\mathrm{{~m}}$ opposite $q_t$ as shown in Fig. $25-2$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-25-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The electric field due to the first charge is
        $\\displaystyle E_1=\\frac{{1}}{{4 \\pi \\varepsilon_o}} \\frac{{q}}{{r^2}}=9.0 \\times 10^9 \\frac{{\\mathrm{{Nm}}^2}}{{\\mathrm{{C}}^2}} \\frac{{4.0 \\times 10^{{-6}} \\mathrm{{C}}}}{{18 \\mathrm{{~m}}^2}}=2.0 \\times 10^3 \\mathrm{{~N}} / \\mathrm{{C}}$
        The electric field due to the second charge is
        $\\displaystyle E_2=\\frac{{1}}{{4 \\pi \\varepsilon_o}} \\frac{{q}}{{r^2}}=9.0 \\times 10^9 \\frac{{\\mathrm{{Nm}}^2}}{{\\mathrm{{C}}^2}} \\frac{{3.0 \\times 10^{{-6}} \\mathrm{{C}}}}{{9.0 \\mathrm{{~m}}^2}}=3.0 \\times 10^3 \\mathrm{{~N}} / \\mathrm{{C}}$

        The direction of the fields due to these two charges is shown on the diagram. The vectors are written in component form
        $\\displaystyle E_1=\\left(2.0 \\times 10^3 \\cos 45^{{\\circ}} i+2.0 \\times 10^3 \\sin 45^o j\\right) \\mathrm{{N}} / \\mathrm{{C}}=\\left(1.4 \\times 10^3 i+1.4 \\times 10^3 j\\right) \\mathrm{{N}} / \\mathrm{{C}}$
        $\\displaystyle \\hat{{E}}_2=\\left(0 i+-3.0 \\times 10^3 j\\right) \\mathrm{{N}} / \\mathrm{{C}}$
        with resultant $\\boldsymbol{{E}}_{{\\mathrm{{R}}}}=\\left(1.4 \\times 10^3 i+-1.6 \\times 10^3 j \\mathrm{{j}}\\right) \\mathrm{{N}} / \\mathrm{{C}}$

        As an exercise verify that in number plus angle form this vector is $E_R=2.1 \\times 10^3 \\angle-49^{{\\circ}} \\mathrm{{N}} / \\mathrm{{C}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_25_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An electric dipole is depicted as equal and opposite charges separated by a distance $2 a$ as shown in Fig. 25-1. Find the electric field at a distance $r$ along the bisector of the center line of the charges.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-25-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        First draw the diagram and note the direction of the electric field vectors. The horizontal components of $E_1$ and $E_2$ add to zero. The vertical component is made up of two vectors of equal magnitude, $E_1 \\cos \\theta$. The magnitude of the vertical vector is $2 E_1 \\cos \\theta$.
        From the geometry, $\\displaystyle \\cos \\theta=\\frac{{a}}{{\\sqrt{{a^2+r^2}}}} \\quad \\text {{ and }} \\quad E_1=\\frac{{q}}{{4 \\pi \\varepsilon_o\\left(a^2+r^2\\right)}}$
        $\\displaystyle E=\\frac{{1}}{{4 \\pi \\varepsilon_o}} \\frac{{2 a q}}{{\\left(a^2+r^2\\right)^{{3 / 2}}}}$

        This is the magnitude of the electric vector. The direction of the vector is down or parallel to a line drawn from $+q$ to $-q$. If $r$ is large compared to $a$ (often the case) then
        $\\displaystyle E \\approx \\frac{{1}}{{4 \\pi \\varepsilon_o}} \\frac{{2 a q}}{{r^3}}$
        and for large distances the field falls off as $1 / r^3$. The product $2 a q$ is called the dipole moment $p=2 a q$. It is interesting to note that for a point charge the $E$ field is radial (in or out) and falls off as $1 / r^2$ while for the dipole $E$ is tangential to the radius and falls off as $1 / r^3$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_24_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the force of attraction between sodium and chlorine ions in salt molecules if each ion carries one electronic charge and the separation is $2.8 \\times 10^{{-10}} \\mathrm{{~m}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle F=\\frac{{1}}{{4 \\pi \\varepsilon_o}} \\frac{{q^2}}{{r^2}}=9.0 \\times 10^9 \\frac{{\\mathrm{{Nm}}^2}}{{\\mathrm{{C}}^2}} \\frac{{\\left(1.6 \\times 10^{{-19}} \\mathrm{{C}}\\right)^2}}{{\\left(2.8 \\times 10^{{-10}} \\mathrm{{~m}}\\right)^2}}=2.9 \\times 10^{{-9}} \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_24_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider two conducting balls both of mass $m$ and equal charge $q$ suspended by nonconducting cords of equal length, $l$ as shown in Fig 24-4. How does the separation of the balls depend on charge, mass, and length?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-25-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The arrangement is shown in the accompanying drawing along with the vector diagram. This is a force balance problem. The mechanical force is due to gravity and is directed down. The electrical force is the Coulomb force and is directed along the line of centers of the conducting balls, which for this geometry is horizontal.

        On the vector diagram the $m g$ force is balanced by the vertical component of the tension in the cord, or $T$ $\\cos \\theta=m g$. The horizontal component of the tension is $T \\sin \\theta$, and replacing $T$ with $m g / \\cos \\theta$ the horizontal component is
        $\\displaystyle T \\sin \\theta=\\frac{{m g \\sin \\theta}}{{\\cos \\theta}}=m g \\tan \\theta$
        But $\\tan \\theta$ can be written in terms of the geometry as $\\displaystyle \\tan \\theta=\\frac{{x / 2}}{{\\ell}}=\\frac{{x}}{{2 \\ell}}$ and $\\displaystyle T \\sin \\theta=\\frac{{m g x}}{{2 \\ell}}$
        The horizontal component of the tension in the cord is balanced by the Coulomb force of repulsion 
        $\\displaystyle F=\\frac{{1}}{{4 \\pi \\epsilon_o}} \\frac{{q^2}}{{x^2}}=T \\sin \\theta=\\frac{{m g x}}{{2 \\ell}}$
        or $\\displaystyle x=\\left[\\frac{{q^2 \\ell}}{{2 \\pi \\varepsilon_o m g}}\\right]^{{1 / 3}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_24_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Arrange three charges in the form of an equilateral triangle as shown in Fig. $24-3$ and find the force on $q_3$ due to the other two.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-24-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        First Solution: This is a more complicated vector problem. First calculate the force on $q_3$ due to $q_1$ and then due to $q_2$. Adding these vector forces is called superposition. The forces can be added as vectors as long as the force due to one charge does not affeet the force due to another charge. In this case the force on $q_3$ due to $q_1$ does not influence the force on $q_3$ due to $q_2$.
        The force on $q_3$ due to $q_1$ is
        $\\displaystyle F_{{31}}=9.0 \\times 10^9 \\frac{{\\mathrm{{Nm}}^2}}{{\\mathrm{{C}}^2}} \\frac{{2.0 \\times 10^{{-6}} \\mathrm{{C}} \\cdot 5.0 \\times 10^{{-6}} \\mathrm{{C}}}}{{4.0 \\mathrm{{~m}}^2}}=0.0225 \\mathrm{{~N}}$
        and is directed as shown in Fig. 24-3. Likewise the force on $q_3$ due to $q_2$ is
        $\\displaystyle F_{{32}}=9.0 \\times 10^9 \\frac{{\\mathrm{{Nm}}^2}}{{\\mathrm{{C}}^2}} \\frac{{3.0 \\times 10^{{-6}} \\mathrm{{C}} \\cdot 5.0 \\times 10^{{-6}} \\mathrm{{C}}}}{{4.0 \\mathrm{{~m}}^2}}=0.0338 \\mathrm{{~N}}$
        and is directed as shown. Referring to Fig. 24-3, the $x$ and $y$ components of the resultant force can be written. The $x$ component of these two forces is
        $F_x=F_{{31}} \\cos 60^\\alpha+F_{{32}} \\cos 60^{{\\circ}}=(0.0112+0.0169) \\mathrm{{N}}=0.0281 \\mathrm{{~N}}$
        The $y$-component of these two forces is
        $F_y=F_{{31}} \\sin 60^{{\\circ}}-F_{{32}} \\sin 60^{{\\circ}}=(0.0195-0.0292) \\mathrm{{N}}=0.0097 \\mathrm{{~N}}$

        Second Solution: Using unit vector notation the forces are written as
        $F_{{31}}=F_{{31}} \\cos 60^{{\\circ}} i+F_{{31}} \\sin 60^{{\\circ}} j=(0.0112 i+0.0195 j) \\mathrm{{N}}$
        $F_{{32}}=F_{{32}} \\cos 60^{{\\circ}} i-F_{{32}} \\sin 60^{{\\circ}} j=(0.0169 i-0.0292 j) \\mathrm{{N}}$
        and combine to$\\quad F_3=(0.0281 i-0.0097 j) \\mathrm{{N}}$
        Rather than worry about the algebraic signs of angles other than $0^{{\\circ}}$ to $90^{{\\circ}}$ draw the picture and keep the angles under $90^{{\\circ}}$. This way it is easy to "see" the sign of the various components. Most mistakes in problems like this are algebraic sign mistakes. The way to avoid these mistakes is to draw and label a vector diagram, more than one if necessary.
        As an exercise find the magnitude and direction of the resultant.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_24_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider a line of charges, $q_1=8.0 \\mu \\mathrm{{C}}$ at the origin, $q_2=-12 \\mu \\mathrm{{C}}$ at $2.0 \\mathrm{{~cm}}$, and $q_3=10 \\mu \\mathrm{{C}}$ at $4.0 \\mathrm{{~cm}}$, as shown in Fig. 24-2. What is the force on $q_3$ due to the other two charges?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-24-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The force on $q_3$, due to $q_1$ is repulsive as shown in Fig. 24-2. The magnitude of $F_{{31}}$ is
        $\\displaystyle F_{{31}}=9.0 \\times 10^9 \\frac{{\\mathrm{{Nm}}^2}}{{\\mathrm{{C}}^2}} \\frac{{80 \\times 10^{{-12}} \\mathrm{{C}}^2}}{{16 \\times 10^{{-4}} \\mathrm{{~m}}^2}}=450 \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        file_name = 'how-to-solve-physics-problems-fig-24-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        It is not necessary to be overly concerned about the sign of the force. The vector makes the direction clear. Drawing a vector diagram is a more sure way of getting the final direction correct than using algebraic signs in the force calculations.

        The force on $q_3$ due to $q_2$ is attractive, and the magnitude is
        $\\displaystyle F_{{32}}=9.0 \\times 10^9 \\frac{{\\mathrm{{Nm}}^2}}{{\\mathrm{{C}}^2}} \\frac{{120 \\times 10^{{-12}} \\mathrm{{C}}^2}}{{4.0 \\times 10^{{-4}} \\mathrm{{~m}}^2}}=2700 \\mathrm{{~N}}$
        The resultant force is $2250 \\mathrm{{~N}}$ toward $q_1$ and $q_2$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_24_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Two charges, $q_1=3.0 \\times 10^{{-8}} \\mathrm{{C}}$ and $q_2=-4.0 \\times 10^{{-8}} \\mathrm{{C}}$, are separated by $6.0 \\times 10^{{-3}} \\mathrm{{~m}}$ as shown in Fig $24-1$. What is the force of one on the other?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-24-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        $\\displaystyle F=\\frac{{1}}{{4 \\pi \\varepsilon_o}} \\frac{{q_1 q_2}}{{r^2}}=9.0 \\times 10^9 \\frac{{\\mathrm{{Nm}}^2}}{{\\mathrm{{C}}^2}} \\frac{{12 \\times 10^{{-16}} \\mathrm{{C}}^2}}{{36 \\times 10^{{-6}} \\mathrm{{~m}}^2}}=0.30 \\mathrm{{~N}}$
        This is the force each charge experiences due to the other. The fact that $\\mathrm{{q}}_1$ is positive and $\\mathrm{{q}}_2$ is negative means that the force is directed from one to the other along the line joining their centers.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_23_11(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Submarines use $1000 \\mathrm{{~Hz}}$ signals to avoid collisions. A submarine traveling at $5.0 \\mathrm{{~m}} / \\mathrm{{s}}$ observes another submarine on the same course behind it and detects the collision avoidance signal from this submarine at $1005 \\mathrm{{~Hz}}$. What is the speed of the other submarine? The velocity of sound in water is $1500 \\mathrm{{~m}} / \\mathrm{{s}}$.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This is a Doppler effect problem with both source and observer moving. The hard part of the problem is getting the signs correct. First note that the frequency is higher than emitted so the submarines are closing. If the observer were stationary the formula would be
        $\\displaystyle f^{{\\prime}}=f\\left[v /\\left(v-v_s\\right)\\right]$. If the observer were moving away from a stationary source the formula would be $f^{{\\prime}}=f\\left[\\left(v-v_0\\right) / v\\right]$.
        So combining, the correct formula is $\\displaystyle f^{{\\prime}}=f \\frac{{v-v_0}}{{v-v_s}}$
        and putting in the numbers $\\displaystyle 1005 \\mathrm{{~Hz}}=1000 \\mathrm{{~Hz}}\\left(\\frac{{1500-5}}{{1500-v_8}}\\right)$ or $v_s=12 \\mathrm{{~m}} / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_23_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        $\\mathrm{{A}}$ railroad train traveling at $30 \\mathrm{{~m}} / \\mathrm{{s}}$ emits a note at $400 \\mathrm{{~Hz}}$. What frequency is heard by an observer a) on the train, b) standing beside the track, and c) on another train approaching at $35 \\mathrm{{~m}} / \\mathrm{{s}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        
        soln_s = f"""
        a) On the train the observer hears the $400 \\mathrm{{~Hz}}$.
        b) Beside the track with the train approaching the observer hears a higher frequency
        $\\displaystyle f^{{\\prime}}=f \\frac{{v}}{{v-v_{{\\mathrm{{s}}}}}}=400 \\mathrm{{~Hz}} \\frac{{340}}{{310}}=439 \\mathrm{{~Hz}}$
        And with the train moving away the observer hears a lower frequency
        $f^{{\\prime}}=f \\frac{{v}}{{v+v_{{\\mathrm{{s}}}}}}=400 \\mathrm{{~Hz}} \\frac{{340}}{{370}}=368 \\mathrm{{~Hz}}$
        c) On the other train the observer hears a higher frequency due to the source approaching, $(v-v_{{\\mathrm{{s}}}})$ term, and observer approaching, $\\left(v+v_{{\\circ}}\\right)$ term.
        $f^{{\\prime}}=f \\frac{{v+v_0}}{{v-v_{{\\mathrm{{s}}}}}}=400 \\mathrm{{~Hz}} \\frac{{375}}{{310}}=484 \\mathrm{{~Hz}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
    
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_23_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the apparent frequency of the siren in problem 23-8 after you have passed it and are moving away?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        When you are moving away the frequency is lower because you are intercepting fewer wave fronts than it you were stationary so use the minus sign in equation 23-7.
        $f^{{\\prime}}=300 \\mathrm{{~Hz}}(1-30 / 340)=274 \\mathrm{{~Hz}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_23_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the apparent frequency of a siren emitting a $300 \\mathrm{{~Hz}}$ sinusoidal signal that you are approaching at $30 \\mathrm{{~m}} / \\mathrm{{s}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Use equation 23-8 with the plus sign because you are approaching the source and intercepting more waves than if you were stationary.
        $f^{{\\prime}}=300 \\mathrm{{~Hz}}(1+30 / 340)=326 \\mathrm{{~Hz}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_23_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the $\\mathrm{{dB}}$ loss when you double your distance from a spherically symmetric constantoutput power source?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Since the source is constant power output and spherically symmetric the inverse square law applies. Take $r_1$ and $r_2=2 r_1$ and apply power equals power.
        $4 \\pi r_1^2 I_1=4 \\pi r_2^2 I_2=16 \\pi r_1^2 I_2$ and $I_1=4 I_2$
        The $\\mathrm{{dB}}$ levels at $r_1$ and $r_2$ are
        $\\displaystyle \\mathrm{{dB}}_1=10 \\log \\frac{{I_1}}{{I_0}} \\quad \\text {{ and }} \\quad \\mathrm{{dB}}_2=10 \\log \\frac{{I_2}}{{I_{{\\mathrm{{o}}}}}}$
        and the difference is
        $\\displaystyle \\mathrm{{dB}}_1-\\mathrm{{dB}}_2=10 \\log \\frac{{I_1}}{{I_{{\\mathrm{{o}}}}}}-10 \\log \\frac{{I_2}}{{I_{{\\mathrm{{o}}}}}}=10\\left[\\log I_1-\\log I_{{\\mathrm{{o}}}}-\\log I_2+\\log I_{{\\mathrm{{o}}}}\\right]=10 \\log \\frac{{I_1}}{{I_2}}=10 \\log 4=6.0$
        Doubling the distance from the source results in a $6.0 \\mathrm{{~dB}}$ loss.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_23_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Most people experience pain at sound levels above $100 \\mathrm{{~dB}}$. What is the intensity associated with $100 \\mathrm{{~dB}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Equation 23-6 needs to be solved for $I$, and this requires switching from a logarithmic to an exponential equation. See the Introduction, Mathematical Background for a review of logarithmic and exponential equations.
        First write $\\beta / 10=\\log \\left(I / I_{{\\mathrm{{o}}}}\\right)$ and remember that $x=\\log y$ has the solution $y=10^x$. The use of "log" implies base 10.
        Switch from a logarithmic equation to an exponential one and write $I / I_o=10^{{\\beta / 10}}$.
        For $\\beta=100$
        Practice switching logarithmic equations to exponential equations and vice versa and solving for the various variables. This is not a common algebraic operation, and you need sufficient practice to develop proficiency with exponents and logarithms.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_23_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Ilow far away can a $50 \\mathrm{{~W}}$ tornado siren be heard? Assume the minimum detectable intensity is $5.0 \\times 10^{{-6}}$ $\\mathrm{{W}} / \\mathrm{{m}}^2$
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The power is $P=4 \\pi^2$ I so $\\displaystyle r=\\sqrt{{\\frac{{P}}{{4 \\pi I}}}}=\\sqrt{{\\frac{{50 \\mathrm{{~W}} / \\mathrm{{m}}^2}}{{4 \\pi 5.0 \\times 10^{{-6}} \\mathrm{{~W}}}}}}=890 \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_23_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        If the intensity from a source is $4.0 \\times 10^{{-3}} \\mathrm{{~W}} / \\mathrm{{m}}^2$ at $12 \\mathrm{{~m}}$, what will the intensity be at $4.0 \\mathrm{{~m}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Assume a spherically symmetric radiator and equate the powers at the two radii.
        $P_1 =P_2$ requires $4 \\pi r_1^2 I_1=4 \\pi r_2^2 I_2$
        $\\displaystyle I_1 =\\left(\\frac{{r_2}}{{r_1}}\\right)^2 I_2=\\left(\\frac{{12}}{{4.0}}\\right)^2 4.0 \\times 10^{{-3}} \\mathrm{{~W}} / \\mathrm{{m}}^2=3.6 \\times 10^{{-2}} \\mathrm{{~W}} / \\mathrm{{m}}^2$
        Be careful on problems like this that you don't get the ratio upside down. Develop the habit of checking for the obvious. When you get closer to a source the intensity increases and vice versa.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_23_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the intensity of a $4.0 \\mathrm{{~W}}$ source at $2.0 \\mathrm{{~m}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The power is $P=4 \\pi^2$ I so
        $\\displaystyle P=4 \\pi r^2 I \\text {{ so }} I=\\frac{{P}}{{4 \\pi r^2}}=\\frac{{4.0 \\mathrm{{~W}}}}{{4 \\pi \\cdot 4.0 \\mathrm{{~m}}^2}}=8.0 \\times 10^{{-2}} \\mathrm{{~W}} / \\mathrm{{m}}^2$
        Consider a spherically symmetric source and intensity measurements on two different concentric spheres. The total power is the same so the power collected over one sphere is the same as the power collected over any other sphere. In equation form this statement is
        $P_1=P_2 \\text {{ or }} 4 \\pi r_1^2 I_1=4 \\pi r_2^2 I_2$
        leading to
        $\\displaystyle \\frac{{I_1}}{{I_2}}=\\left(\\frac{{r_2}}{{r_1}}\\right)^2$
        which is the inverse square law.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_23_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the total power output of a source with intensity $0.050 \\mathrm{{~W}} / \\mathrm{{m}}^2$ at a distance of $3.0 \\mathrm{{~m}}$ from the source?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        
        soln_s = f"""
        The total power is the intensity times the area of the sphere with radius $3.0 \\mathrm{{~m}}$.
        $P_{{\\mathrm{{t}}}}=I\\left(4 \\pi r^2\\right)=\\left(0.050 \\mathrm{{~W}} / \\mathrm{{m}}^2\\right) 4 \\pi\\left(9.0 \\mathrm{{~m}}^2\\right)=5.6 \\mathrm{{~W}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-23-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_23_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        At some arbitrary position $x$ graph displacement and pressure as a function of time on the same axis and explain the relationship.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-23-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        

        soln_s = f"""
        Pick a point $x$ where the displacement versus time is a sine curve (equation 23-1). The pressure curve (equation 23-2) is a cosine curve. Maximum pressure occurs when $y$ is moving to the right with maximum speed. This corresponds to maximum positive slope or the point $t=0$ on the graph. Minimum pressure (a rarefaction) occurs $180^{{\\circ}}$ later where the slope of $y$ versus $t$ has its largest negative value.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_22_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An open at both ends organ pipe has two successive harmonics of $567 \\mathrm{{~Hz}}$ and $850 \\mathrm{{~Hz}}$. What is the length of the pipe? Take the speed as $340 \\mathrm{{~m}} / \\mathrm{{s}}$.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        For an open at both ends pipe the wavelengths are given by equation 22-6. For successive harmonics $(n$ and $n+1)$ two equations can be written in the form $v=\\lambda f$.
        $340 \\mathrm{{~m}} / \\mathrm{{s}}=(2 L / n) 567 \\mathrm{{~Hz}}$ and $340 \\mathrm{{~m}} / \\mathrm{{s}}=(2 L / n+1) 850 \\mathrm{{~Hz}}$
        $\\displaystyle \\frac{{567}}{{n}}=\\frac{{850}}{{n+1}} \\text {{ or }} n=2$

        Refer to Fig. 22-4. These frequencies correspond to the second and third harmonics. The lowest frequency (567 $\\mathrm{{Hz}})$ corresponds to the longest wavelength and using $v=\\lambda f$
        $\\displaystyle \\lambda=\\frac{{v}}{{f}}=\\frac{{340 \\mathrm{{~m}} / \\mathrm{{s}}}}{{567 \\mathrm{{~Hz}}}}=0.60 \\mathrm{{~m}}$
        The pipe then is $0.60 \\mathrm{{~m}}$ long.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_22_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the lowest possible frequency of a $5.0 \\mathrm{{~m}}$ long organ pipe?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The lowest frequency corresponds to the longest wavelength, and the longest wavelength is for a closed at one end open at the other pipe (see Figs. 22-3 and 22-4). In this situation the wavelength is $20 \\mathrm{{~m}}$ corresponding to a frequency of
        $\\displaystyle f=\\frac{{v}}{{\\lambda}}=\\frac{{340 \\mathrm{{~m}} / \\mathrm{{s}}}}{{20 \\mathrm{{~m}}}}=17 \\mathrm{{~Hz}}$
        This frequency is below what most people can hear.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_22_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Standing waves are excited in a 1.0m long pipe open at one end, closed at the other. Take the speed of the waves as $340 \\mathrm{{~m}} / \\mathrm{{s}}$ and calculate the frequency for the fundamental and the first harmonic.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The wavelength is determined by the geometry, so look to Fig. 22-3 where the left diagram shows the fundamental wave. If the pipe is $1.0 \\mathrm{{~m}}$ then the wavelength is $4.0 \\mathrm{{~m}}$ and the frequency is
        $\\displaystyle f=\\frac{{v}}{{\\lambda}}=\\frac{{340 \\mathrm{{~m}} / \\mathrm{{s}}}}{{4.0 \\mathrm{{~m}}}}=85 \\mathrm{{~Hz}}$
        The next wavelength is shown in the right diagram and requires a $\\pi=(4 / 3) \\mathrm{{m}}$ and
        $\\displaystyle f=\\frac{{v}}{{\\lambda}}=\\frac{{340 \\mathrm{{~m}} / \\mathrm{{s}}}}{{4 / 3 \\mathrm{{~m}}}}=255 \\mathrm{{~Hz}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_22_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        One string on a cello is tightened to produce concert A $(440 \\mathrm{{~Hz}})$ when the string is not touched (shortened by placing a finger on it). The length of the vibrating string is $60 \\mathrm{{~cm}}$ and mass is $2.0 \\mathrm{{gm}}$. How much must the player shorten the string to play a $660 \\mathrm{{~Hz}}$ note?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The tension (velocity) is adjusted so that the frequency is $440 \\mathrm{{~Hz}}$ for a wavelength of $1.2 \\mathrm{{~m}}$. The velocity is
        $v=\\lambda f=(1.2 \\mathrm{{~m}})(440 \\mathrm{{~Hz}})=528 \\mathrm{{~m}} / \\mathrm{{s}}$
        If the tension is not changed, the velocity is not changed so the wavelength to produce $660 \\mathrm{{II}}$ is
        $528 \\mathrm{{~m}} / \\mathrm{{s}}=\\lambda .660 \\mathrm{{~Hz}} \\quad \\text {{ or }} \\quad \\lambda=0.80 \\mathrm{{~m}}$
        The length of the string to produce this frequency is one-half of a wavelength or $40 \\mathrm{{~cm}}$. The player needs to shorten the string by $20 \\mathrm{{~cm}}$ (from $60 \\mathrm{{~cm}}$ to $40 \\mathrm{{~cm}}$ ). This is done by pressing the string onto the fingerboard.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_22_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation described in problem 22-2, add that the amplitude (maximum displacement from the center line of the string) is $2.0 \\mathrm{{~mm}}$. What is the displacement at $0.20,0.40$, and $0.60 \\mathrm{{~mm}}$, and the time for one of these oscillations?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The oscillation is described in both space and time by equation $22-1$ at $x=0.20 \\mathrm{{~m}}$. The equation is
        $\\displaystyle y=(2 A \\cos 2 \\pi f t) \\sin 2 \\pi \\frac{{0.20}}{{1.6}}=(2 A \\cos 2 \\pi f t) \\sin \\frac{{\\pi}}{{4}}$
        The $\\sin \\pi / 4=0.71$ term specifies the fraction of the $2.0 \\mathrm{{~mm}}$ (amplitude) of the oscillation at $x=0.20 \\mathrm{{~m}}$. This maximum displacement is $1.4 \\mathrm{{~mm}}$. This is the displacement measured from the center line of the string. The height of the envelope or total (up and down) excursion of the piece of string is $2.8 \\mathrm{{~mm}}$.
        At $x=0.40 \\mathrm{{~mm}}$ the sine term becomes $\\sin \\pi / 2=1$ and the total excursion is $4.0 \\mathrm{{~mm}}$.
        Because of symmetry the maximum excursion at $0.60 \\mathrm{{~m}}$ is the same as at $0.20 \\mathrm{{~m}}$.
        The stretched string goes up and down together, that is, all points on the string are vibrating with the same frequency of $40 \\mathrm{{~Hz}}$, so the time for one oscillation is $1 / 40 \\mathrm{{~s}}=0.025 \\mathrm{{~s}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_22_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An $0.040 \\mathrm{{~kg}}$ string $0.80 \\mathrm{{~m}}$ long is stretched and vibrated in a fundamental mode with a frequency of $40 \\mathrm{{~Hz}}$. What is the speed (of propagation) of the wave and the tension in the string?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        
        soln_s = f"""
        The relationship $v=\\lambda f$ is used to calculate the speed of the wave. Since the string is vibrating in fundamental mode, the wavelength is $1.6 \\mathrm{{~m}}$ (see Fig. 22-1), so
        $v=\\lambda f=(1.6 \\mathrm{{~m}})(40 \\mathrm{{~Hz}})=64 \\mathrm{{~m}} / \\mathrm{{s}}$
        Having determined the velocity from this basic relation, the tension can be determined from $v=\\sqrt{{T / \\mu}}$. The linear density of the string is $\\mu=0.040 \\mathrm{{~kg}} / 0.80 \\mathrm{{~m}}=0.050 \\mathrm{{~kg}} / \\mathrm{{m}}$ so
        $T=v^2 \\mu=(64 \\mathrm{{~m}} / \\mathrm{{s}})^2 0.050 \\mathrm{{~kg}} / \\mathrm{{m}}=205 \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_22_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The third harmonic is excited on a $2.4 \\mathrm{{~m}}$ length of stretched string. What are the positions of the loops and nodes?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        
        soln_s = f"""
        To produce the third harmonic there are two nodes at 0.80 and $1.6 \\mathrm{{~m}}$ so as to divide the string into three segments. The loops are half way between the nodes at $0.40,1.2$, and $2.0 \\mathrm{{~m}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-22-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_21_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What happens to the power if the frequency is increased by a factor of 10 ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Since the frequency appears in the formula as the square, the power increases by a factor of one hundred.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_21_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A steel wire of length $1.0 \\mathrm{{~m}}$ and mass $4.0 \\mathrm{{gm}}$ is stretched with a force of $50 \\mathrm{{~N}}$. Waves of amplitude $1.2 \\times 10^{{-3}} \\mathrm{{~m}}$ and frequency $1000 \\mathrm{{~Hz}}$ are traveling on the wire. What is the average power of these waves?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The lowest frequency corresponds to the longest wavelength, and the longest wavelength is for a closed at one end open at the other pipe (see Figs. 22-3 and 22-4). In this situation the wavelength is $20 \\mathrm{{~m}}$ corresponding to a frequency of
        $\\displaystyle f=\\frac{{v}}{{\\lambda}}=\\frac{{340 \\mathrm{{~m}} / \\mathrm{{s}}}}{{20 \\mathrm{{~m}}}}=17 \\mathrm{{~Hz}}$
        This frequency is below what most people can hear.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_21_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the speed and wavelength of a $20 \\mathrm{{~Hz}}$ frequency wave on a string of $0.20 \\mathrm{{~kg}} / \\mathrm{{m}}$ under tension of 80 $\\mathrm{{N}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The velocity of the wave is
        $\\displaystyle v=\\sqrt{{F / \\mu}}=\\sqrt{{80 \\mathrm{{~N}} / 0.20 \\mathrm{{~kg}} / \\mathrm{{m}}}}=20 \\mathrm{{~m}} / \\mathrm{{s}}$
        Use the basic relation $v=\\lambda f$ to find
        $\\displaystyle \\lambda=\\frac{{v}}{{f}}=\\frac{{20 \\mathrm{{~m}} / \\mathrm{{s}}}}{{20 \\mathrm{{~Hz}}}}=1.0 \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_21_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation of problem 21-2, what is the expression for the space variation of the wave at $t=0.25 \\mathrm{{~s}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $y=(0.050 \\mathrm{{~m}}) \\sin 2 \\pi\\left(10-\\frac{{x}}{{0.20 \\mathrm{{~m}}}}\\right)$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_21_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation described in problem 21-2, what is the expression for the time variation of a point on the striug at $x=0.010 \\mathrm{{~m}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The expression for $y$ with $x=0.10 \\mathrm{{~m}}$ is
        $\\displaystyle y=(0.050 \\mathrm{{~m}}) \\sin 2 \\pi\\left(\\frac{{t}}{{0.025 \\mathrm{{~s}}}}-0.50\\right)$
        If at $t=0$, a wave is initiated from the point $x=0$ in the positive direction, then according to this expression, it takes $t=(0.025 \\mathrm{{~s}})(0.50)=0.0125 \\mathrm{{~s}}$ for the wave to reach $0.010 \\mathrm{{~m}}$. After this $0.0125 \\mathrm{{~s}}$ time interval, the point $x=$ $0.010 \\mathrm{{~m}}$ oscillates in a sinusoidal manner. Look back at the numbers for wavelength and period in problem 21-2 and notice that it takes $1 / 2$ of a period tor the wave to travel $1 / 2$ of a wavelength.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_21_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A sinusoidal wave is propagated on a string with amplitude $0.050 \\mathrm{{~m}}$, wavelength $0.20 \\mathrm{{~m}}$, and frequency of $40 \\mathrm{{~Hz}}$. Write an expression for the displacement of the string as a function of the time and position on the string.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        
        soln_s = f"""
        The period of the wave is $1 / 40 s=0.025 s$. The displacement is according to equation 21-2
        $\\displaystyle y=(0.050 \\mathrm{{~m}}) \\sin 2 \\pi\\left(\\frac{{t}}{{0.025 \\mathrm{{~s}}}}-\\frac{{x}}{{0.20 \\mathrm{{~m}}}}\\right)$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-23-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_21_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The end of a stretched rope is being vibrated up and down at $120 \\mathrm{{H}} z$. A "snapshot" of the wave indicates a wavelength of $0.50 \\mathrm{{~m}}$. What is the speed of the wave?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        
        soln_s = f"""
        The wave has velocity given by $v=\\lambda f=0.50 \\mathrm{{~m}}(120 \\mathrm{{~Hz}})=60 \\mathrm{{~m}} / \\mathrm{{s}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_20_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the change in entropy for $200 \\mathrm{{~kg}}$ of water slowly heated from $20^{{\\circ}} \\mathrm{{C}}$ to $80^{{\\circ}} \\mathrm{{C}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        
        soln_s = f"""
        The basic definition of entropy is $\\Delta S=\\Delta Q_r / T$. The heating process is surely reversible. Switching to a calculus definition
        $\\Delta S=\\frac{{d Q_r}}{{T}}=\\frac{{m c d T}}{{T}}$
        and integrating over the temperature range
        $\\Delta S=\\int_T^{{T_f}} m c \\frac{{d T}}{{T}}=\\left.m c \\ln T\\right|_{{T_f}} ^{{T_f}}=200 \\mathrm{{~kg}} \\frac{{4200 \\mathrm{{~J}}}}{{\\mathrm{{~kg}} \\cdot{{ }}^{{\\circ}} \\mathrm{{C}}}} \\ln 4=1.16 \\times 10^6 \\mathrm{{~J}} / \\mathrm{{K}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
    
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_20_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the change in entropy for $50 \\mathrm{{~kg}}$ of water going from liquid to vapor at $100^{{\\circ}} \\mathrm{{C}}$ ?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The conversion of water from liquid to vapor is reversible, and the heat required is the heat of vaporization of water $\\left(3.3 \\times 10^5 \\mathrm{{~J}} / \\mathrm{{kg}}\\right)$, so
        $\\displaystyle \\Delta S=\\frac{{3.3 \\times 10^5 \\mathrm{{~J}} / \\mathrm{{kg}}(50 \\mathrm{{~kg}})}}{{373 \\mathrm{{~K}}}}=4.4 \\times 10^4 \\mathrm{{~J}} / \\mathrm{{K}}$
        Water vapor has more disorder than water (liquid).
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_20_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Assuming Carnot efficiency calculate the coefficient of performance of a heat pump operating between $20^{{\\circ}} \\mathrm{{C}}$ and $12^{{\\circ}} \\mathrm{{C}}$. Find the minimum power requirements to move thermal energy from the $12^{{\\circ}} \\mathrm{{C}}$ area to the $20^{{\\circ}} \\mathrm{{C}}$ area at the rate of $20 \\mathrm{{~kW}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The coefficient of performance for a heat pump is
        $\\displaystyle C O P_C(\\text {{ heat pump }})=\\frac{{T_H}}{{T_H-T_C}}=\\frac{{293}}{{8}}=37$
        The coefficient of performance is the ratio $Q_H / W$ so $Q_H / W=37$, or
        $\\displaystyle W=\\frac{{Q_H}}{{37}}=\\frac{{20 \\mathrm{{~kW}}}}{{37}}=0.54 \\mathrm{{~kW}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_20_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the coefficient of performance of a refrigerator operating as a Carnot engine between $5^{{\\circ}} \\mathrm{{C}}$ and $25^{{\\circ}} \\mathrm{{C}}$ ?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\operatorname{{COP}}_c$ (refrigerator) $=278 / 20=14$
        Most refrigerators run at $1 / 2$ or less of their Carnot efficiencies.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_20_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An electric generating plant has an output of $50 \\mathrm{{MW}}$. Steam enters the turbines at $550^{{\\circ}} \\mathrm{{C}}$ and exits at $110^{{\\circ}} \\mathrm{{C}}$. The efficiency of the plant is $80 \\%$ of the theoretical Carnot efficiency. What is the rate of heat consumption and heat rejection?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The Carnot efficiency is $e=1-T_d T_n=1-383 \\mathrm{{~K}} / 823 \\mathrm{{~K}}=0.53$
        Eighty percent of this is $43 \\%$ efficiency. This means that $43 \\%$ of the input is equal to the $50 \\mathrm{{MW}}$ output, giving an input requirement of $116 \\mathrm{{MW}}$. The rate of heat rejection is $57 \\%$ of the $116 \\mathrm{{MW}}$ input or $66 \\mathrm{{MW}}$ (rejected).
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_20_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A heat engine takes in $2000 \\mathrm{{~J}}$ of heat and performs $400 \\mathrm{{~J}}$ of work each cycle. What is the efficiency, amount of heat rejected, and the power output for a cycle rate of 100 cycles/s?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The amount of heat rejected in cach cycle is $2000 \\mathrm{{~J}}-400 \\mathrm{{~J}}=1600 \\mathrm{{~J}}$.
        The efficiency is
        $e=1-\\frac{{\\left|Q_C\\right|}}{{\\left|Q_H\\right|}}=1-\\frac{{1600 \\mathrm{{~J}}}}{{2000 \\mathrm{{~J}}}}=1-0.8=0.20 \\implies 20 \\%$
        The power output for the engine is
        $P=\\frac{{100 \\mathrm{{cycle}}}}{{\\mathrm{{s}}}} \\frac{{400 \\mathrm{{~J}}}}{{\\text {{ cycle }}}}=40 \\mathrm{{~kW}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_20_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation of problem 20-3 increase the pressure in the boiler so that the temperature of the steam is $450 \\mathrm{{~K}}$. What is the efficiency of the engine?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $e=1-(308 / 450)=0.32 \\implies 32 \\%$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_20_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A steam engine is operated with steam at atmospheric pressure and $100^{{\\circ}} \\mathrm{{C}}$. Steam leaves the engine at slightly above outside temperature, $35^{{\\circ}} \\mathrm{{C}}$. What is the efficiency of the engine?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $e=1-308 / 373=0.17 \\implies 17 \\%$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_20_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        If an (gasoline) automobile engine performs $10 \\mathrm{{~kW}}$ of work and is $28 \\%$ efficient, what is the rate at which heat is taken in and expelled?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        
        soln_s = f"""
        It doesn't matter whether we take the heat in and the work performed for one cycle or the rates for heat in and work performed. The efficiency is $e=W / Q_H$ so the rate of heat in is
        $\\displaystyle Q_H=\\frac{{W}}{{e}}=\\frac{{10 \\mathrm{{~kW}}}}{{0.28}}=36 \\mathrm{{~kW}}$
        From the efficiency equation $20-2, \\quad 20-2, \\frac{{Q_C}}{{Q_H}}=1-e$ so the heat expelled is
        $Q_C=Q_H(1-e)=36 \\mathrm{{~kW}}(1-0.28)=26 \\mathrm{{~kW}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-23-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_20_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the efficiency of an engine that takes in $300 \\mathrm{{~J}}$ and expels $100 \\mathrm{{~J}}$ over each cycle?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $e=1-\\left|Q_C\\right| /\\left|Q_H\\right|=1-(100 / 300)=0.67 \\implies 67 \\%$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_19_12(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An ideal monatomic gas at $p a=3.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2, V a=0.060 \\mathrm{{~m}}^3$, and $T=27^{{\\circ}} \\mathrm{{C}}$ expands adiabatically to $p b$ $=2.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2, V b=0.085 \\mathrm{{~m}}^3$ and then isothermally to $V c=0.100 \\mathrm{{~m}}^3$. What is the final temperature, pressure, and work performed by the gas? Show these paths on a $p-V$ diagram.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First calculate the temperature at the end of the adiabatic expansion using the ideal gas law $\\displaystyle  \\frac{{p_a V_a}}{{T_a}}=\\frac{{p_b V_b}}{{T_b}}$ or $\\displaystyle  T_b=T_a \\frac{{p_b}}{{p_a}} \\frac{{V_b}}{{V_a}}=300 \\mathrm{{~K}} \\frac{{2.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2}}{{3.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2}} \\frac{{0.060 \\mathrm{{~m}}^3}}{{0.085 \\mathrm{{~m}}^3}}=140 \\mathrm{{~K}}$
        The gas has moved along an adiabat from the $300 \\mathrm{{~K}}$ isotherm to the $140 \\mathrm{{~K}}$ isotherm (point $a$ to point $b$ on Fig. 19-5).
        The pressure at the end of the isothermal expansion can be calculated from the ideal gas law with the temperature a constant, $P b V b=P c V c$, or
        $\\displaystyle  p_c=p_b\\left(V_b / V_c\\right)=2.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2\\left(0.085 \\mathrm{{~m}}^3 / 0.100 \\mathrm{{~m}}^3\\right)=1.7 \\times 10^5 \\mathrm{{~Pa}}$
        The temperature of the gas at point $c$ is $140 \\mathrm{{~K}}$, and the pressure is $1.7 \\times 10^5 \\mathrm{{~Pa}}$.
        The work performed is positive in both parts of the expansion. During the adiabatic portion of the expansion, the work is the area under the $a-b$ portion of the curve (see problem 19-11).
        $W_{{a-b}}=\\frac{{1}}{{\\gamma-1}}\\left[p_a V_a-p_b V_b\\right]$
        $W_{{a-b}}=\\frac{{1}}{{0.40}}\\left[3.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2\\left(0.060 \\mathrm{{~m}}^3\\right)-2.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2\\left(0.085 \\mathrm{{~m}}^3\\right)\\right]$
        $W_{{a-b}}=\\frac{{1}}{{0.40}}\\left[1.8 \\times 10^4 \\mathrm{{~J}}-1.7 \\times 10^4 \\mathrm{{~J}}\\right]=2500 \\mathrm{{~J}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-19-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        In the isothermal portion of the expansion the work is the area under the $b-c$ curve (equation 19-3).
        $\\displaystyle W=\\int_{{V_b}}^{{V_c}} p d V=n R T \\ln \\frac{{V_c}}{{V_b}}$
        The $n R T$ is equal to $p b V b$, so
        $\\displaystyle W=p_b V_b \\ln \\left(V_c / V_b\\right)$
        This is the work performed in the isothermal expansion. Along the isotherm $p b V b / T b=p c V c / T c$ reduces to $p b V b=p c V c$ because $T b=T c$.
        $\\displaystyle W_{{b-c}}=2.0 \\times 10^5 \\frac{{\\mathrm{{N}}}}{{\\mathrm{{m}}^2}}\\left(0.085 \\mathrm{{~m}}^3\\right) \\ln \\frac{{0.100 \\mathrm{{~m}}^3}}{{0.085 \\mathrm{{~m}}^3}}=2760 \\mathrm{{~J}}$
        The total work performed by the gas is $5260 \\mathrm{{~J}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_19_11(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An ideal gas at $3.0 \\times 10^5 \\mathrm{{~Pa}}$ and $0.080 \\mathrm{{~m}}^3$ is compressed adiabatically to $0.020 \\mathrm{{~m}}^3$. What is the final pressure and the work performed?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The final pressure is from $-W=n C_V \\Delta T=n C_V\\left(T_2-T_1\\right)$, or
        $p_2=p_1\\left(y_1 / V_2\\right)^y=3.0 \\times 10^5 \\mathrm{{~Pa}}(4)^{{14}}-2.1 \\times 10^6 \\mathrm{{~Pa}}$
        The work performed is from $-W=n C_r \\Delta T=n C_r\\left(T_2-T_1\\right)$. Substituting from the ideal gas law $p V=n R T$ and $R / C v$ $=\\gamma-1$
        $\\displaystyle W=\\frac{{C_V}}{{R}}\\left(p_1 V_1-p_2 V_2\\right)=\\frac{{1}}{{\\gamma-1}}\\left(p_1 V_1-p_2 V_2\\right)$
        $\\displaystyle W=\\frac{{1}}{{1.40-1}}\\left[3.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2\\left(0.080 \\mathrm{{~m}}^3\\right)-2.1 \\times 10^6 \\mathrm{{~N}} / \\mathrm{{m}}^2\\left(0.020 \\mathrm{{~m}}^3\\right)\\right]$
        $\\displaystyle W=\\frac{{1}}{{0.40}}\\left[2.4 \\times 10^4 \\mathrm{{~N}} \\cdot \\mathrm{{m}}-4.2 \\times 10^4 \\mathrm{{~N}} \\cdot \\mathrm{{m}}\\right]=-4.5 \\times 10^4 \\mathrm{{~J}}$
        Work is performed on the gas in the amount of $4.5 \\times 10^4 \\mathrm{{~J}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_19_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Air and fuel at $300 \\mathrm{{~K}}$ and 1.0atm is compressed in an automobile engine with a compression ratio of 8 to 1. Take $\\gamma=1.40$ and calculate the final pressure and temperature of the mixture.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The pressure is from $p_1 V_1^\\gamma=p_2 V_2^\\gamma$ or
        $p_2=p_1\\left(V_1 / V_2\\right)^\\gamma=1.0 \\mathrm{{~atm}}(8)^{{1.4}}=18.4 \\mathrm{{~atm}}$
        $T_1 V_1^{{\\gamma-1}}=T_2 V_2^{{\\gamma-1}}$
        The temperature is from
        or
        $T_2=T_1\\left(V_1 / V_2\\right)^{{\\gamma-1}}=300 \\mathrm{{~K}}(8)^{{0.40}}=689 \\mathrm{{~K}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_19_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Two moles of an ideal gas expand isothermally at $295 \\mathrm{{~K}}$ from $0.60 \\mathrm{{~m}}^3$ to $0.80 \\mathrm{{~m}}^3$. What would be the temperature of the gas if the same expansion were adiabatic?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The temperature is from equation 19-7.
        $T_2=T_1\\left(V_1 / V_2\\right)^{{\\gamma-1}}=295 \\mathrm{{~K}}(0.60 / 0.80)^{{1.40-1}}=263 \\mathrm{{~K}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_19_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Two moles of an ideal gas at a pressure of $1.0 \\mathrm{{~atm}}$ has $400 \\mathrm{{~J}}$ of heat added in expanding from a volume of $4.7 \\times 10^{{-3}} \\mathrm{{~m}}^3$ to $7.0 \\times 10^{{-3}} \\mathrm{{~m}}^3$. What is the change in internal energy and temperature? What are the specific heats?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The work performed is $W=p \\Delta V=1.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2\\left(2.3 \\times 10^{{-3}} \\mathrm{{~m}}^3\\right)=2.30 \\mathrm{{~J}}$.
        The change in internal energy is from the first law $\\Delta U=\\Delta Q-W=400 \\mathrm{{~J}}-230 \\mathrm{{~J}}=170 \\mathrm{{~J}}$.
        The change in temperature is from $\\Delta U=(3 / 2) n R \\Delta T$, or
        $\\displaystyle \\Delta T=\\frac{{2 \\Delta U}}{{3 n R}}=\\frac{{2 \\cdot 170 \\mathrm{{~J}}}}{{3(2 \\mathrm{{~mol}})(8.3 \\mathrm{{~J}} / \\mathrm{{mol}} \\cdot \\mathrm{{K}})}}=6.8 \\mathrm{{~K}}$
        The heat capacity at constant pressure is from $\\Delta Q=n C_p \\Delta T$, or
        $\\displaystyle C_p=\\frac{{\\Delta Q}}{{n \\Delta T}}=\\frac{{400 \\mathrm{{~J}}}}{{2 \\mathrm{{~mol}} \\cdot 6.8 \\mathrm{{~K}}}}=\\frac{{29 \\mathrm{{~J}}}}{{\\mathrm{{~mol}} \\cdot \\mathrm{{K}}}}$
        The heat capacity at constant volume is
        $\\displaystyle C_V=C_p-R=29 \\mathrm{{~J}} / \\mathrm{{mol}} \\cdot \\mathrm{{K}}-8.3 \\mathrm{{~J}} / \\mathrm{{mol}} \\cdot \\mathrm{{K}}=20.7 \\mathrm{{~J}} / \\mathrm{{mol}} \\cdot \\mathrm{{K}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_19_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The heat necessary to raise the temperature of 10 moles of an ideal gas at a constant pressure of $1.0 \\times 10^5$ Pa 10 degrees is $2100 \\mathrm{{~J}}$. What is the volume change and the heat capacity at constant pressure and constant volume?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The work performed in the volume change is $\\Delta W=p \\Delta V=n R \\Delta T$ so
        $\\displaystyle \\Delta V=\\frac{{n R \\Delta T}}{{p}}=\\frac{{10 \\mathrm{{~mol}}(8.3 \\mathrm{{~J}} / \\mathrm{{mol}} \\cdot \\mathrm{{K}}) 10 \\mathrm{{~K}}}}{{1.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2}}=8.3 \\times 10^{{-3}} \\mathrm{{~m}}^3$
        Starting with the basic statement $\\Delta Q={{ }}_n C_p \\Delta T$ the heat capacity at constant pressure is
        $\\displaystyle C_p=\\frac{{1}}{{n}} \\frac{{\\Delta Q}}{{\\Delta T}}=\\frac{{1}}{{10 \\mathrm{{~mol}}}} \\frac{{2100 \\mathrm{{~J}}}}{{10 \\mathrm{{~K}}}}=\\frac{{21 \\mathrm{{~J}}}}{{\\mathrm{{~mol}} \\cdot \\mathrm{{K}}}}$
        The gas constant $R$ is $8.3 \\mathrm{{~J}} / \\mathrm{{mol}}$. $\\mathrm{{K}}$ so
        $\\displaystyle C_V=C_p-R=21 \\mathrm{{~J}} / \\mathrm{{mol}} \\cdot \\mathrm{{K}}-8.3 \\mathrm{{~J}} / \\mathrm{{mol}} \\cdot \\mathrm{{K}}=12.7 \\mathrm{{~J}} / \\mathrm{{mol}} \\cdot \\mathrm{{K}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_19_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the total internal energy of 10 moles of an ideal monatomic gas at $20^{{\\circ}} \\mathrm{{C}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        A monatomic gas is composed of single atoms that behave as mass points, and as such, they can only translate. They eannot rotate or vibrate. See problem 18-5 and the associated discussion to confirm that the internal energy is $(3 / 2) n R T$.
        $U=n R T=(3 / 2) 10 \\mathrm{{~mol}}(8.3 \\mathrm{{~J}} / \\mathrm{{mol}} \\cdot \\mathrm{{K}}) 293 \\mathrm{{~K}}=3.6 \\times 10^4 \\mathrm{{~J}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_19_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An ideal gas initially at $3.0 \\times 10^5 \\mathrm{{~Pa}}$, volume of $0.030 \\mathrm{{~m}}^3$, and temperature $20^{{\\circ}} \\mathrm{{C}}$ is heated at constant pressure to a volume of $0.120 \\mathrm{{~m}}^3$. How much work was performed by the gas? If $80 \\mathrm{{KJ}}$ of heat energy was supplied to the gas, what was the change in internal energy?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The work performed is
        $W=p \\Delta V=3.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2(0.120-0.030) \\mathrm{{m}}^3=2.7 \\times 10^4 \\mathrm{{~J}}$
        The increase in internal energy is
        $\\Delta U=Q-W=8.0 \\times 10^4 \\mathrm{{~J}}-2.7 \\times 10^4 \\mathrm{{~J}}=5.3 \\times 10^4 \\mathrm{{~J}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_19_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        In a certain process $1.5 \\times 10^5 \\mathrm{{~J}}$ of heat is added to an ideal gas to keep the pressure at $2.0 \\times 10^5 \\mathrm{{~Pa}}$ while the volume expands from $6.3 \\mathrm{{~m}}^3$ to $7.1 \\mathrm{{~m}}^3$. What is the change in internal energy for the gas?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Apply the first law $\\Delta U=\\Delta Q-\\Delta W$. Since the pressure is constant
        $\\Delta W=p \\Delta V=2.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2\\left(0.8 \\mathrm{{~m}}^3\\right)=1.6 \\times 10^5 \\mathrm{{~J}}$
        $\\Delta U=1.5 \\times 10^5 \\mathrm{{~J}}-1.6 \\times 10^5 \\mathrm{{~J}}=-1.0 \\times 10^4 \\mathrm{{~J}}$
        The internal energy of the gas has been decreased by $1.0 \\times 10^4 \\mathrm{{~J}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_19_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        You eat 100 food calories worth of nuts. How high would you, at $65 \\mathrm{{~kg}}$, have to climb to "work off" these 100 caluries?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The 100 food calories is $100 \\mathrm{{kcal}}$ into your (thermodynamic) system. The work to reduce your total internal energy back to what it was before eating the nuts is
        $1.0 \\times 10^5 \\mathrm{{cal}}(4.2 \\mathrm{{~J}} / \\mathrm{{cal}})=65 \\mathrm{{~kg}}\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) h \\text {{ or }} h=660 \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_19_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A gas at constant pressure of $4.0 \\times 10^5 \\mathrm{{~Pa}}$ is cooled so that its volume decreases from $1.6 \\mathrm{{~m}}^3$ to $1.2 \\mathrm{{~m}}^3$. What work is performed by the gas?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Since pressure is a constant the work performed is from equation 19-1.
        $\\Delta W=p \\Delta V=4.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2\\left(-0.4 \\mathrm{{~m}}^3\\right)=-1.6 \\times 10^5 \\mathrm{{~J}}$
        The negative sign indicates that work was done on the system by an outside agent.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_19_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Two moles of an ideal gas maintained at $20^{{\\circ}} \\mathrm{{C}}$ expand until the pressure is one-half the original. How much work is done by the gas?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Since the gas remains at constant temperature, it goes from one state to the other along an isotherm (see Fig. 19-1).
        $\\displaystyle W=n R T \\ln \\frac{{p_1}}{{p_2}}=2 \\mathrm{{~mol}} \\frac{{8.3 \\mathrm{{~J}}}}{{\\mathrm{{~mol}} \\cdot \\mathrm{{K}}}} 293 \\mathrm{{~K}} \\cdot \\ln 2=3370 \\mathrm{{~J}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-19-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_18_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the previous problem find the heat required if the gas were diatomic.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Assume that at room temperature the diatomic molecule is a translator and rotator (see Fig. 18-6) so that $C_V=5 R / 2$. Then
        $\\displaystyle \\Delta Q=n \\frac{{5 R}}{{2}} \\Delta T=2.0 \\mathrm{{~mol}} \\frac{{5(8.3 \\mathrm{{~J}} / \\mathrm{{mol}} \\cdot \\mathrm{{K}})}}{{2}} 25 \\mathrm{{~K}}=1040 \\mathrm{{~J}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-18-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_18_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        How much heat is required to raise the temperature of 2.0 mole of monatomic gas $25 \\mathrm{{~K}}$ starting at room temperature.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Use $\\Delta Q=n C_v \\Delta T$ with $C_v$ for a diatomic gas as $3 R / 2$ so
        $\\displaystyle \\Delta Q=n \\frac{{3 R}}{{2}} \\Delta T=2.0 \\mathrm{{~mol}} \\frac{{3(8.3 \\mathrm{{~J}} / \\mathrm{{mol}} \\cdot \\mathrm{{K}})}}{{2}} 25 \\mathrm{{~K}}-620 \\mathrm{{~J}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_18_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Isotopes of uranium are sometimes separated by gaseous diffusion. The lighter isotope, having a higher speed diffuses more rapidly than the heavier isotope. What is the ratio of the rms speeds of $\\mathrm{{U}} 235$ and $\\mathrm{{U}} 238$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The rms speed is $v_{{r m s}}=\\sqrt{{3 k T / m}}$. The ratio of these speeds is
        $\\displaystyle \\frac{{\\left(v_{{r m s}}\\right)_{{235}}}}{{\\left(v_{{r m s}}\\right)_{{238}}}}=\\frac{{\\sqrt{{3 k T / 235}}}}{{\\sqrt{{3 k T / 238}}}}=\\sqrt{{\\frac{{238}}{{235}}}}=1.006$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_18_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Two moles of helium are in a tank at $25^{{\\circ}} \\mathrm{{C}}$. Find the total translational kinetic energy, the kinetic energy per molecule, and the rms speed of the atoms.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The total translational kinetic energy is
        $\\displaystyle E=\\frac{{3}}{{2}} n R T=\\frac{{3}}{{2}} 2.0 \\mathrm{{~mole}} \\frac{{8.3 \\mathrm{{~J}}}}{{\\mathrm{{~mole}} \\cdot \\mathrm{{K}}}} 298 \\mathrm{{~K}}=7420 \\mathrm{{~J}}$
        The energy per molecule is
        $\\displaystyle \\frac{{m\\left(v^2\\right)_{{a v g}}}}{{2}}=\\frac{{3}}{{2}} k T=\\frac{{3}}{{2}}\\left[\\frac{{1.4 \\times 10^{{-23}} \\mathrm{{~J}}}}{{\\mathrm{{~K}}}}\\right] 298 \\mathrm{{~K}}=6.26 \\times 10^{{-21}} \\mathrm{{~J}}$
        The rms speed is
        $\\displaystyle v_{{r m s}}=\\sqrt{{\\frac{{3 R T}}{{M}}}}=\\left[\\frac{{3(8.3 \\mathrm{{~J}} / \\mathrm{{mol}} \\cdot \\mathrm{{K}}) 298 \\mathrm{{~K}}}}{{4.0 \\times 10^{{-3}} \\mathrm{{~kg}} / \\mathrm{{mol}}}}\\right]^{{1 / 2}}=1360 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_18_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the pressure at $3000 \\mathrm{{~m}}$ elevation above the earth?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Assume $T=0^{{\\circ}} \\mathrm{{C}}=273 \\mathrm{{~K}}$. $T$ decreases with height because the air is heated, primarily by radiation from the earth, so consider this an average temperature over the $3000 \\mathrm{{~m}}$. Air is mostly oxygen and nitrogen and has an average molecular weight of $28.8 \\mathrm{{~g}} / \\mathrm{{mole}}$ or $28.8 \\times 10^{{-3}} \\mathrm{{~kg}} / \\mathrm{{mole}}$. The gravitational constant varies only slightly over this distance.
        $\\displaystyle \\frac{{M g}}{{R T}} y=\\frac{{28.8 \\times 10^{{-3}} \\mathrm{{~kg}} / \\mathrm{{mol}}\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)}}{{8.3 \\mathrm{{~J}} / \\mathrm{{mol}} \\cdot \\mathrm{{K}}(273 \\mathrm{{~K}})}} 3000 \\mathrm{{~m}}=0.374$
        First calculate the exponent
        then the pressure $p=p_d e^{{-0.374}}=1.0 \\mathrm{{~atm}}(0.69)=0.69 \\mathrm{{~atm}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_18_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A weather balloon is inflated with helium at just over atmospheric pressure of $1.2 \\times 10^5 \\mathrm{{~Pa}}$, temperature. $20^{{\\circ}} \\mathrm{{C}}$, and volume $2.5 \\mathrm{{~m}}^3$. When the balloon rises to where the pressure is $0.50 \\times 10^5 \\mathrm{{~Pa}}$ and temperature is $58^{{\\circ}} \\mathrm{{C}}$, what is the volume?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Using
        $\\displaystyle \\frac{{p_1 V_1}}{{T_1}}=\\frac{{p_2 V_2}}{{T_2}}, \\quad \\frac{{1.2 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2 2.5 \\mathrm{{~m}}^3}}{{293 \\mathrm{{~K}}}}-\\frac{{0.5 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2 V_2}}{{215 \\mathrm{{~K}}}}$ or $\\displaystyle V_2=\\frac{{215 \\mathrm{{~K}}}}{{293 \\mathrm{{~K}}}} \\frac{{1.2 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2}}{{0.5 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2}} 2.5 \\mathrm{{~m}}^3=4.4 \\mathrm{{~m}}^3$

        As an exercise find the number of moles of this gas.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_18_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An air bubble at the bottom of a lake has a volume of $20 \\mathrm{{~cm}}^3$, pressure of $4.9 \\mathrm{{~Pa}}$, and temperature $4^{{\\circ}} \\mathrm{{C}}$. The bubble rises to the surface where the temperature is $20^{{\\circ}} \\mathrm{{C}}$ and the pressure $1.0 \\mathrm{{~Pa}}$. Find the volume as the bubble reaches the surface. Also find the number of moles of the gas.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-18-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        
        soln_s = f"""
        Use $\\displaystyle \\frac{{p_1 V_1}}{{T_1}}=\\frac{{p_2 V_2}}{{T_2}}$
        $\\displaystyle \\frac{{\\left(4.9 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2\\right) 20 \\mathrm{{~cm}}^3}}{{277 \\mathrm{{~K}}}}=\\frac{{\\left(1.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2\\right)}}{{293 \\mathrm{{~K}}}} / 2$
        or
        $\\displaystyle V_2=\\frac{{293 \\mathrm{{~K}}}}{{277 \\mathrm{{~K}}}} \\frac{{4.9 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2}}{{1.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2}} 20 \\mathrm{{~cm}}^3=104 \\mathrm{{~cm}}^3$
        To find the number of moles use $p V / T=n R$ and the given conditions at the bottom of the lake.
        $\\displaystyle n=\\frac{{p V}}{{R T}}=\\frac{{\\left(4.9 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2\\right) 20 \\mathrm{{~cm}}^3}}{{(8.3 \\mathrm{{~J}} / \\mathrm{{mol}} \\cdot \\mathrm{{K}})(277 \\mathrm{{~K}})}}\\left(\\frac{{\\mathrm{{m}}}}{{10^2 \\mathrm{{~cm}}}}\\right)^3=0.0043 \\mathrm{{~mol}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_18_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the volume of a mole of ideal gas at standard temperature and pressure (STP)?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Standard temperature is $0^{{\\circ}} \\mathrm{{C}}=273 \\mathrm{{~K}}$. Standard pressure is latm $=1.0 \\times 10^5 \\mathrm{{~Pa}}$. From the ideal gas law
        $\\displaystyle V=\\frac{{n R T}}{{p}}=\\frac{{1 \\mathrm{{~mol}}(8.3 \\mathrm{{~J}} / \\mathrm{{mol}} \\cdot \\mathrm{{K}}) 273 \\mathrm{{~K}}}}{{1.0 \\times 10^5 \\mathrm{{~Pa}}}}=0.023 \\mathrm{{~m}}^3$
        This is a 23 liter container or a box $0.28 \\mathrm{{~m}}$ on a side Remember that a Pascal is a N$/ \\mathrm{{m}}^2$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_17_13(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $5.0 \\mathrm{{~kg}}$ bullet traveling $800 \\mathrm{{~m}} / \\mathrm{{s}}$ passes through a $5.0 \\mathrm{{~kg}}$ copper block at rest on a frictionless table. Twenty percent of the kinetic energy of the bullet goes into heating the block. Thirty percent goes into kinetic energy of the block. Fifty percent remains with the bullet. The block and bullet are initially at $25^{{\\circ}} \\mathrm{{C}}$. What is the temperature and speed of the block and the speed of the bullet? $\\left(c_{{\\mathrm{{cu}}}}=390 \\mathrm{{~J}} / \\mathrm{{kg}} . \\mathrm{{C}}\\right)$

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First find the initial kinetic energy of the bullet.
        $K E=(1 / 2) m v^2=(1 / 2) 5.0 \\times 10^{{-3}} \\mathrm{{~kg}}(800 \\mathrm{{~m}} / \\mathrm{{s}})^2=1600 \\mathrm{{~J}}$
        Twenty percent of this, or $320 \\mathrm{{~J}}$, goes to heating the copper block.
        $320 \\mathrm{{~J}}=5.0 \\mathrm{{~kg}}\\left(390 \\mathrm{{~J}} / \\mathrm{{kg}} \\cdot \\mathrm{{C}}^o\\right) \\Delta T \\quad \\text {{ or }} \\quad \\Delta T=0.16^{{\\circ}} \\mathrm{{C}}$
        Thirty percent, or 480J, goes to kinetic energy of the block.
        $480 \\mathrm{{~J}}=(\\mathrm{{l}} / 2)(5.0 \\mathrm{{~kg}}) v^2 \\quad \\text {{ or }} \\quad v=14 \\mathrm{{~m}} / \\mathrm{{s}}$
        Fifty percent, or $800 \\mathrm{{~J}}$, stays with the bullet.
        $800 \\mathrm{{~J}}-(1 / 2)\\left(5.0 \\times 10^{{-3}} \\mathrm{{~kg}}\\right) v^2 \\quad \\text {{ or }} \\quad v=570 \\mathrm{{~m}} / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_17_12(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A block of ice initially at $50 \\mathrm{{~kg}}$ and traveling at $5.4 \\mathrm{{~m}} / \\mathrm{{s}}$ slides along a horizontal surface until it comes to rest. Assume all the heat generated due to friction is used to melt the ice. How much ice is melted?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        All the kinetic energy in the block Goesinto work expended to overcome friction, and all this energy Goesinto melting the ice. The initial kinetic energy is
        $K E=(1 / 2) m v^2=(1 / 2) 50 \\mathrm{{~kg}}(5.4 \\mathrm{{~m}} / \\mathrm{{s}})^2=730 \\mathrm{{~J}}$
        All this energy is used to melt the ice.
        $730 \\mathrm{{~J}}=\\left(3.3 \\times 10^5 \\mathrm{{~J}} / \\mathrm{{kg}}\\right) m \\quad \\text {{ or }} \\quad m=0.0022 \\mathrm{{~kg}}=2.2 \\mathrm{{~g}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_17_11(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Two rods with cross sections of $4.0 \\mathrm{{~cm}}^2$, one copper of $1.2 \\mathrm{{~m}}$ length, and the other steel of $1.0 \\mathrm{{~m}}$ length, are connected together with their opposite ends held at $100^{{\\circ}} \\mathrm{{C}}$ and $0^{{\\circ}} \\mathrm{{C}}$ as shown. What is the heat flow and the temperature of the junction at equilibrium? $k_{{\\mathrm{{cu}}}}-380 \\mathrm{{~W}} / \\mathrm{{m}} . \\mathrm{{C}}^{{\\circ}}$ and $k_{{\\mathrm{{st}}}}-50 \\mathrm{{~W}} / \\mathrm{{m}} . \\mathrm{{C}}^{{\\circ}}$ )

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The key to this problem is that the heat flow in the two rods must be the same. If the heat flow in the copper is greater than the heat flow in the steel then the joint will get hot. This does not happen so the heat flow must be the same.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-17-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        Take the temperature of the junction as $T_{{\\mathrm{{j}}}}$, and set the heat flows equal
        $\\displaystyle \\frac{{k_{{\\mathrm{{cu}}}} A\\left(100-T_{{\\mathrm{{j}}}}\\right)^o \\mathrm{{C}}}}{{1.2 \\mathrm{{~m}}}}=\\frac{{k_{{\\mathrm{{st}}}} A\\left(T_{{\\mathrm{{j}}}}-0\\right)^o \\mathrm{{C}}}}{{1.0 \\mathrm{{~m}}}}$
        Putting in the numbers
        $1.0 \\mathrm{{~m}}\\left(380 \\mathrm{{~W}} / \\mathrm{{m}} \\cdot \\mathrm{{C}}^a\\right)\\left(100-T_{{\\mathrm{{j}}}}\\right)^a \\mathrm{{C}}=1.2 \\mathrm{{~m}}\\left(50 \\mathrm{{~W}} / \\mathrm{{m}} \\cdot \\mathrm{{C}}^a\\right)\\left(T_{{\\mathrm{{j}}}}-0\\right)^{{\\circ}} \\mathrm{{C}}$
        $38000-380 T_{{\\mathrm{{j}}}}=60 T_{{\\mathrm{{j}}}} \\text {{ or }} \\quad T_{{\\mathrm{{j}}}}=86^{{\\circ}} \\mathrm{{C}}$
        The heat flow is from either formula
        $\\displaystyle H_{{\\mathrm{{cu}}}}=\\frac{{380 \\mathrm{{~W}} / \\mathrm{{m}} \\cdot \\mathrm{{C}}^o 4.0 \\times 10^{{-4}} \\mathrm{{~m}}^2}}{{1.2 \\mathrm{{~m}}}}(100-86)^{{\\circ}} \\mathrm{{C}}=1.8 \\mathrm{{~W}}$
        Calorimetry combined with other energy analysis techniques can be used to solve other types of problems.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_17_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the heat flow through a wooden door of area $2.0 \\mathrm{{~m}}^2$ and thickness $5.0 \\mathrm{{~cm}}$ when the temperature difference is $20 \\mathrm{{~K}}$. The $k$ for wood is $0.10 \\mathrm{{~W}} / \\mathrm{{m}} . \\mathrm{{K}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The heat flow is $\\displaystyle H=\\frac{{k A \\Delta T}}{{L}}=\\frac{{0.10 \\mathrm{{~W}} / \\mathrm{{m}} \\cdot \\mathrm{{K}}\\left(2.0 \\mathrm{{~m}}^2\\right) 20 \\mathrm{{~K}}}}{{5.0 \\times 10^{{-2}} \\mathrm{{~m}}}}=80 \\mathrm{{~W}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_17_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        If in the previous problem the amount of ice is reduced to $0.50 \\mathrm{{~kg}}$ what is the final temperature of the composition?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Based on the previous calculation there is not enough ice to lower the temperature of the water to $0^{{\\circ}} \\mathrm{{C}}$. One way of doing the problem is to calculate the drop in temperature of the $3.0 \\mathrm{{~kg}}$ of water at $25^{{\\circ}} \\mathrm{{C}}$ due to the melting ice, then treat the problem as a mixture of two amounts of water at different temperatures.
        However, since we already know that all the ice melts and the final mixture is between $0^{{\\circ}} \\mathrm{{C}}$ and $25^{{\\circ}} \\mathrm{{C}}$, the equation tor the tinal temperature can be set up with the statement: The energy in the $3.0 \\mathrm{{~kg}}$ of water at $25^{{\\circ}} \\mathrm{{C}}$ Goesinto melting the ice and raising the temperature of the $0^{{\\circ}} \\mathrm{{C}}$ water.
        $3.0 \\mathrm{{~kg}}\\left(4200 \\mathrm{{~J}} / \\mathrm{{kg}} \\cdot \\mathrm{{C}}^o\\right)(25-T)^o \\mathrm{{C}}=0.50 \\mathrm{{~kg}}\\left(3.3 \\times 10^5 \\mathrm{{~J}} / \\mathrm{{kg}}\\right)+0.50 \\mathrm{{~kg}}\\left(4200 \\mathrm{{~J}} / \\mathrm{{kg}} \\cdot \\mathrm{{C}}^o\\right)(T-0)^{{\\circ}} \\mathrm{{C}}$
        $315000-12600 T=165000+2100 T \\quad \\text {{ or }} \\quad T=10^{{\\circ}} \\mathrm{{C}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_17_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An insulated container holds $3.0 \\mathrm{{~kg}}$ of water at $25^{{\\circ}} \\mathrm{{C}}$. One kilogram of ice at $0^{{\\circ}} \\mathrm{{C}}$ is added to the mixture. What is the final temperature and composition (ice and water) of the mixture?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First calculate the energy to bring all the water to $0^{{\\circ}} \\mathrm{{C}}$.
        $Q=3.0 \\mathrm{{~kg}}(4200 \\mathrm{{~J}} / \\mathrm{{kg}} \\cdot \\mathrm{{K}}) 25 \\mathrm{{~K}}=315000 \\mathrm{{~J}}$
        Now ask how much ice could be melted with this amount of energy.
        $315000 \\mathrm{{~J}}=\\left(3.3 \\times 10^5 \\mathrm{{~J}} / \\mathrm{{kg}}\\right) m \\quad \\text {{ or }} \\quad m=0.95 \\mathrm{{~kg}}$
        All the ice will not be melted so the final mixture will be $3.95 \\mathrm{{~kg}}$ of water and $0.05 \\mathrm{{~kg}}$ of ice all at $0^{{\\circ}} \\mathrm{{C}}$. There will be no further heat transfer because both the water and the ice are at the same temperature.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_17_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $150 \\mathrm{{~g}}$ cup of coffee (water) at $80^{{\\circ}} \\mathrm{{C}}$ has added to it $20 \\mathrm{{~g}}$ of sugar (carbon) at $25^{{\\circ}} \\mathrm{{C}}$. What is the final temperature of the insulated mixture? $\\left(c_c=0.12 \\mathrm{{cal}} / \\mathrm{{g}} \\cdot \\mathrm{{C}}^{{\\circ}}\\right)$

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        In this problem heat from the water raises the temperature of the carbon until they are both at the same temperature. Saying it simply, this is a Goesinto problem. The heat in the water Goesinto heating the carbon until they both reach an equilibrium temperature, $T$.
        The heat leaving the water is $150 \\mathrm{{~g}}\\left(1.0 \\mathrm{{cal}} / \\mathrm{{g}} . \\mathrm{{C}}^{{\\circ}}\\right)\\left(80-T^{{\\circ}} \\mathrm{{C}}\\right.$.
        The final temperature is between 80 and 25 , so writing the temperature difference as $80-T$ produces a positive number (of calories).
        This heat goes to raise the temperature of the carbon $20 \\mathrm{{~g}}\\left(0.12 \\mathrm{{cal}} / \\mathrm{{g}} . \\mathrm{{C}}^{{\\circ}}\\right)(T-25)^{{\\circ}} \\mathrm{{C}}$.
        Notice that the temperature difference in parentheses is written so as to produce a positive number.
        The equation will read: The heat from the water Goesinto heating the carbon.
        $150 \\mathrm{{~g}}\\left(1.0 \\mathrm{{cal}} / \\mathrm{{g}} \\cdot \\mathrm{{C}}^o\\right)(80-T)^o \\mathrm{{C}}=20 \\mathrm{{~g}}\\left(0.12 \\mathrm{{cal}} / \\mathrm{{g}} \\cdot \\mathrm{{C}}^o\\right)(T-25)^o \\mathrm{{C}}$
        or
        $12000-150 T=2.4 T-60 \\quad \\text {{ or }} \\quad T=79.1^{{\\circ}} \\mathrm{{C}}$
        These problems can be complicated with three or more components at different temperatures. If you guess wrong as to where the equilibrium temperature lies, the calculations will produce a temperature higher or lower than the original temperatures. If this occurs you know that the initial assumption was wrong, so go back with another (better) assumption of the equilibrium temperature and do the problem over again. The main problem with calorimetry problems is keeping the algebraic signs correct. These problems can be done with a positive heat minus negative heat approach, but the Goesinto approach cuts down on the number of negative signs and reduces the possible number of places where you can get a sign wrong.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_17_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        How much heat is required to raise a $1.8 \\mathrm{{~kg}}$ copper tea kettle containing $2.0 \\mathrm{{~kg}}$ of water from $20^{{\\circ}} \\mathrm{{C}}$ to $100^{{\\circ}} \\mathrm{{C}}$ ? The specific heat of copper is $0.092 \\mathrm{{cal}} / \\mathrm{{g}} . \\mathrm{{C}}^{{\\circ}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The total heat is
        $Q=\\left[m_{{\\mathrm{{cu}}}} c_{{\\mathrm{{cu}}}}+m_{{\\text {{wat }}}} c_{{\\text {{wat }}}}\\right] \\Delta I$
        $Q=\\left[1800 \\mathrm{{~g}}\\left(0.092 \\mathrm{{cal}} / \\mathrm{{g}} \\cdot \\mathrm{{C}}^o\\right)+2000 \\mathrm{{~g}}\\left(1.0 \\mathrm{{cal}} / \\mathrm{{g}} \\cdot \\mathrm{{C}}^o\\right)\\right] 80^{{\\circ}} \\mathrm{{C}}=1.73 \\times 10^5 \\mathrm{{cal}}$
        The specific heat is sometimes expressed in $\\mathrm{{J}} / \\mathrm{{kg}} . \\mathrm{{K}}$ rather than cal/g. $\\mathrm{{C}}^{{\\circ}}$. For the previous problem the calculations would read
        $Q=[1.8 \\mathrm{{~kg}}(390 \\mathrm{{~J}} / \\mathrm{{kg}} \\cdot \\mathrm{{K}})+2.0 \\mathrm{{~kg}}(4200 \\mathrm{{~J}} / \\mathrm{{kg}} \\cdot \\mathrm{{K}})] 80 \\mathrm{{~K}}=728000 \\mathrm{{~J}}$
        Use the calorie to joule equivalence to verify the numbers in this problem.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_17_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the fractional change in period for a brass pendulum going through a $5 \\mathrm{{C}}^{{\\circ}}$ change from summer to winter (not unusual).
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\displaystyle \\frac{{\\Delta t}}{{t}}=\\alpha \\frac{{\\Delta T}}{{2}}=19 \\times 10^{{-6}} \\frac{{1}}{{C^o}} \\frac{{5 C^o}}{{2}}=47.5 \\times 10^{{-6}}=0.0000475 \\text {{ or }} 0.00475 \\%$
        As the temperature increases, the length of the pendulum gets longer, and the time for one oscillation gets longer. Pendulum clocks run slower in the summer.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_17_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the change in volume of a brass sphere of $8.0 \\mathrm{{~cm}}$ in radius on going from $0^{{\\circ}} \\mathrm{{C}}$ to $100^{{\\circ}} \\mathrm{{C}}$. $\\left(\\alpha_{{\\text {{tras }}}}=19 \\times 10^{{-6}} / \\mathrm{{C}}^0\\right)$
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The change in volume is
        $\\Delta V-3 \\alpha V \\Delta T-3\\left(19 \\times 10^{{-6}} 1 / \\mathrm{{C}}^{{\\circ}}\\right)(4 / 3) \\pi(8.0 \\mathrm{{~cm}})^3 100 \\mathrm{{C}}^{{\\circ}}-12.2 \\mathrm{{~cm}}^3$
        The fractional change in volume $\\Delta V / V$ is
        $\\displaystyle \\frac{{\\Delta V}}{{V}}=\\frac{{3 \\alpha V \\Delta T}}{{V}}=3 \\alpha \\Delta T=3\\left(19 \\times 10^{{-6}} 1 / \\mathrm{{C}}^{{\\circ}}\\right) 100 \\mathrm{{C}}^{{\\circ}}=0.0057$
        The percentage change in volume is $0.57 \\%$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_17_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A steel bridge is set in concrete pillars on opposite sides of a river. The bridge is $300 \\mathrm{{~m}}$ long, and the work is performed on a day when the temperature is $18^{{\\circ}} \\mathrm{{C}}$. What clearance is required for the bridge not to buckle at $45^{{\\circ}} \\mathrm{{C}} ?\\left(\\alpha_{{\\text {{steel }}}}=11 \\times 10^{{-6}} / \\mathrm{{C}}^{{\\circ}}\\right)$
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The change in length is
        $\\Delta \\ell=\\alpha \\ell \\Delta T=\\left(11 \\times 10^{{-6}} 1 / \\mathrm{{C}}^\\rho\\right) 300 \\mathrm{{~m}}\\left(27 \\mathrm{{C}}^o\\right)=8.9 \\times 10^{{-2}} \\mathrm{{~m}}$
        A minimum of $8.9 \\mathrm{{~cm}}$ would be required for expansion.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_17_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Two objects have an initial temperature difference of $20 \\mathrm{{C}}^{{\\circ}}$. In two minutes the temperature difference is $18 \\mathrm{{C}}^{{\\circ}}$. Calculate the time for the temperature difference to be $10 \\mathrm{{C}}^{{\\circ}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This is a classic type of problem where the physical law is known, data is given to calculate the specific relationship, and this relationship is used to predict something. Keep this process in mind as you go through this problem. It is used in the solution of many problems in physics.
        The basic law is $\\Delta T=\\Delta T_{{0}} e^{{-k t}}$
        Now, use the given data to determine $k$. Units are left off in the exponent to avoid clutter, but keep in mind that the 2.0 is in minutes, so $k$ will be in reciprocal minutes.
        $\\displaystyle 18 \\mathrm{{C}}^o=\\left(20 \\mathrm{{C}}^o\\right) e^{{-2.0 k}} \\text {{ so }} \\ln \\frac{{18}}{{20}}=-2.0 k \\text {{ or }} k=-\\frac{{1}}{{2}} \\ln 0.90=0.053\\left(\\frac{{1}}{{\\min }}\\right)$
        This gives $k$ for this specific situation, and the law for this specific situation is
        $\\Delta T=\\Delta T_{{0}} e^{{-0.053 t}}$
        Now ask, what time is required for the temperature difference to decline to $10^{{\\circ}} \\mathrm{{C}}, \\Delta T=10^{{\\circ}} \\mathrm{{C}}$ for an initial difference of $20^{{\\circ}} \\mathrm{{C}}$.
        $10=20 \\mathrm{{e}}^{{-0.053 t}}$ and switching to logarithms $\\displaystyle \\ln \\frac{{1}}{{2}}=-0.053 \\text {{ or }} t=-\\frac{{1}}{{0.053}} \\ln \\frac{{1}}{{2}}=13 \\mathrm{{~min}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_17_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider a linear temperature device such as the resistance of a piece of metal. If the resistance of the device is $800 \\Omega$ at $20^{{\\circ}} \\mathrm{{C}}$ and $900 \\Omega$ at $60^{{\\circ}} \\mathrm{{C}}$. what is the temperature when the device reads $835 \\Omega$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-17-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        soln_s = f"""
        Figure 17-2 shows the linear relationship of the device. Since the slope is a constant, the slope of the graph on the right is equal to the slope of the graph on the left (similar triangles).
        So $\\displaystyle \\frac{{100 \\Omega}}{{40^{{\\circ}} \\mathrm{{C}}}}=\\frac{{35 \\Omega}}{{\\Delta T}}$ or $\\Delta T=14^{{\\circ}} \\mathrm{{C}}$ and the measured temperature is $34^{{\\circ}} \\mathrm{{C}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_16_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A cylindrical container of radius $20 \\mathrm{{~cm}}$ contains water to a height of $3.0 \\mathrm{{~m}}$. Two meters down from the water level there is a hole $0.50 \\mathrm{{~cm}}$ in radius. Find the velocity and volume flow rate of water leaving the hole. What is the shape of the stream, and where does it strike the ground?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Apply the Bernoulli equation $p_1+\\rho g y_1+\\frac{{1}}{{2}} \\rho v_1^2=p_2+\\rho g y_2+\\frac{{1}}{{2}} \\rho v_2^2$
        The pressure is the same: $p_1=p_2$. The velocity $v_1$ is so small compared to $v_2$ that it can be neglected: set $v_1=0$. The distance from the top of the water to the hole is $2.0 \\mathrm{{~m}}$.
        $g h=\\frac{{1}}{{2}} v_2^2, \\quad v_2=\\sqrt{{2 g h}}=\\sqrt{{2\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 2.0 \\mathrm{{~m}}}}=6.3 \\mathrm{{~m}} / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-16-10.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        The differential volume of liquid flowing out of the hole is $d V=A_2 v_2 d t$
        $\\displaystyle \\frac{{d V}}{{d t}}=A_2 v_2=A_2 \\sqrt{{2 g h}} \\quad \\text {{ and }} \\quad \\frac{{d V}}{{d t}}=\\pi\\left(0.5 \\times 10^{{-2}} \\mathrm{{~m}}\\right)^2 6.3 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}=4.9 \\times 10^{{-4}} \\frac{{\\mathrm{{m}}^3}}{{\\mathrm{{~s}}}}$
        To analyze the motion look at a differential piece of the stream. Set up a right-handed coordinate system with origin at the exit hole. This coordinate system is similar to ones used in some projectile problems. The differential piece of liquid has initial velocity in the $+y$ direction of $6.3 \\mathrm{{~m}} / \\mathrm{{s}}$ and is acted upon by gravity in the $+x$ direction.
        The six equations of motion for acceleration, velocity, and position in the $x$ and $y$ directions are
        $a_x=9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2 \\qquad a_y=0$
        $v_x=\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t \\qquad v_y=6.3 \\mathrm{{~m}} / \\mathrm{{s}}$
        $x=\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2 \\qquad y=(6.3 \\mathrm{{~m}} / \\mathrm{{s}}) t$
        These are the equations of a parabola or half of a parabola in this case.
        The differential piece of water falls $1.0 \\mathrm{{~m}}$ so set $x=1.0 \\mathrm{{~m}}$ in the equation for $x$
        $1.0 \\mathrm{{~m}}=\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2$ or $t=0.45 \\mathrm{{~s}}$
        This is the time for the differential piece of the stream to reach the ground.
        Then $y$ at $t=0.45 \\mathrm{{~s}}$ is $y \\mid t=0.45=2.8 \\mathrm{{~m}}$. The stream strikes the ground at $2.8 \\mathrm{{~m}}$ from the tank.
        The approximation of $v_1$ as zero is justified with the continuity equation, $A_1 v_1=A_2 v_2$.
        $\\displaystyle \\pi(20 \\mathrm{{~cm}})^2 v_1=\\pi(0.50 \\mathrm{{~cm}})^2 v_2 \\quad \\text {{ or }} \\quad v_1=\\left(\\frac{{0.50}}{{20}}\\right)^2 v_2=\\left(\\frac{{1}}{{40}}\\right)^2 v_2=\\frac{{1}}{{1600}} v_2$
        The $v_2$ is 1600 times $v_1$ !
        As a side exercise solve the equation $y=6.3 t$ for $t$ and substitute into the equation for $x$, obtaining an equation in the form $x=a y_2$, the equation of a parabola. Then substitute $x=1.0 \\mathrm{{~m}}$ to ohtain $y$.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_16_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation of problem $16-7$, find the volume flow rate, that is $d V / d t$, for a hole of radius $3.0 \\mathrm{{~mm}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The flow rate is the instantaneous rate of liquid flowing out of the container. This is most easily obtained from the statement of the differential volume.
        $\\displaystyle d V=A_1 d l=A_2 v_2 d t \\quad \\text {{ so that }} \\quad \\frac{{d V}}{{d t}}=A_2 v_2=A_2 \\sqrt{{2 g h}}$
        For a hole of radius $3.0 \\mathrm{{~mm}}$ and liquid depth $1.0 \\mathrm{{~m}}$
        $\\displaystyle \\frac{{d V}}{{d t}}=\\pi\\left(3.0 \\times 10^{{-3}} \\mathrm{{~m}}\\right)^2 4.4 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}=125 \\times 10^{{-6}} \\frac{{\\mathrm{{m}}^3}}{{\\mathrm{{~s}}}}=1.25 \\times 10^{{-4}} \\frac{{\\mathrm{{m}}^3}}{{\\mathrm{{~s}}}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_16_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the speed at which liquid flows out a small hole in the bottom of a large tank containing liquid to a depth of $1.0 \\mathrm{{~m}}$.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-16-9.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        soln_s = f"""
        Apply the Rernoulli equation $\\displaystyle p_1+\\rho g y_1+\\frac{{1}}{{2}} \\rho v_1^2=p_2+\\rho g y_2+\\frac{{1}}{{2}} \\rho v_2^2$
        Note that $p_1=p_z$, so remove these terms from the equation. Because $A_1 \\gg A_2, v_1$ is very small and is taken as zero. Take $y_1=0$, so $y_2$ is the depth of the liquid. The Bernoulli equation for this case reduces to
        $\\displaystyle \\rho g h=\\frac{{1}}{{2}} \\rho v_2^2 \\quad \\text {{ or }} \\quad v_2=\\sqrt{{2 g h}}$
        The velocity of liquid from a small hole in a container depends only on the height of the column of liquid.
        For a liquid height of $1.0 \\mathrm{{~m}}$,
        $v_2=\\sqrt{{2\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 1.0 \\mathrm{{~m}}}}=4.4 \\mathrm{{~m}} / \\mathrm{{s}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_16_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An iceberg floats in seawater. What percentage of the iceberg is submerged?
        $\\left(\\rho_{{\\text {{ice }}}}=0.92 \\times 10^{{-3}} \\mathrm{{~kg}} / \\mathrm{{m}}^3\\right)\\left(\\rho_{{S W}}=1.0 \\times 10^3 \\mathrm{{~kg}} / \\mathrm{{m}}^3\\right)$
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\Lambda$ ssume a cubical iceberg of side $s$. You can assume any shape you wish. Cubical, however, seems convenient. The mass of the cubical iceberg is $pice_{{s3}}$, and the weight is $pice_{{s3}}g$. Since the density of the iceberg is less than the density of the seawater the iceberg will float. The buoyant force of the seawater is $p S W g$ times the volume submerged. Since the iceberg is cubical the submerged volume must be $s_2$, the area of the bottom of the iceberg, times a fraction of the side $k$ times $s$.
        This buoyant force must equal the weight of the iceberg. $p i c e_{{s3}} g=p S w g_{{s2}}(\\mathrm{{Ks}})$.
        So K, the fraction of the iceberg submerged is $\\displaystyle K=\\frac{{\\rho_{{\\text {{ice }}}}}}{{\\rho_{{\\mathrm{{SW}}}}}}=\\frac{{0.92}}{{1.0}}=0.92$
        Thus $92 \\%$ of the iceberg is submerged.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_16_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A rectangular-shaped, open-top steel barge is $10 \\mathrm{{~m}}$ by $3.0 \\mathrm{{~m}}$ and has sides $1.0 \\mathrm{{~m}}$ high. The mass of the barge is $8.0 \\times 10^3 \\mathrm{{~kg}}$. Ilow much mass can the barge hold if it can safely sink $0.75 \\mathrm{{~m}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The volume of water displaced is length times width times alluwed depth, $10 \\mathrm{{~m}} \\times 3.0 \\mathrm{{~m}} \\times 0.75 \\mathrm{{~m}}=22.5 \\mathrm{{~m}}^3$. The buoyant force is the weight of the water displaced (Archimedes' principle)
        $F_B=\\rho g V=1000 \\mathrm{{~kg}} / \\mathrm{{m}}^3\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 22.5 \\mathrm{{~m}}^3=2.2 \\times 10^5 \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        file_name = 'how-to-solve-physics-problems-fig-16-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        This buoyant force must support the barge $\\left(m_B\\right)$ plus the load $\\left(m_L\\right.$
        $2.2 \\times 10^5 \\mathrm{{~N}}=g\\left(m_{{\\mathrm{{B}}}}+m_{{\\mathrm{{L}}}}\\right), \\quad m_{{\\mathrm{{B}}}}+m_{{\\mathrm{{L}}}}=22 \\times 10^3 \\mathrm{{~kg}}, \\quad$ and $\\quad m_{{\\mathrm{{L}}}}=14 \\times 10^3 \\mathrm{{~kg}}$
        As an additional exercise calculate how deep the unloaded barge will sink in the water.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_16_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A U-tube originally containing mercury has water added to one arm to a depth of $20 \\mathrm{{~cm}}$. What is the pressure at the water-mercury interface? What is the height of the mercury column as measured from the watermercury level?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First Solution: Assume that the difference in air pressure can be neglected. The pressure at the watermercury interface is duc to the column of water plus the atmospheric pressure $\\rho_0$, or $p=p_0+p w g .20 \\mathrm{{~cm}}$. Neglecting $p_0$ 。 gives the gauge pressure as
        $p=1000 \\mathrm{{~kg}} / \\mathrm{{m}}^3\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 20 \\times 10^{{-2}} \\mathrm{{~m}}=1960 \\mathrm{{~Pa}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-16-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        At this level (the level of the interface) the pressure on the mercury arm must be the same. (If it were not the liquids would move!) The height of the mercury level is
        $\\displaystyle 1960 \\mathrm{{~Pa}}=\\rho_{{\\mathrm{{M}}}} g h_{{\\mathrm{{M}}}} \\quad$ or $\\displaystyle \\quad h_{{\\mathrm{{M}}}}=\\frac{{1960 \\mathrm{{~N}} / \\mathrm{{m}}^2}}{{14 \\times 10^3 \\mathrm{{~kg}} / \\mathrm{{m}}^3\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)}}=1.4 \\mathrm{{~cm}}$
        As we expect, the same pressure is produced by a much shorter column of mercury.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        soln_s = f"""
        Second Solution:Analytically the problem can be approached by writing expressions for the pressure at the level of the water-mercury interface.
        In the left arm $p W=p_0+p W g h W$, where the subscript $\\mathrm{{W}}$ refers to the water column.
        In the right arm the pressure is $p M=p_0+p M g h M$, where the subscript $\\mathrm{{M}}$ refers to the mercury column.
        Since these pressures must be equal at the water-mercury interface
        $\\rho_{{\\mathrm{{W}}}} g h_{{\\mathrm{{W}}}}+p_o=\\rho_{{\\mathrm{{M}}}} g h_{{\\mathrm{{M}}}}+p_o \\quad$ or $\\quad \\rho_{{\\mathrm{{W}}}} h_{{\\mathrm{{W}}}}=\\rho_{{\\mathrm{{M}}}} h_{{\\mathrm{{M}}}}$
        For this problem $\\displaystyle h_{{\\mathrm{{M}}}}=h_W \\frac{{\\rho_{{\\mathrm{{W}}}}}}{{\\rho_{{\\mathrm{{M}}}}}}=20 \\mathrm{{~cm}} \\frac{{1}}{{14}}=1.4 \\mathrm{{~cm}}$
        Practice variations of this problem with different liquids (different densities).
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_16_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The classic problem in buoyancy is to determine if a "gold" statue is really gold by measuring its weight in and out of water. A certain "gold" statue weighs $70 \\mathrm{{~N}}$ when out of water and $64 \\mathrm{{~N}}$ when immersed in water.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        With the statue suspended from a scale the weight is $m g$, or density times $g$, times the volume of the statue. In equation form $m g=\\rho ? g V=70 \\mathrm{{~N}}$.
        In water the statue weighs less because of the buoyant force: $\\rho$ ? $g V-\\rho w g V=64 \\mathrm{{~N}}$
        Subtracting one equation from the other we can determine the volume from
        $\\rho_{{\\mathrm{{W}}}} g V=6 \\mathrm{{~N}}, \\quad 1000 \\mathrm{{~kg}} / \\mathrm{{m}}^3\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) V=6 \\mathrm{{~N}}, \\quad V=6.1 \\times 10^{{-4}} \\mathrm{{~m}}^3$

        With this value for $V$, calculate the unknown density from the first equation
        $\\displaystyle \\rho_{{!}} g V=70 \\mathrm{{~N}}, \\quad \\rho ?=\\frac{{70 \\mathrm{{~N}}}}{{6.1 \\times 10^{{-4}} \\mathrm{{~m}}^3\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)}}=12 \\times 10^3 \\frac{{\\mathrm{{kg}}}}{{\\mathrm{{m}}^3}}$
        This density is much less than the density of gold $\\rho_{{\\text {{gold }}}}=19 \\times 10^3 \\mathrm{{~kg}} / \\mathrm{{m}}^3$ ) and closer to the density of lead ( $\\rho_{{\\text {{lead }}}}=$ $11 \\times 10^3 \\mathrm{{~kg}} / \\mathrm{{m}}^3$ ) suggesting that the statue is lead with a thin covering of gold.

        This same experiment can be conducted by lowering the statue into a beaker of water on a scale. The volume is determined by the increase in the level of water in the beaker. The drop in tension in the cord is offset by the increase in the scale reading when the statue is lowered into the beaker and the calculations proceed as before.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_16_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider the classic problem of a hydraulic lift. In a typical service station $12 \\mathrm{{~atm}}$ is applied to the small area with the lifting column $9.0 \\mathrm{{~cm}}$ in radius. Find the force transferred to the large area and the mass of vehicle that can be lifted.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-16-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        soln_s = f"""
        The hydraulic fluid is at the same level so $\\rho_1=\\rho_2$. or
        $\\displaystyle \\frac{{F_1}}{{A_1}}=\\frac{{F_2}}{{A_2}}$
        A force $F_1$ applied at $A_1$ is multiplied by the ratio of the arcas so $F_2=\\left(A_2 / A_1\\right) F_1$
        The lifting force $F_2$ can alsu be rewritlen as $F_2=A_2\\left(F_1 / A_1=A_2 \\mu_1\\right.$, and pultiing in the numbers
        $\\displaystyle F_2=\\pi(9.0 \\mathrm{{~cm}})^2 12 \\mathrm{{~atm}} \\frac{{1.0 \\times 10^5 \\mathrm{{~N}} / \\mathrm{{m}}^2}}{{\\mathrm{{~atm}}}}\\left(\\frac{{\\mathrm{{m}}}}{{100 \\mathrm{{~cm}}}}\\right)^2=3.0 \\times 10^4 \\mathrm{{~N}}$
        And the equivalent mass (from $F=m g$ ) is $3.1 \\times 10^3 \\mathrm{{~kg}}$. This number is somewhat larger than the mass of typical cars.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_16_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the pressure at the bottom of a $3.0 \\mathrm{{~m}}$ deep pool.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $\\Delta p=\\rho g h=1000 \\mathrm{{~kg}} / \\mathrm{{m}}^3\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 3.0 \\mathrm{{~m}}=2.9 \\times 10^4 \\mathrm{{~Pa}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_15_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the restoring constant (torsional constant) for a disk of mass $3.0 \\mathrm{{~kg}}$ and radius $0.20 \\mathrm{{~m}}$ oscillating with a period of $2.0 \\mathrm{{~s}}$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The moment of inertia for a disk is
        $\\displaystyle I=\\frac{{m r^2}}{{2}}=\\frac{{3.0 \\mathrm{{~kg}}(0.20 \\mathrm{{~m}})^2}}{{2}}=0.060 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2$
        The restoring constant is
        $\\displaystyle \\kappa=\\frac{{4 \\pi^2 l}}{{T^2}}=\\frac{{4 \\pi^2 \\cdot 0.060 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2}}{{4.0 \\mathrm{{~s}}^2}}=0.59 \\mathrm{{~N}} \\cdot \\mathrm{{m}}$
        Note that these are the correct units since torque has the dimensions N.m, and in Torque $=-k \\theta$ the angle is in (dimensionless) radians.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_15_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $0.20 \\mathrm{{~kg}}$ block traveling at $20 \\mathrm{{~m}} / \\mathrm{{s}}$ slides into and sticks to an $0.80 \\mathrm{{~kg}}$ block resting on a frictionless surface and connected to a spring with force constant $80 \\mathrm{{~N}} / \\mathrm{{m}}$. What is the angular frequency, frequency, and displacement as a function of time? Also what fraction of the original energy in the moving block appears in the system?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The collision is inelastic; the blocks stick together, so applying conservation of momentum $0.20 \\mathrm{{~kg}}(20 \\mathrm{{~m}} / \\mathrm{{s}})=(1.0 \\mathrm{{~kg}}) V$. The velocity of the $1.0 \\mathrm{{~kg}}$ combination is $V=4.0 \\mathrm{{~m}} / \\mathrm{{s}}$. This is the velocity of the system at zero displacement. The analysis of this collision is similar to the ballistic pendulum problem in the chapter on collisions.
        The total energy of the oscillating system is
        $E=\\frac{{m V^2}}{{2}}=\\frac{{1.0 \\mathrm{{~kg}}}}{{2}} \\frac{{16 \\mathrm{{~m}}^2}}{{\\mathrm{{~s}}^2}}=8.0 \\mathrm{{~J}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-15-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        $\\omega=\\sqrt{{\\frac{{k}}{{m}}}}=\\sqrt{{\\frac{{80 \\mathrm{{~N}} / \\mathrm{{m}}}}{{1.0 \\mathrm{{~kg}}}}}}=8.9 \\frac{{1}}{{\\mathrm{{~s}}}}$
        The angular frequency is
        The frequency is from $\\omega=2 \\pi f$ or $f=\\omega / 2 \\pi=1.4(1 / \\mathrm{{s}})$
        The general expression for the displacement is $y=A \\sin \\omega t$. The sine function without a phase angle completely describes the motion because it starts $(t=0)$ at the equilibrium position with maximum velocity. The amplitude of the motion is obtained from
        $k A^2 / 2=E \\quad \\text {{ or }} \\quad A=\\sqrt{{\\frac{{2 E}}{{k}}}}=\\sqrt{{\\frac{{2 \\cdot 8.0 \\mathrm{{~J}}}}{{80 \\mathrm{{~N}} / \\mathrm{{m}}}}}}=0.45 \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_15_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For a $3.0 \\mathrm{{~kg}}$ mass-spring system executing simple harmonic motion according to $y=(4.0 \\mathrm{{~cm}}) \\cos (\\pi t / 4-\\pi / 6)$, what is the total energy and the $y$ position for equal division of the energy between kinetic and potential?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The general expression for displacement is $y=A \\cos (a t+\\theta)$. Ionking at the two expressions,
        $\\displaystyle \\omega=\\sqrt{{\\frac{{k}}{{m}}}}=\\frac{{\\pi}}{{4}} \\frac{{1}}{{\\mathrm{{~s}}}} \\text {{ so }} k=\\omega^2 m=\\frac{{\\pi^2}}{{4^2}} \\frac{{1}}{{\\mathrm{{~s}}^2}} 3.0 \\mathrm{{~kg}}\\left(\\frac{{\\mathrm{{m}}}}{{\\mathrm{{m}}}}\\right)=1.85 \\frac{{\\mathrm{{N}}}}{{\\mathrm{{m}}}}$
        Knowing $k$, the total energy is $\\displaystyle E=\\frac{{k A^2}}{{2}}=\\frac{{1.85 \\mathrm{{~N}}}}{{\\mathrm{{~m}}}}\\left(16 \\times 10^{{-4}} \\mathrm{{~m}}^2\\right)=29.6 \\times 10^{{-4}} \\mathrm{{~J}}$
        The potential energy due to position is $k y^2 / 2$ and $y=A \\cos (\\omega t+\\theta)$ so that the energy due to position is $k A^2 \\cos ^2(\\omega t+\\theta) / 2$. The energy due to position is $1 / 2$ the total when the factor
        $\\cos ^2(\\omega t+\\vartheta)=1 / 2 \\text {{ or }} \\cos (\\omega t+\\vartheta)=1 / \\sqrt{{2}}$
        For this particular system, the time for equal division of the energy is when
        $\\cos \\left(\\frac{{\\pi}}{{4}} t-\\frac{{\\pi}}{{6}}\\right)=\\frac{{1}}{{\\sqrt{{2}}}}, \\quad \\frac{{\\pi}}{{4}} t-\\frac{{\\pi}}{{6}}=\\cos ^{{-1}} \\frac{{1}}{{\\sqrt{{2}}}}, \\quad \\frac{{\\pi}}{{4}} t - \\frac{{\\pi}}{{6}}=\\frac{{\\pi}}{{4}}, \\quad \\frac{{\\pi}}{{4}} t=\\frac{{5 \\pi}}{{12}}, \\quad \\text {{ or }} \\quad t=\\frac{{5}}{{3}}\\mathrm{{~s}}$
        The position of the mass for equal division of the energy is
        $\\left.y\\right|_{{E / 2}}=(4.0 \\mathrm{{~cm}}) \\cos \\left(\\frac{{\\pi}}{{4}} \\frac{{5}}{{3}}-\\frac{{\\pi}}{{6}}\\right)=(4.0 \\mathrm{{~cm}}) \\cos \\left(\\frac{{3 \\pi}}{{12}}\\right)=2.83 \\mathrm{{~cm}}$
        As an exercise, calculate the velocity at this time.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_15_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Find the amplitude, angular frequency, and frequency for a mass-spring system with mass $2.0 \\mathrm{{~kg}}$ that is oscillating according to $x=(2.0 \\mathrm{{~m}}) \\cos (6 \\pi / / \\mathrm{{s}})$. Write and graph the expressions for position, velocity, and acceleration. Finally, calculate the total energy of the system.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The amplitude, angular frequency, and frequency are determined by comparing $x=(2.0 \\mathrm{{~m}}) \\cos (6 \\pi / \\mathrm{{s}})$ to the standard form $x=A \\cos \\omega t=A \\cos 2 \\pi f t$.
        The amplitude is $2.0 \\mathrm{{~m}}$. The angular frequency is $6 \\pi(1 / \\mathrm{{s}})$. The frequency is $3(1 / \\mathrm{{s}})$.
        The expressions for position, velocity (first derivative), and acceleration (second derivative) are shown in Fig. 15-5 associated with the graphs.
        In Fig. 15-5, the units are dropped in the argument of trigonometric functions, and the maximum values for the velocity and acceleration are placed on the graphs.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-15-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        The conservation of energy statement is $\\displaystyle \\frac{{m v^2}}{{2}}+\\frac{{k x^2}}{{2}}=E$ For a 2.0kg mass the total energy can be evaluated when $x$ is zero and the velocity maximum.
        $\\displaystyle E=\\frac{{m v_{{\\max }}^2}}{{2}}=\\frac{{2.0 \\mathrm{{~kg}}(12 \\pi \\mathrm{{m}} / \\mathrm{{s}})^2}}{{2}}=1420 \\mathrm{{~J}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_15_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Find the maximum velocity of a mass-spring system with mass $2.0 \\mathrm{{~kg}}$, spring constant $0.80 \\mathrm{{~N}} / \\mathrm{{m}}$, and amplitude of oscillation $0.36 \\mathrm{{~m}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        All the energy of the system is stored in the spring at maximum displacement so the total energy is
        $\\displaystyle E=\\frac{{k x_{{\\max }}^2}}{{2}}=\\frac{{0.80 \\mathrm{{~N}}}}{{\\mathrm{{~m}}}} \\frac{{(0.36 \\mathrm{{~m}})^2}}{{2}}=0.052 \\mathrm{{~J}}$
        This is the total energy in the system, and at zero displacement all this energy is in kinetic energy so
        $\\displaystyle \\frac{{m v_{{\\max }}^2}}{{2}}=0.052 \\mathrm{{~J}} \\quad \\text {{ or }} \\quad v_{{\\max }}=\\left[\\frac{{2 \\cdot 0.052 \\mathrm{{~J}}}}{{2.0 \\mathrm{{~kg}}}}\\right]^{{1 / 2}}=0.23 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_15_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An oscillating mass-spring system has displacement $10 \\mathrm{{~cm}}$, velocity $-12 \\mathrm{{~m}} / \\mathrm{{s}}$, and acceleration $-20 \\mathrm{{~m}} / \\mathrm{{s}}^2$. What is the period of the system?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Since the displacement is not zero at $t=0$. the motion cannot be described with a sin $\\omega t$ function. Since the velocity is not zero at $t=0$, the motion in not at an extreme, and cannot be described with a cos $\\omega t$ function. Because of the initial conditions, a sine or cosine function of $\\omega t$ plus a phase angle must be used. It makes no difference whether a sine or cosine function is used since the phase angle can absorb the $90^{{\\circ}}$ difference between a sine and a cosine function. Look at Fig. 15-4 and go over the logic in this paragraph until it is clear in your mind.
        Take $x=A \\cos (\\omega t+\\theta)$ then $v=-A \\omega \\sin (\\omega t+\\theta)$ and $a=-A \\omega^2 \\cos (\\omega t+\\theta)$
        Using the equations for $x$ and $a$, at $t=0$,
        $\\displaystyle \\omega^2=-\\frac{{a(0)}}{{x(0)}} \\text {{ and }} \\omega=\\sqrt{{\\frac{{20 \\mathrm{{~m}} / \\mathrm{{s}}^2}}{{0.10 \\mathrm{{~m}}}}}}=14 \\frac{{1}}{{\\mathrm{{~s}}}}$
        The period is from
        $\\displaystyle \\omega=\\frac{{2 \\pi}}{{T}} \\quad$ or $\\quad T-\\frac{{2 \\pi}}{{\\omega}}-\\frac{{2 \\pi}}{{\\sqrt{{200}}}} \\mathrm{{~s}}=0.44 \\mathrm{{~s}}$
        The phase angle comes from $v(0)$ and $a(0)$.
        $\\frac{{v(0)}}{{a(0)}}=\\frac{{\\sin \\vartheta}}{{\\omega \\cos \\vartheta}}=\\frac{{1}}{{\\omega}} \\tan \\vartheta \\quad \\text {{ or }} \\quad \\tan \\vartheta=\\omega \\frac{{v(0)}}{{a(0)}}=\\sqrt{{200}} \\frac{{1}}{{\\mathrm{{~s}}}} \\frac{{-12 \\mathrm{{~m}} / \\mathrm{{s}}}}{{-20 \\mathrm{{~m}} / \\mathrm{{s}}^2}}=8.48 \\quad \\text {{ so }} \\vartheta=83^0$
        The specific functional relations are
        $x=A \\cos \\left(\\omega t+83^{{\\circ}}\\right), \\quad v=-A \\omega \\sin \\left(\\omega t+83^{{\\circ}}\\right) \\text {{ and }} a=-A \\omega^2 \\cos \\left(\\omega t+83^{{\\circ}}\\right)$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        file_name = 'how-to-solve-physics-problems-fig-15-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        Look at these relations and relate them to what the system is doing at $t=0$. The amplitude is past its maximum positive excursion and on the way down corresponding to the cosine function at $83^{{\\circ}}$ The velocity is negative and heading toward its most negative value as is the negative sine function at $83^{{\\circ}}$. The acceleration is negative and tending toward zero just as is the negative cosine function at $83^{{\\circ}}$.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_15_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An $0.80 \\mathrm{{~kg}}$ mass hangs from a spring. When an additional $0.20 \\mathrm{{~kg}}$ mass is added the spring elongates another $3.0 \\mathrm{{~cm}}$ What is the period of oscillation of the spring?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First determine the force constant of the spring from $F=-k x$
        $\\displaystyle 0.20 \\mathrm{{~kg}} \\frac{{9.8 \\mathrm{{~m}}}}{{\\mathrm{{~s}}^2}}=k \\cdot 3.0 \\times 10^{{-2}} \\mathrm{{~m}} \\quad$ or $\\displaystyle \\quad k=\\frac{{0.20 \\mathrm{{~kg}} \\cdot 9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2}}{{0.03 \\mathrm{{~m}}}}=65 \\frac{{\\mathrm{{N}}}}{{\\mathrm{{m}}}}$
        $\\displaystyle \\omega=\\sqrt{{\\frac{{k}}{{m}}}}=2 \\pi f=\\frac{{2 \\pi}}{{T}}$
        The period is the reciprocal of the frequency
        $\\displaystyle T=2 \\pi \\sqrt{{\\frac{{m}}{{k}}}}=2 \\pi \\sqrt{{\\frac{{0.80 \\mathrm{{~kg}}}}{{65 \\mathrm{{~N}} \\mathrm{{~m}}}}}}=0.70 \\mathrm{{~s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_15_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A mass-spring system has a mass of $0.50 \\mathrm{{~kg}}$ and period of $1.0 \\mathrm{{~s}}$. The maximum velocity is $0.20 \\mathrm{{~m}} / \\mathrm{{s}}$. Find the frequency, angular frequency, constant of the spring, and displacement for this velocity and period.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The frequency is the reciprocal of the period or $f=1.0(1 / \\mathrm{{s}})$, and the angular frequency is $2 \\pi f$ or $2 \\pi(1 / \\mathrm{{s}})$. The spring constant comes from the relation
        $\\displaystyle \\omega=\\sqrt{{\\frac{{k}}{{m}}}} \\quad \\text {{ or }} \\quad k=\\omega^2 m=[2 \\pi(1 / \\mathrm{{s}})]^2 0.50 \\mathrm{{~kg}}=20 \\frac{{\\mathrm{{kg}}}}{{\\mathrm{{s}}^2}}\\left(\\frac{{\\mathrm{{m}}}}{{\\mathrm{{m}}}}\\right)=20 \\frac{{\\mathrm{{N}}}}{{\\mathrm{{m}}}}$
        The general expression for velocity is $v=A \\omega \\cos \\omega t$ where $A \\omega$ is the maximum velocity. It makes no difference whether the function is a sine or cosine. The maximum velocity is still $A \\omega$. The amplitude for this particular motion then is
        $\\displaystyle A=\\frac{{v}}{{\\omega}}=\\frac{{0.20 \\mathrm{{~m}} / \\mathrm{{s}}}}{{2 \\pi(1 / \\mathrm{{s}})}}=0.032 \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_15_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A piece of metal rests on top of a piston executing simple harmonic motion in the vertical plane with an amplitude of $0.20 \\mathrm{{~m}}$. At what frequency will the piece of metal "tloat" at the top of the cycle?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Take the motion to be described by $y=(0.20 \\mathrm{{~m}}) \\cos \\omega t$.
        Two derivatives produces the acceleration $a=-(0.20 \\mathrm{{~m}}) \\omega^2 \\cos \\omega$.
        The maximum acceleration, which is the constant $(0.20 \\mathrm{{~m}}) \\omega^2$, is at the extreme of the motion where the cosine term is one. $\\Lambda \\mathrm{{t}}$ the bottom of the cycle, the piece of metal will experience an acceleration greater than $g$ and at the top an acceleration less than $g$. The piece will "float" when that acceleration is greater than $g$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-15-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        $\\displaystyle (0.20 \\mathrm{{~m}}) \\omega^2=g=9.8 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}^2}} \\quad$ or $\\quad \\omega=\\sqrt{{\\frac{{9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2}}{{0.20 \\mathrm{{~m}}}}}}=7.0 \\frac{{1}}{{\\mathrm{{~s}}}}$
        or $\\displaystyle f=\\frac{{1}}{{2 \\pi}} \\sqrt{{\\frac{{9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2}}{{0.20 \\mathrm{{~m}}}}}}=1.1 \\frac{{1}}{{\\mathrm{{~s}}}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_15_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A mass-spring system has mass $2.0 \\mathrm{{~kg}}$, spring with constant $3.0 \\mathrm{{~N}} / \\mathrm{{m}}$, and amplitude of vibration $0.10 \\mathrm{{~m}}$. Calculate the angular frequency and frequency. Write expressions for the position, velocity, and acceleration as a function of time. Take $t=0$ at the equilibrium position. Graph position, velocity, and acceleration as a function of time.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The angular frequency is
        $\\displaystyle \\omega=\\sqrt{{\\frac{{k}}{{m}}}}=\\sqrt{{\\frac{{3.0 \\mathrm{{~N}}}}{{2.0 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}}}}}=\\sqrt{{\\frac{{3}}{{2}}}} \\frac{{1}}{{\\mathrm{{~s}}}}=1.2 \\frac{{1}}{{\\mathrm{{~s}}}}$
        The frequency is
        $\\displaystyle f=\\frac{{1}}{{2 \\pi}} \\sqrt{{\\frac{{k}}{{m}}}}=\\frac{{1}}{{2 \\pi}} \\sqrt{{\\frac{{3.0 \\mathrm{{~N}}}}{{2.0 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}}}}}=\\frac{{1}}{{2 \\pi}} \\sqrt{{\\frac{{3}}{{2}} \\frac{{1}}{{\\mathrm{{~s}}}}}}=0.19 \\frac{{1}}{{\\mathrm{{~s}}}}$
        If $t=0$ at the equilibrium position, then the displacement must be given by $x=B \\sin \\omega t$.
        At $t=0$ the sine function is zero, so this is the appropriate function to describe the motion. For an amplitude of $0.10 \\mathrm{{~m}}$ this expression should read
        $\\displaystyle x=(0.10 \\mathrm{{~m}}) \\sin \\sqrt{{\\frac{{3}}{{2}}}} \\frac{{1}}{{\\mathrm{{~s}}}} t \\quad$ or $\\quad x=(0.10 \\mathrm{{~m}}) \\sin 2 \\pi 0.19(1 / \\mathrm{{s}}) t$ 
        The velocity is the time derivative of $x$ or $\\displaystyle v=0.10 \\sqrt{{\\frac{{3}}{{2}}}} \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}} \\cos \\sqrt{{\\frac{{3}}{{2}} t}}$
        At $t=0$ the velocity, as the mass passes through the equilibrium position, is a maximum. The acceleration is another time derivative of the velocity or
        $\\displaystyle a=-0.10 \\frac{{3}}{{2}} \\frac{{\\mathrm{{m}}}}{{2}} \\sin \\sqrt{{\\frac{{3}}{{2}}}} t$
        At $t=0$ the acceleration is zero; the spring is neither elongated nor compressed (at the equilibrium position), so the force (and acceleration) of the mass is zero.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-15-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        These three curves graphed one under the other in Fig. 15-2 give a good picture of the motion from the position, velocity, and acceleration point of view.

        Look at these graphs and visualize the motion. At the equilibrium position, displacement is zero, the velocity is a maximum, and acceleration is zero. As the mass moves to the positive maximum displacement, the velocity decreases to zero and the acceleration (down) increases to a maximum. At the maximum displacement, the velocity is zero and the acceleration is a maximum. As the mass starts down, the velocity is negative reaching maximum again as the mass passes through equilibrium, Go over this picture of the motion until the relationship between position, velocity, and acceleration is clear in your mind.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_14_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The orbit radius of the satellite of problem $14-4$ is $6.8 \\times 10^6 \\mathrm{{~m}}$, and the period is $93 \\mathrm{{~min}}$. The orbit radius of the moon is $3.8 \\times 10^8 \\mathrm{{~m}}$, and the period is 27.3 days. Is this consistent with Kepler's third law?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Kepler's third law states that $T$ is a constant times the $3 / 2$ power of the radius, so
        $T_m=k r_m^{{3 / 2}}$ and $T_s=k r_s^{{3 / 2}}$ If the constant is the same then $\\displaystyle \\frac{{T_m}}{{T_s}}=\\left(\\frac{{r_m}}{{r_s}}\\right)^{{3 / 2}}$
        Putting in the numbers
        $\\displaystyle \\frac{{T_m}}{{T_s}}=\\frac{{27.3 \\mathrm{{day}}}}{{93 \\mathrm{{~min}}}} \\frac{{24 \\mathrm{{hr}}}}{{\\text {{ day }}}} \\frac{{60 \\mathrm{{~min}}}}{{\\mathrm{{hr}}}}=420 \\text {{ and }}\\left(\\frac{{r_m}}{{r_s}}\\right)^{{3 / 2}}=\\left(\\frac{{3.8 \\times 10^8 \\mathrm{{~m}}}}{{6.8 \\times 10^6 \\mathrm{{~m}}}}\\right)^{{3 / 2}}=420$
        The satellite and moon orbits verify Kepler's third law for circular orbits.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_14_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the speed, period, and radial acceleration of a satellite placed in orbit $400 \\mathrm{{~km}}$ above the eartl
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The orbit radius is $6.4 \\times 10^6 \\mathrm{{~m}}+0.4 \\times 10^6 \\mathrm{{~m}}=6.8 \\times 10^6 \\mathrm{{~m}}$
        The velocity is
        $\\displaystyle v=\\sqrt{{\\frac{{G m_E}}{{r}}}}=\\left[\\frac{{6.7 \\times 10^{{-11}} \\mathrm{{~N}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{kg}}^2\\left(6.0 \\times 10^{{24}} \\mathrm{{~kg}}\\right)}}{{6.8 \\times 10^6 \\mathrm{{~m}}}}\\right]^{{1 / 2}}=7690 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}$
        The period is
        $\\displaystyle T=2 \\pi \\sqrt{{\\frac{{r^3}}{{G m_E}}}}-2 \\pi\\left[\\frac{{\\left(6.8 \\times 10^6 \\mathrm{{~m}}\\right)^3}}{{6.7 \\times 10^{{-11}} \\mathrm{{~N}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{kg}}^2\\left(6.0 \\times 10^{{24}} \\mathrm{{~kg}}\\right)}}\\right]^{{1 / 2}}=93 \\mathrm{{~min}}$
        The radial acceleration is
        $\\displaystyle a_{{r a d}}=\\frac{{v^2}}{{r}}=\\frac{{G m_E}}{{r^2}}=\\frac{{6.7 \\times 10^{{-11}} \\mathrm{{~N}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{kg}}^2\\left(6.0 \\times 10^{{24}} \\mathrm{{~kg}}\\right)}}{{\\left(6.8 \\times 10^6 \\mathrm{{~m}}\\right)^2}}=8.7 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}^2}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_14_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What initial vertical speed is necessary to shoot a satellite to $300 \\mathrm{{~km}}$ above the earth?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Use energy analysis and the same energy-balance equation as in the previous problem with the velocity at $300 \\mathrm{{~km}}$ equal to zero.
        $\\displaystyle \\frac{{1}}{{2}} m v^2-\\frac{{G m m_E}}{{r_E}}=-\\frac{{G m m_E}}{{r_E+3.0 \\times 10^5 \\mathrm{{~m}}}} \\quad \\text {{ or }} \\quad v^2=2 G m_E\\left[\\frac{{1}}{{r_E}}-\\frac{{1}}{{r_E+3.0 \\times 10^5 \\mathrm{{~m}}}}\\right]$
        Putting in the numbers
        $v^2=2 \\cdot 6.7 \\times 10^{{-11}} \\frac{{\\mathrm{{N}} \\cdot \\mathrm{{m}}^2}}{{\\mathrm{{~kg}}^2}}\\left(6.0 \\times 10^{{24}} \\mathrm{{~kg}}\\right)\\left[\\frac{{1}}{{6.4 \\times 10^6 \\mathrm{{~m}}}}-\\frac{{1}}{{6.7 \\times 10^6 \\mathrm{{~m}}}}\\right]$
        or
        $\\displaystyle v=2.4 \\times 10^3 \\mathrm{{~m}} / \\mathrm{{s}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_14_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the escape velocity on the surface of the earth.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The escape velocity is the minimum velocity for a mass to escape the gravitational attraction of the earth. Envision a mass being shot vertically up from the surface of the earth and write the energy statement for it at two levels, the surface of the earth and some height $r$ further out from the center of the earth. The conservation of energy statement will read: The kinetic plus potential energy on the surface of the earth equals the kinetic plus potential energy at some point $r$.
        $\\displaystyle \\frac{{1}}{{2}} m v_1^2-G \\frac{{m m_E}}{{r_E}}=\\frac{{1}}{{2}} m_2 v_2^2-G \\frac{{m m_E}}{{r}}$
        The escape velocity is that velocity that will produce zero velocity $\\left(v_2=0\\right)$ when the mass is infinitely far away (when $\\mathrm{{r}}=\\infty, G m m_{{\\varepsilon}} / r=0$ ). This reduces the equation to
        $\\displaystyle \\frac{{v_1^2}}{{2}}=\\frac{{G m_E}}{{r_E}} \\quad \\text {{ or }} \\quad v_1=\\sqrt{{\\frac{{2 G m_E}}{{r_E}}}}$
        The mass and radius of the earth are on the constants page of your text so
        $\\displaystyle v_1=\\sqrt{{\\frac{{2\\left(6.7 \\times 10^{{-11}} \\mathrm{{~N}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{kg}}^2\\right) 6.0 \\times 10^{{24}} \\mathrm{{~kg}}}}{{6.4 \\times 10^6 \\mathrm{{~m}}}}}}=1.1 \\times 10^4 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}$
        As an exercise calculate the escape velocity for the moon of radius $1.7 \\times 10^6 \\mathrm{{~m}}$ and mass $7.4 \\times 10^{{22}} \\mathrm{{~kg}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_14_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the gravitational force of attraction between two $7.5 \\mathrm{{~kg}}$ bowling balls with $0.50 \\mathrm{{~m}}$ between centers?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Use the gravitational force equation
        $F= G \\frac{{m_1 m_2}}{{r^2}}=\\left(6.7 \\times 10^{{11}} \\mathrm{{~N}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{kg}}^2\\right) \\frac{{7.5 \\mathrm{{~kg}} \\cdot 7.5 \\mathrm{{~kg}}}}{{(0.50 \\mathrm{{~m}})^2}}=1.5 \\times 10^8 \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_13_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider a non-uniform bar of $200 \\mathrm{{~N}}$ weight and $4.0 \\mathrm{{~m}}$ length suspended horizontally between two walls where the angles between the walls and cables are $37^{{\\circ}}$ and $50^{{\\circ}}$. What are the tensions in the cables?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The extensions of the cables specify the center of mass. Looking just to the center of mass and using the common side of the triangle write
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-13-8.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        $\\displaystyle \\tan 37^{{\\circ}}=\\frac{{x}}{{c}} \\quad$ and $\\displaystyle \\quad \\tan 50^{{\\circ}}=\\frac{{4.0-x}}{{c}}$ Solve for $C$ and set equal $\\displaystyle \\quad \\frac{{4.0-x}}{{\\tan 50^{{\\circ}}}}=\\frac{{x}}{{\\tan 37^{{\\circ}}}}$
        and $4.0 \\tan 37^{{\\circ}}-\\mathrm{{x}} \\tan 37^{{\\circ}}=\\mathrm{{x}} \\tan 50^{{\\circ}}$ or $\\displaystyle \\quad x=\\frac{{4.0 \\tan 37^{{\\circ}}}}{{\\tan 37^{{\\circ}}+\\tan 50^{{\\circ}}}}-1.55 \\mathrm{{~m}}$
        Draw a vector diagram at the center of mass line. Write the equilibrium conditions as
        $T_1 \\cos 53^{{\\circ}}=T_2 \\cos 40^{{\\circ}}$ and $T_1 \\sin 53^{{\\circ}}+T_2 \\sin 40^{{\\circ}}=200 \\mathrm{{~N}}$
        Rewriting
        $T_1 \\cos 53^{{\\circ}}-T_2 \\cos 40^{{\\circ}}=0 \\quad$ and $\\quad T_1 \\sin 53^{{\\circ}}+T_2 \\sin 40^{{\\circ}}=200 \\mathrm{{~N}}$
        Multiplying and adding
        $T_1 \\cos 53^{{\\circ}} \\sin 40^{{\\circ}}-T_2 \\cos 40^{{\\circ}} \\sin 40^{{\\circ}}=0$
        $T_1 \\sin 53^{{\\circ}} \\cos 40^{{\\circ}}+T_2 \\sin 40^{{\\circ}} \\cos 40^{{\\circ}}=200 \\mathrm{{~N}} \\cos 40^{{\\circ}}$
        and
        $\\displaystyle T_1=\\frac{{200 \\mathrm{{~N}} \\cos 40^{{\\circ}}}}{{\\cos 53^{{\\circ}} \\sin 40^{{\\circ}}+\\sin 53^{{\\circ}} \\cos 40^{{\\circ}}}}=\\frac{{200 \\mathrm{{~N}} \\cos 40^{{\\circ}}}}{{\\sin 93^{{\\circ}}}}=153 \\mathrm{{~N}}$
        A sum of two angles trigonometric identity (see the Introduction, Mathematical Background) was used in the denominator.
        $\\displaystyle T_2=\\frac{{153 \\mathrm{{~N}} \\cos 53^{{\\circ}}}}{{\\cos 40^{{\\circ}}}}=120 \\mathrm{{~N}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_13_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Set up a $300 \\mathrm{{~N}}$ strut $14 \\mathrm{{~m}}$ long with a cable as shown in Fig. 13-7. Find the tension in the cable and the reaction force at the pivot.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The reaction force at the pivot is set up with vertical and horizontal components. The weight of the strut is taken at the center of the strut with the $2000 \\mathrm{{~N}}$ weight at the end. The tension is written along with all the components needed to write the force and torque equations.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-13-7.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        Apply the sum of the forces laws.
        $\\sum F_x=0: \\quad F_H=T \\cos 40^{{\\circ}} \\quad \\sum F_y=0: \\quad F_V=2000 \\mathrm{{~N}}+300 \\mathrm{{~N}}+T \\sin 40^{{\\circ}}$
        Apply the sum of the torques law about the pivot point.
        $\\sum \\tau=0:\\left(300 \\mathrm{{~N}} \\cos 60^{{\\circ}}\\right) 7.0 \\mathrm{{~m}}+\\left(2000 \\mathrm{{~N}} \\cos 60^{{\\circ}}\\right) 14 \\mathrm{{~m}}=\\left(\\mathrm{{T}} \\cos 70^{{\\circ}}\\right) 14 \\mathrm{{~m}}$
        Note that the length of the strut does not influence the calculation.
        Solve the torque equation for $T$.
        $2150 \\mathrm{{~N}} \\cos 60^{{\\circ}}=T \\cos 70^{{\\circ}} \\text {{ or }} T=3140 \\mathrm{{~N}}$
        Use this value to find $F_H$ and $F_V$
        $F_H=3140 \\mathrm{{~N}} \\cos 40^{{\\circ}}=2410 \\mathrm{{~N}}\\quad$ and $\\quad F_V=2300 \\mathrm{{~N}}+3140 \\mathrm{{~N}} \\sin 40^{{\\circ}}=4320 \\mathrm{{~N}}$
        The reaction at the floor can be written in terms of magnitude and angle.
        $F=\\sqrt{{2410^2+4320^2}}=4950 \\mathrm{{~N}} \\quad \\theta=\\tan ^{{-1}}(4320 / 2410)=61^{{\\circ}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_13_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $20 \\mathrm{{~kg}}$ beam 4.0m long is pivoted at a wall and supported by a cable as shown in Fig. 13-6. What is the force at the wall and the tension in the cable?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        There are two points of possible confusion in this problem. First, the mass of the beam is taken as down at the center of the beam. Second, the reaction of the wall, at the pivot, has two components, one up the wall (holding the end of the beam up) and one out of the wall (keeping the beam from sinking into the wall).
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-13-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        Set up the vector diagram at the pivot and the point where the cable is attached. The reaction at the wall is shown with components up the wall and normal to the wall. The weight of the beam is shown as down and from the center of the beam with a component at a right angle to the beam. The tension in the cable is broken up into vertical and horizontal components (for the sum of the forces laws) and a component at right angles to the beam (for the sum of the torques laws). Applying the sum of the forces laws means looking at the vector diagram and adding up all the vectors in the horizontal or vertical directions.
        $\\displaystyle \\sum F_y=0: R \\cos \\theta+T \\cos 30^{{\\circ}}=196 \\mathrm{{~N}} \\quad \\sum F_x=0: \\quad R \\sin \\theta=T \\sin 30^{{\\circ}}$
        In applying the sum of the torques, it is convenient to take the pivot point at the wall. Since the angle $\\theta$ is unknown, the least complicated sum of the torques statement will come from the tension in the cable and the weight of the beam.
        Using the components at right angles $\\displaystyle \\sum \\tau=0: 4.0 \\mathrm{{~m}}=160 \\mathrm{{~N}} .2 .0 \\mathrm{{~m}}$
        Solve the last equation for the tension in the cable
        $\\displaystyle T=\\frac{{160 \\mathrm{{~N}}}}{{2 \\cos 65^{{\\circ}}}}=189 \\mathrm{{~N}}$
        Use this value to solve for
        $\\displaystyle R \\sin \\theta=189 \\mathrm{{~N}} \\sin 30^{{\\circ}}=95 \\mathrm{{~N}}$ and $R \\cos \\theta=196 \\mathrm{{~N}}-189 \\mathrm{{~N}} \\cos 30^{{\\circ}}=32 \\mathrm{{~N}}$
        Now divide these two equations $\\displaystyle \\frac{{R \\sin \\theta}}{{R \\cos \\theta}}=\\tan \\theta=\\frac{{95}}{{32}}=3.0$ or $\\quad \\theta=71^{{\\circ}}$
        and finally, $R$ is obtained from one of the original equations
        $\\displaystyle R=\\frac{{189 \\mathrm{{~N}} \\sin 30^{{\\circ}}}}{{\\sin 71^{{\\circ}}}}=100 \\mathrm{{~N}}$
        The main points in this problem are the detailed vector diagram and the choice of the wall as the pivot point in calculating the torques.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_13_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Place an $80 \\mathrm{{~N}}$ ladder $8.0 \\mathrm{{~m}}$ long on a floor and against a frictionless roller on a $5.0 \\mathrm{{~m}}$ high wall. The minimum angle for equilibrium is $70^{{\\circ}}$. Find the coefficient of friction between the ladder and the floor.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The roller is free to rotate so the reaction force is normal to the ladder. This reaction force can be viewed as having vertical and horizontal components. The reaction force at the floor is written as vertical and horizontal components. The geometry and vector diagram are shown in
        Fig. 13-5. The length to the roller is obtained from the geometry: $\\sin 70^{{\\mathrm{{n}}}}=5.0 \\mathrm{{~m}} / L$ or $L=5.3 \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        file_name = 'how-to-solve-physics-problems-fig-13-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        Write the force equilibrium equations.
        $\\sum F_x=0: \\quad f=R \\sin 70^{{\\circ}} \\quad \\sum F_y=0: \\quad N+R \\cos 70^{{\\circ}}=80 \\mathrm{{~N}}$
        The torque about the point on the floor is produced by the reaction force at the roller and the weight of the ladder taken at its midpoint.
        $\\sum \\tau=0:\\left(80 \\mathrm{{~N}} \\cos 70^{{\\circ}}\\right) 4.0 \\mathrm{{~m}}=R \\cdot 5.3 \\mathrm{{~m}}$
        Rewriting with $f=\\mu N$
        $\\mu N=R \\sin 70^{{\\circ}} \\quad N+R \\cos 70^{{\\circ}}=80 \\mathrm{{~N}} \\quad(320 \\mathrm{{~N}} \\cdot \\mathrm{{m}}) \\cos 70^{{\\circ}}=(5.3 \\mathrm{{~m}}) R$
        Solve these equations from last to first to obtain
        $R=20.6 \\mathrm{{~N}} \\quad N=73.0 \\mathrm{{~N}} \\quad \\mu=0.26$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_13_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Place a $7.0 \\mathrm{{~m}}$ uniform, $150 \\mathrm{{~N}}$ ladder against a frictionless wall at an angle of $75^{{\\circ}}$. What are the reaction forces at the ground and wall and the minimum cocfficient of friction of the ground?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Figure 13-4 shows the ladder with the $150 \\mathrm{{~N}}$ actimg duwn at the center of the ladder and the sides of the triangle formed by the ladder, wall, and ground. The vector portion of the figure shows the two components of the reaction force at the ground. The reaction force at the ground is not necessarily in the direction of the ladder. This (vector) force can be written in terms of a force (magnitude) at an angle (with horizontal and vertical components) or directly in component form as done here. Writing the force in component force helps to avoid the temptation to place the force at the same angle as the ladder. Either way requires two variables. Since the wall is frictionless, the reaction at the wall is entirely horizontal. Refer to Fig. 13-4 and write the equilibrium conditions.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-13-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        $\\displaystyle \\sum F_x=0: \\quad F_H=R_H \\quad \\sum F_y=0: \\quad F_V=150 \\mathrm{{~N}}$
        $\\sum \\tau=0$ The torque on the ladder is taken about the point where the ladder contacts the ground. Note that this choice eliminates two variables from the torque statement. Torque is the (component of the) force at right angles to the lever arm times that lever arm. The $150 \\mathrm{{~N}}$ weight has a component $150 \\mathrm{{Ncos}} 75^{{\\circ}}$ times $3.5 \\mathrm{{~m}}$. The reaction at the wall has a component $R_{{\\mathrm{{H}}}} \\cos 15^{{\\circ}}$ times $7.0 \\mathrm{{~m}}$ so the sum of the torques equal zero statement is
        $\\left(150 \\mathrm{{~N}} \\cos 75^{{\\circ}}\\right) 3.5 \\mathrm{{~m}}=\\left(R_H \\cos 15^{{\\circ}}\\right) 7.0 \\mathrm{{~m}} \\quad \\text {{ or }} \\quad 150(0.90) \\mathrm{{N}} \\cdot \\mathrm{{m}}=R_H(6.8) \\mathrm{{N}} \\cdot \\mathrm{{m}}$
        so
        $R_H=(75 \\mathrm{{~N}})\\left(\\cos 75^{{\\circ}} / \\cos 15^{{\\circ}}\\right)=20 \\mathrm{{~N}} \\quad \\text {{ and }} \\quad F_H=20 \\mathrm{{~N}}$
        $F_{{\\mathrm{{v}}}}$ and $F_{{\\mathrm{{H}}}}$ are related via the coefficient of friction, $F_n=\\mu F_v$ so $\\mu=20 / 150=0.13$
        Second Solution: There is another way to calculate the torques. Notice that the $150 \\mathrm{{~N}}$ force is horizontally $0.90 \\mathrm{{~m}}$ away from the pivot point. If the $150 \\mathrm{{~N}}$ force vector is moved vertically so its tail is on a horizontal line from the pivot point, then the torque is easily written as $(150 \\mathrm{{~N}})(0.90 \\mathrm{{~m}})$. A similar operation can be performed on $R_H$ yielding a torque $R_H(6.8 \\mathrm{{~m}})$. Vectors maintain their length (magnitude) and orientation (angle) but can be moved about a vector diagram for (our) convenience. If the torque statement is written in this way it is identical to the torque statement in the first solution.
        $150 \\mathrm{{~N}} \\cdot 0.90 \\mathrm{{~m}}=R_H 6.8 \\mathrm{{~m}}$
        As an exercise redo the problem with a person of known weight positioned a specific distance up the ladder.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_13_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $500 \\mathrm{{~N}}$ diver is on the end of a 4.0m diving board of negligible mass. The board is on pedestals as shown in Fig. 13-3. What are the forces that each pedestal exerts on the diving board?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        There are no forces in the $x$-direction, but in the $y$-direction the pedestal at $A$ and the diver are acting down while the force at pedestal $B$ is acting up. Pedestal $B$ is in compression while pedestal $A$ is in tension. Draw the vector diagram and write the equations.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-13-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        $\\displaystyle \\sum F_y=0: \\quad F_A+500 \\mathrm{{~N}}=F_B$
        In this problem apply $\\sum \\tau=0$ about pedestal $A$. Taking the rotation point at pedestal $A$, the diver produces a clockwise torque, and pedestal $B$ an anti-clockwise torque that are in equilibrium. Note that this is not the only possible rotation point.
        $\\displaystyle \\sum \\tau=0 \\quad 500 \\mathrm{{~N}} \\cdot 4.0 \\mathrm{{~m}}=F_B \\cdot 1.5 \\mathrm{{~m}}$
        Solve for $F_B$ :
        $\\displaystyle F_B: F_B=\\frac{{500 \\mathrm{{~N}} \\cdot 4.0 \\mathrm{{~m}}}}{{1.5 \\mathrm{{~m}}}}=1330 \\mathrm{{~N}}$
        Solve for $\\sum \\tau=0$
        As an exercise write the $\\sum \\tau-0$ about pedestal $B$ and the diver.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_13_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        In the arrangement shown in Fig. 13-2, what is the minimum coefficient of friction to prevent the 8.0kg mass from sliding?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-13-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        soln_s = f"""
        The forces come together where the cords come together. This point is not moving so apply the equilibrium conditions here. The vector diagram at this point is also in Fig. 13-2.
        The equilibrium conditions are:
        $\\displaystyle\\sum F_x=0: \\quad T_1 \\cos 60^{{\\circ}}=F \\quad \\sum F_y=0 \\quad T_1 \\sin 60^{{\\circ}}=39 \\mathrm{{~N}}$
        Solving for $T_1: \\quad T_1=\\frac{{39 \\mathrm{{~N}}}}{{\\sin 60^{{\\circ}}}}-45 \\mathrm{{~N}}$ :
        The force is: $F=45 \\mathrm{{~N}} \\cos 60^{{\\circ}}=22.5 \\mathrm{{~N}}$
        This force is supplied by the frictional force $\\mu \\mathrm{{mg}}$, so
        $\\displaystyle \\mu=\\frac{{22.5 \\mathrm{{~N}}}}{{8.0 \\mathrm{{~kg}} \\cdot 9.8\\left(\\mathrm{{~m}} / \\mathrm{{s}}^2\\right)}}=0.29$
        This is the minimum coefficient of friction that will prevent me system from moving.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_13_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Hang a 50kg mass with ropes making angles of $30^{{\\circ}}$ and $45^{{\\circ}}$ as shown in Fig. 13-1. Calculate the tension in the ropes.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-13-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        Note that all the forces come together at the knot in the rope so draw a force diagram about this point. The only laws to apply are for equilibrium in the $x$ and $y$ directions.
        $\\sum F_x=0: \\quad T_1 \\cos 30^{{\\circ}}=T_2 \\cos 45^{{\\circ}} \\quad \\sum F_y=0: \\quad T_1 \\sin 30^{{\\circ}}+T_2 \\sin 45^{{\\circ}}=294 \\mathrm{{~N}}$
        This provides two equations in two unknowns. Because $\\sin 45^{{\\circ}}=\\cos 45^{{\\circ}}$ rewrite
        $T_1\\left(\\sin 30^{{\\circ}}+\\cos 30^{{\\circ}}\\right)=294 \\mathrm{{~N}} \\quad \\text {{ or }} \\quad T_1=215 \\mathrm{{~N}}$
        and
        $T_2 \\cos 45^{{\\circ}}=(215 \\mathrm{{~N}}) \\cos 30^{{\\circ}} \\quad \\text {{ or }} \\quad T_2=263 \\mathrm{{~N}}$
        As an exercise work through this problem with different angles.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_12_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For problem 12-9 take the mass of the sphere as $m=2.0 \\mathrm{{~kg}}, r=0.30 \\mathrm{{~m}}$, the hanging mass as $M=0.80 \\mathrm{{~kg}}$, and the height the mass falls through as $h=1.5 \\mathrm{{~m}}$. Find the velocity of the hanging mass, and the angular momentum of the sphere.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The velocity of the mass, the rope, and a point on the equator of the sphere (all the same velocity) is 
        $\\displaystyle 0.80 \\mathrm{{~kg}} \\frac{{9.8 \\mathrm{{~m}}}}{{\\mathrm{{~s}}^2}} 1.5 \\mathrm{{~m}}=\\left(\\frac{{2.0 \\mathrm{{~kg}}}}{{5}}+\\frac{{0.80 \\mathrm{{~kg}}}}{{2}}\\right) v^2 \\quad$ or $\\displaystyle \\quad v=3.8 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}$ from the energy statement
        The angular velocity of the sphere is
        The angular momentum of the sphere is
        $\\displaystyle L=I \\omega=\\left[\\frac{{2 \\cdot 2.0 \\mathrm{{~kg}}(0.30 \\mathrm{{~m}})^2}}{{5}}\\right]\\left[\\frac{{13 \\mathrm{{rad}}}}{{\\mathrm{{s}}}}\\right]=0.94 \\frac{{\\mathrm{{kg}} \\cdot \\mathrm{{m}}^2}}{{\\mathrm{{~s}}}}$
        As an added exercise give the pulley in this problem a mass and a radius and go through the problem again. Rememher that I and $(\\mathrm{{t}}$ will be different for the pulley.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_12_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A solid sphere is constrained to rotate about a vertical axis passing through the center of the sphere. A cord is wrapped around what would be the equator, passes over a pulley of negligible mass, and is attached to a mass that is allowed to fall under the influence of gravity. Write a conservation of energy statement for the system.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-12-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        soln_s = f"""
        If the mass falls a distance $h$, then the energy gained by the system is $M g h$. This energy Goesinto translational kinetic energy of the mass and rotational kinetic energy of the sphere. A conservation of energy statement is
        $\\displaystyle M g h=\\frac{{1}}{{2}} I \\omega^2+\\frac{{1}}{{2}} M v^2$
        The velocity of a point on the string and the velocity of the mass are the same and are related to the angular velocity through $v=r \\omega$, so the energy statement can be written in terms of $v$.
        Additionally, the formula for $I=2 \\mathrm{{~m}} r^2 / 5$ can be substituted so
        $\\displaystyle M g h=\\frac{{1}}{{2}} I \\frac{{v^2}}{{r^2}}+\\frac{{1}}{{2}} M v^2=\\left(\\frac{{m}}{{5}}+\\frac{{M}}{{2}}\\right) v^2$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_12_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        When the $30 \\mathrm{{~kg}}$ mass in problem $12-7$ has fallen (starting from rest) through $4.0 \\mathrm{{~m}}$, what is the linear velocity, angular velocity of the drum, and the time for this to occur?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This part of the problem is approached from an energy point of view. The $30 \\mathrm{{~kg}}$ mass falls through a distance of $4.0 \\mathrm{{~m}}$. The work performed by gravity Goesinto translational
        $K E$ of the $30 \\mathrm{{~kg}}$ mass and rotational $K E$ of the cylinder. The main difficulty in problems like this is remembering that some of the work performed results in rotational $K E$ and some in translational $K E$.
        $\\displaystyle M g h=\\frac{{1}}{{2}} M v^2+\\frac{{1}}{{2}} I \\omega^2=\\frac{{1}}{{2}} M v^2+\\frac{{1}}{{2}}\\left(\\frac{{m r^2}}{{2}}\\right) \\frac{{v^2}}{{r^2}} \\quad$ or $\\displaystyle \\quad M g h=v^2\\left(\\frac{{M}}{{2}}+\\frac{{m}}{{4}}\\right)$
        Putting in the numbers
        $\\displaystyle 30 \\mathrm{{~kg}} \\frac{{9.8 \\mathrm{{~m}}}}{{\\mathrm{{~s}}^2}} 4.0 \\mathrm{{~m}}=v^2\\left(\\frac{{30 \\mathrm{{~kg}}}}{{2}}+\\frac{{10 \\mathrm{{~kg}}}}{{2}}\\right) \\quad$ or $\\displaystyle \\quad v^2=59 \\frac{{\\mathrm{{m}}^2}}{{\\mathrm{{~s}}^2}} \\quad$ or $\\displaystyle \\quad v=7.7 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}$
        From $v=r \\omega$ angular velocity is $\\displaystyle \\omega=\\frac{{\\nu}}{{r}}=\\frac{{7.7 \\mathrm{{~m}} / \\mathrm{{s}}}}{{1.0 \\mathrm{{~m}}}}=7.7 \\frac{{\\mathrm{{rad}}}}{{\\mathrm{{s}}}}$
        The time for this system to reach this velocity is from $\\omega=\\omega_0+\\tau 1$.
        $\\displaystyle t=\\frac{{\\omega}}{{\\alpha}}=\\frac{{7.7(\\mathrm{{rad}} / \\mathrm{{s}})}}{{8.4\\left(\\mathrm{{rad}} / \\mathrm{{s}}^2\\right)}}=0.92 \\mathrm{{~s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_12_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Now consider a $10 \\mathrm{{~kg}}$ solid cylindrical drum with radius $1.0 \\mathrm{{~m}}$ rotating about its cylindrical axis under the influence of a force produced by a $30 \\mathrm{{~kg}}$ mass attached to a cord wound around the drum. What is the torque, moment of inertia, and angular acceleration?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-12-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        soln_s = f"""
        The moment of inertia for a solid drum rotating about its axis is $m r^2 / 2$.
        $I=m r^2 / 2=10 \\mathrm{{~kg}}(1.0 \\mathrm{{~m}})^2 / 2=5.0 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2$
        The torque on the drum is the tension in the cable times the radius of the drum. This tension is much like the tension encountered in the problems of multiple hlocks sliding on tables as found in Chapter 5, Force. The system accelerates clockwise, so the unbalanced force on $M$ makes it accelerate according to
        $M g-F=M a \\quad \\text {{ or }} \\quad F=M(g-a)$
        The torque on the drum makes it angularly accelerate according to $\\mathrm{{r}} F=1 \\alpha$
        The acceleration of $M$ is the same as the tangential acceleration, $a=r \\tau$ so
        $\\displaystyle r F=\\left[\\frac{{m r^2}}{{2}}\\right] \\frac{{a}}{{r}} \\quad \\text {{ or }} \\quad F=\\frac{{m a}}{{2}} \\quad \\text {{ leading to }} \\quad \\frac{{m a}}{{2}}=M g-M a \\quad \\text {{ or }} \\quad a=\\frac{{g}}{{1+m / 2 M}}$
        Evaluating
        $\\displaystyle a=\\frac{{9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2}}{{1+10 / 60}}=8.4 \\mathrm{{~m}} / \\mathrm{{s}}^2$
        The angular acceleration is
        $\\displaystyle \\alpha=\\frac{{a}}{{r}}=\\frac{{8.4 \\mathrm{{~m}} / \\mathrm{{s}}^2}}{{1.0 \\mathrm{{~m}}}}=8.4 \\mathrm{{rad}} / \\mathrm{{s}}^2$
        The torque is
        $\\displaystyle \\tau=I \\alpha=\\left(5.0 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2\\right) 8.4 \\mathrm{{rad}} / \\mathrm{{s}}^2=42 \\mathrm{{~N}} \\cdot \\mathrm{{m}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_12_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An amusement park game consists of a paddle wheel arrangement where you shoot at the paddles with a pellet gun, thereby turning the wheel. The paddle wheel is set in motion with an initial angular momentum of $200 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{s}}$ and angular frequency of 4.0rad / $\\mathrm{{s}}$. You shoot eight $40 \\mathrm{{~g}}$ pellets at a speed of $200 \\mathrm{{~m}} / \\mathrm{{s}}$ at the paddles. The pellets hit the paddles at $0.80 \\mathrm{{~m}}$ radius, stick, and impart all their momentum to the wheel. Find the new angular momentum and the new angular frequency.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First calculate the initial moment of inertia of the paddle wheel from $I=I w$.
        $200 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{s}}=I(4.0 \\mathrm{{rad}} / \\mathrm{{s}}) \\quad$ or $\\quad I=50 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-12-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        soln_s = f"""
        Now calculate the $w$ after the wheel has absorbed the (momentum in the) eight pellets. The angular momentum of each pellet is the linear momentum times the $0.80 \\mathrm{{~m}}$ radius.
        $l_{{\\text {{each pellet }}}}=m v r=0.040 \\mathrm{{~kg}}(200 \\mathrm{{~m}} / \\mathrm{{s}}) 0.80 \\mathrm{{~m}}=6.4 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{s}}$
        For eight pellets the "absorbed" angular momentum is $\\mathrm{{L}}_{{\\text {{all pellet }}}}=51.2 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2 / 2$
        The new angular momentum is the original plus this, or $251.2 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{s}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_12_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A hoop of mass $1.0 \\mathrm{{~kg}}$ and radius $0.25 \\mathrm{{~m}}$ is rotating in a horizontal plane with angular momentum $4.0 \\mathrm{{~kg}}$.m ${{ }}^2$ / s. A lump of clay of mass $0.20 \\mathrm{{~kg}}$ is placed (gently) on the hoop. What happens to the angular velocity of the hoop?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The moment of inertia for a hoop (from the table in your text) is $I=m r^2$.
        $I=1.0 \\mathrm{{~kg}}(0.25 \\mathrm{{~m}})^2=0.0625 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2$
        The angular momentum, $L$, is $I w$, , so the initial angular velocity can be calculated from $L=I w_{{\\mathrm{{i}}}}$
        $4.0 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{s}}=\\left(0.0625 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2\\right) \\omega_i \\quad$ or $\\quad \\omega_i=64 \\mathrm{{rad}} / \\mathrm{{s}}$
        Since the lump of clay is placed (gently) on the hoop (The hoop is rotating in the horizontal plane, and the clay is placed on the hoop vertically.), placing the clay only adds mass to the hoop. This additional mass adds
        "another" moment of inertia $m r^2$, so with the angular momentum the same (no external torque has acted to change it)
        $4.0 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{s}}=\\left[1.0 \\mathrm{{~kg}}(0.25 \\mathrm{{~m}})^2+0.2 \\mathrm{{~kg}}(0.25 \\mathrm{{~m}})^2\\right] \\omega_f \\quad$ or $\\quad \\omega_f=53.3 \\mathrm{{rad}} / \\mathrm{{s}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_12_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the angular momentum of a $3.0 \\mathrm{{~kg}}$ mass at $3.0 \\mathrm{{~m}}$ in the $x$-direction and $-2.0 \\mathrm{{~m}}$ in the $y$-direction and with velocity components of $20 \\mathrm{{~m}} / \\mathrm{{s}}$ in the $x$-direction and $-30 \\mathrm{{~m}} / \\mathrm{{s}}$ in the $y$-direction?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-12-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        soln_s = f"""
        First Solution: Write $r$ and $\\boldsymbol{{p}}$ in vector form: $r=(3.0 \\mathrm{{i}}-2.0 \\mathrm{{j}}) \\mathrm{{m}} \\quad \\mathrm{{p}}=(60 \\mathrm{{i}}-90 \\mathrm{{j}}) \\mathrm{{kg}} . \\mathrm{{m}} / \\mathrm{{s}}$ The angular momentum is
        $\\displaystyle L=r \\times p=\\left|\\begin{{array}}{{ccc}} i & j & k \\\\ 3.0 & -2.0 & 0 \\\\ 60 & -90 & 0 \\end{{array}}\\right|=(-270+120) k=-150 k \\frac{{\\mathrm{{kg}} \\cdot \\mathrm{{m}}^2}}{{\\mathrm{{~s}}}}$
        The angular momentum vector is $150 \\mathrm{{~kg}} . \\mathrm{{m}}^2 / \\mathrm{{s}}$ pointed into the paper.
        Second Solution: Now redo the problem with angle and magnitude. The angle between $r$ and $p$ is $-22.6^{{\\circ}}$. Figure 12-3 shows the trigonometric relations. Note that the angle between $\\boldsymbol{{r}}$ and $\\boldsymbol{{p}}$ is negative. The fact that the angle is negative can be easily missed. Rather than rely on the sign of the angle, the better way to determine the direction of the angular momentum vector is to use the right-hand rule and cross $r$ into $p$ with the thumb giving the direction of $L$. The magnitude of $L$ is
        $L=|r| \\cdot|p| \\sin \\theta=\\sqrt{{13}} \\sqrt{{11700}} \\sin \\left(-22.6^{{\\circ}}\\right)=-150 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_12_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the rotational inertia for one $2.0 \\mathrm{{~kg}}$ mass at the end of a $3.0 \\mathrm{{~m}}$ rod of negligible mass and then the rotational inertia of a dumbbell consisting of two $2.0 \\mathrm{{~kg}}$ masses on the ends of a $6.0 \\mathrm{{~m}}$ long rod of negligible mass pivoted about the center of the rod.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        For the one mass, the sum of the $m r^2$ 's is just $2.0 \\mathrm{{~kg}}\\left(9.0 \\mathrm{{~m}}^2\\right)=18 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2$.
        For the dumbbell, there are two masses, and the rotational inertia would be $36 \\mathrm{{~kg}} . \\mathrm{{m}}^2$.
        Notice that if the dumbbell were rotated about one mass then the rotational incrtia would be
        $2.0 \\mathrm{{~kg}}\\cdot 36 \\mathrm{{~m}}^2=72 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2$. This is of course neglecting the extent of the mass close around the rotating axis.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_12_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation of problem $12-1$ add that the linear velocity is $2.8 \\mathrm{{~m}} / \\mathrm{{s}}$ and find the angular momentum.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The linear momentum is $m v=2.5 \\mathrm{{~kg}}(2.8 \\mathrm{{~m}} / \\mathrm{{s}})=7.0 \\mathrm{{~kg}} . \\mathrm{{m}} / \\mathrm{{s}}$. The linear velocity is the tangential velocity, which is always at right angles to the radius; so the direction of the angular momentum vector is out of the paper, and the magnitude of the cross product is simplified because the sine of $90^{{\\circ}}$ is one.
        $L=r p=1.6 \\mathrm{{~m}}(7.0 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}})=11.2 \\mathrm{{~kg}} \\cdot \\mathrm{{m}}^2 / \\mathrm{{s}}^2$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_12_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the torque on a $2.5 \\mathrm{{~kg}}$ mass constrained to rotate in a $1.6 \\mathrm{{~m}}$ radius circle with a $250 \\mathrm{{~N}}$ force applied at a (constant) angle of $35^{{\\circ}}$ hetween $r$ and $F$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The problem is depicted in Fig. 12-2. The definition of $\\boldsymbol{{r}} \\times \\boldsymbol{{F}}$. gives the direction of the torque as out of the paper. Application of the crossing of $\\boldsymbol{{r}}$ into $\\boldsymbol{{F}}$ is similar to generating a righthanded coordinate system by rotating $x$ into $y$ with the right hand with the thumb giving the direction of $\\mathrm{{z}}$. The magnitude of the torque is
        $T=|r| F \\mid \\sin \\theta=1.5 \\mathrm{{~m}}(250 \\mathrm{{~N}}) \\sin 35^{{\\circ}}=215 \\mathrm{{~N}} \\cdot \\mathrm{{m}}$
        Torque is $r$ cross force, and similarly, angular momentum is $r$ cross momentum or
        $L=r \\times p$
        Angular momentum is a vector in a direction perpendicular to the plane of $\\boldsymbol{{r}}$ and $\\boldsymbol{{p}}$ and magnitude equal to the product of $r$ and the component of $p$ perpendicular to $r$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_11_11(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A grinding wheel of $30 \\mathrm{{~cm}}$ diameter is rotating with angular velocity of $3.0(\\mathrm{{rad}} / \\mathrm{{s}})$ and slowing under constant acceleration of $-3.0\\left(\\mathrm{{rad}} / \\mathrm{{s}}^2\\right)$. Calculate everything possible about the motion at several different times.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Circular motion problems even more than linear motion problems, often present a challenge as to how to proceed from the data to the specific question. When confused about the specific route to follow in a problem, the best thing to do is calculate something simple and let this first calculation lead you to others and ultimately the answer. After doing the problem once the "hard way" you will learn more direct routes through the problem. Given the data in this problem start calculating some simple things.
        What is the angular velocity at 0.40 s? Use
        $\\omega=\\omega_0+\\alpha t=3.0(\\mathrm{{rad}} / \\mathrm{{s}})-3.0\\left(\\mathrm{{rad}} / \\mathrm{{s}}^2\\right) 0.40 \\mathrm{{~s}}=1.8(\\mathrm{{rad}} / \\mathrm{{s}})$
        What angle has been turned through in this 0.40 s? Use
        $\\theta=\\omega_o t+(1 / 2) \\alpha t^2=3.0(\\mathrm{{rad}} / \\mathrm{{s}}) 0.40 \\mathrm{{~s}}-(1 / 2) 3.0\\left(\\mathrm{{rad}} / \\mathrm{{s}}^2\\right)\\left(0.16 \\mathrm{{~s}}^2\\right)=0.96 \\mathrm{{rad}}$
        What is the velocity of a point on the rim at 0.20 s? First find the angular velocity.
        $\\omega=\\omega_o+\\alpha t=3.0(\\mathrm{{rad}} / \\mathrm{{s}})-3.0\\left(\\mathrm{{rad}} / \\mathrm{{s}}^2\\right)(0.20 \\mathrm{{~s}})=2.4(\\mathrm{{rad}} / \\mathrm{{s}})$
        And then find the linear velocity from $v=r \\omega=15 \\times 10^{{-2}} \\mathrm{{~m}} .2 .4(\\mathrm{{rad}} / \\mathrm{{s}})=0.36 \\mathrm{{~m}} / \\mathrm{{s}}$.
        With this acceleration, how long does it take for the wheel to come to rest? Use $\\omega=\\omega_s+a t$ or $0=30(1 / \\mathrm{{s}})-30\\left(1 / \\mathrm{{s}}^2\\right) t \\quad$ or $\\quad t=10 \\mathrm{{~s}}$
        How much angle is turned through in this time? Use
        $\\omega^2=\\omega_o^2+2 \\alpha \\theta \\text {{ or }} 0=9.0\\left(\\mathrm{{rad}}^2 / \\mathrm{{s}}^2\\right)-2 \\cdot 3.0\\left(\\mathrm{{rad}} / \\mathrm{{s}}^2\\right) \\theta \\text {{ or }} \\theta=1.5 \\mathrm{{rad}}$
        How far does a spot on the rim travel in this time? Use
        $s=r \\theta=15 \\times 10^{{-2}} \\mathrm{{~m}} \\cdot 1.5(\\mathrm{{rad}})=0.225 \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_11_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A glider pilot wishing to fly in a vertical loop dives to attain a speed sufficient to keep the glider from "falling out" of the top of the loop. What is the minimum entry speed for a $200 \\mathrm{{~m}}$ radius loop?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This problem is solved with energy analysis. At the bottom of the loop the glider must have sufficient kinetic energy to climb the diameter of the loop and have enough left to satisfy the $m v^2 / r=m g$ condition. Refer to Fig. 11-5 to write the energy balance statement and the criteria for the glider not "falling out" of the loop.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-11-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        The energy statement is in word form: Kinetic energy at entry equals the potential energy to reach the top of the loop plus the kinetic energy at the top of the loop
        $\\displaystyle \\frac{{1}}{{2}} m v_e^2=2 r m g+\\frac{{1}}{{2}} m v_l^2$
        The minimum velocity condition is
        $\\displaystyle \\frac{{m v_t^2}}{{r}}=m g \\text {{ or }} v_t=\\sqrt{{r g}}$
        Substituting for $v_t$ the minimum energy statement is
        $\\displaystyle \\frac{{1}}{{2}} m v_e^2=2 r m g+\\frac{{1}}{{2}} m r g=\\frac{{5}}{{2}} m r g$ yielding a minimum entry speed of $\\displaystyle v_e=\\sqrt{{5 r g}}$
        For a $200 \\mathrm{{~m}}$ radius loop (very large) the minimum $v_e$ for the glider is
        $v_e=\\sqrt{{5(200 \\mathrm{{~m}}) 9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2}}=99 \\mathrm{{~m}} / \\mathrm{{s}}=356 \\mathrm{{~km}} / \\mathrm{{hr}}$
        For a $100 \\mathrm{{~m}}$ radius loop $v_\\epsilon=70 \\mathrm{{~m}} / \\mathrm{{s}}=252 \\mathrm{{~km}} / \\mathrm{{hr}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_11_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation in problem 11-8 calculate the minimum velocity and period of rotation to keep the mass from "falling out" at the top of the circle.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Zero tension in the cord occurs (at the top) when
        $\\displaystyle \\frac{{m v_{{\\min }}^2}}{{r}}=m g \\quad \\text {{ or }} \\quad v_{{\\min }}=\\sqrt{{r g}}$
        Notice that the mass drops out of the equation. Using the $1.0 \\mathrm{{~m}}$ radius
        $v_{{\\min }}=\\sqrt{{1.0 \\mathrm{{~m}}\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)}}=3.13 \\mathrm{{~m}} / \\mathrm{{s}}$
        The period is calculated by replacing the velocity by $2 \\pi r / T$
        $\\displaystyle \\frac{{2 \\pi r}}{{T}}=\\sqrt{{r g}} \\quad \\text {{ or }} \\quad T=2 \\pi \\sqrt{{\\frac{{r}}{{g}}}}=2 \\pi \\sqrt{{\\frac{{1.0 \\mathrm{{~m}}}}{{9.8\\left(\\mathrm{{~m}} / \\mathrm{{s}}^2\\right)}}}}=2.0 \\mathrm{{~s}}$
        This result can be checked by whirling a mass in a vertical circle and observing that the mass "falls out" of the circle at a period of over $2.0 \\mathrm{{~s}}$. (The length from shoulder to ground is a little over $1.0 \\mathrm{{~m}}$ for most people so this is an easy experiment to perform. For a little adventure try it with a container of liquid!)
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_11_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Take the same mass, radius, and period as in problem 11-7 but with the rotation in a vertical circle. What is the force between the mass and the sling?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        In this case gravity has to be taken into account. The mass exerts a force on the sling radially out and equal to $m v^2 /$ r. Gravity exerts a force $m g$ down. Figure 11-4 shows the relative direction of these forces at four points on the circle.
        The gravitational force is $m g=(2.0 \\mathrm{{~kg}}) 9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2=19.6 \\mathrm{{~N}}$
        The force between the mass and the sling as shown in Fig. 11-4 is $79.0+19.6=98.6 \\mathrm{{~N}}$ at the bottom of the circle and $79.0 \\quad 19.6=59.4 \\mathrm{{~N}}$ at the top of the circle.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_11_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider a mass of $2.0 \\mathrm{{~kg}}$ being whirled in a sling in a horizontal circle. The period of rotation is $1.0 \\mathrm{{~s}}$, and the radius $1.0 \\mathrm{{~m}}$. What is the tension in the cord?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The force that makes the mass travel in a circle must be equal to $m v^2 / r$. The velocity of the mass is the distance traveled (one circumference) divided by the time to travel one circumference (the period)
        $\\displaystyle v=\\frac{{2 \\pi}}{{T}}=\\frac{{2 \\pi \\cdot 1.0 \\mathrm{{~m}}}}{{1.0 \\mathrm{{~s}}}}=2.0 \\pi \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}$
        The force (or tension) in the cord is
        $\\displaystyle F=\\frac{{m v^2}}{{r}}=\\frac{{2.0 \\mathrm{{~kg}}}}{{1.0 \\mathrm{{~m}}}}\\left(2.0 \\pi \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}\\right)^2=79.0 \\mathrm{{~N}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_11_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        If someone of weight $800 \\mathrm{{~N}}$ (on earth) living in the space station described in problem $11-5$ were to move one-half the radius in toward the axis of rotation, what would be their weight at this radius?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Their weight, what a force meter placed hetween them and a "floor" would read, in this environment would be
        $\\displaystyle W-\\frac{{m v^2}}{{r}}=m r w^2=\\frac{{800 \\mathrm{{~N}}}}{{9.8\\left(\\mathrm{{~m}} / \\mathrm{{s}}^2\\right)}} 150 \\mathrm{{~m}}\\left(0.18 \\frac{{1}}{{\\mathrm{{~s}}}}\\right)^2=400 \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_11_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider a space station in the form of a donut with a rectangular cross section connected by spokes to a central axis. The "floor" is the inside of the outer wall. How fast would a $300 \\mathrm{{~m}}$ radius station have to rotate to duplicate the "acceleration due to gravity" on the surface of the earth?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-11-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        soln_s = f"""
        The required acceleration is the $v^2 / r$ acceleration associated with rotational motion. Since $v$ is related to $\\omega$ by $v=r \\omega$, then the required acceleration $a=v^2 / r=r \\omega^2$. Set this acceleration equal to $9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2$ and solve for (i.
        $\\displaystyle \\omega=\\sqrt{{\\frac{{a}}{{r}}}}=\\sqrt{{\\frac{{9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2}}{{300 \\mathrm{{~m}}}}}}=0.181 \\frac{{\\mathrm{{rad}}}}{{\\mathrm{{s}}}}$
        The frequency in revolutions per minute is
        $\\displaystyle f^{{\\prime}}=\\frac{{\\omega}}{{2 \\pi}}=\\frac{{1 \\mathrm{{rev}}}}{{2 \\pi \\mathrm{{rad}}}} \\frac{{0.181 \\mathrm{{rad}}}}{{\\mathrm{{s}}}}=\\frac{{0.0288 \\mathrm{{rev}}}}{{\\mathrm{{s}}}} \\frac{{60 \\mathrm{{~s}}}}{{\\mathrm{{~min}}}}=1.72 \\frac{{\\mathrm{{rev}}}}{{\\mathrm{{min}}}}$
        Notice that in this last calculation "rev" was introduced for clarity.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_11_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the acceleration of the disk in problem 11-3, how many revolutions does it take for the disk to stop?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First, calculate
        $\\displaystyle \\alpha=\\frac{{\\Delta \\omega}}{{\\Delta t}}=\\frac{{-10(\\mathrm{{rad}} / \\mathrm{{s}})}}{{15.1 \\mathrm{{~s}}}}=-0.662 \\frac{{\\mathrm{{rad}}}}{{\\mathrm{{s}}^2}}$
        Now use $\\omega^2=\\omega_0 2+2 \\alpha \\theta$ with $\\omega-0$ and $\\omega=30 \\mathrm{{rad}} / \\mathrm{{s}}$ to find $\\theta$ :
        $\\displaystyle \\omega_o^2=-2 \\alpha \\theta \\quad \\text {{ or }} \\quad \\theta=-\\frac{{\\omega_o^2}}{{2 \\alpha}}=\\frac{{(30 \\mathrm{{rad}} / \\mathrm{{s}})^2}}{{2 \\cdot\\left(0.662 \\mathrm{{rad}} / \\mathrm{{s}}^2\\right)}}=680 \\mathrm{{rad}} \\quad 680 \\mathrm{{rad}} \\frac{{1 \\mathrm{{rev}}}}{{2 \\pi \\mathrm{{rad}}}}=108 \\mathrm{{rev}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_11_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A disk rotating at $30 \\mathrm{{rad}} / \\mathrm{{s}}$ slows to $20 \\mathrm{{rad}} / \\mathrm{{s}}$ while turning through 60 revolutions. How long does this take?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Since the acceleration is constant we can use $\\displaystyle \\theta=\\frac{{\\omega_0+\\omega}}{{2}} t$
        $\\displaystyle t=\\frac{{2 \\theta}}{{\\omega_o+\\omega}}=\\frac{{2 \\cdot 2 \\pi \\cdot 60 \\mathrm{{rad}}}}{{30(\\mathrm{{rad}} / \\mathrm{{s}})+20(\\mathrm{{rad}} / \\mathrm{{s}})}}=\\frac{{240 \\pi}}{{50}} \\mathrm{{~s}}=15.1 \\mathrm{{~s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_11_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        If the automobile of problem $11-1$ accelerates uniformly from $60 \\mathrm{{~km}} / \\mathrm{{hr}}$ to $80 \\mathrm{{~km}} / \\mathrm{{hr}}$ in $3.0 \\mathrm{{~s}}$, what is the angular acceleration?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This acceleration is the acceleration of the point where the wheel meets the road. First calculate the change in velocity.
        $\\displaystyle \\Delta v=(80-60) \\frac{{\\mathrm{{km}}}}{{\\mathrm{{hr}}}}=20 \\frac{{\\mathrm{{km}}}}{{\\mathrm{{hr}}}} \\frac{{1 \\mathrm{{hr}}}}{{60 \\mathrm{{~min}}}} \\frac{{1 \\mathrm{{~min}}}}{{60 \\mathrm{{~s}}}} \\frac{{10^3 \\mathrm{{~m}}}}{{\\mathrm{{~km}}}}=\\frac{{200}}{{36}} \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}$
        This is the change in velocity of the point on the wheel. The linear acceleration, is
        $\\displaystyle a=\\frac{{\\Delta v}}{{\\Delta t}}=\\left[\\frac{{200}}{{36}} \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}\\right] \\frac{{1}}{{3.0 \\mathrm{{~s}}}}=1.85 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}^2}}$
        The angular acceleration is
        $\\displaystyle \\alpha=\\frac{{a}}{{r}}=\\left[\\frac{{1.85 \\mathrm{{~m}}}}{{\\mathrm{{~s}}^2}}\\right] \\frac{{1}}{{0.35 \\mathrm{{~m}}}}=5.29 \\frac{{1}}{{\\mathrm{{~s}}^2}}=5.29 \\frac{{\\mathrm{{rad}}}}{{\\mathrm{{s}}^2}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_11_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An automobile is traveling at $60 \\mathrm{{~km}} / \\mathrm{{hr}}$. Calculate the angular velocity of the $0.35 \\mathrm{{~m}}$ radius wheels.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The linear velocity of the (contact) point on the wheel is
        $\\displaystyle \\frac{{60 \\mathrm{{~km}}}}{{\\mathrm{{hr}}}} \\frac{{1 \\mathrm{{hr}}}}{{3600 \\mathrm{{~s}}}} \\frac{{10^3 \\mathrm{{~m}}}}{{\\mathrm{{~km}}}}=\\frac{{100}}{{6}} \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}$
        The linear velocity of the contact point on the wheel is the same as the velocity of the axle.
        The angular velocity is
        $\\displaystyle \\omega=\\frac{{v}}{{r}}=\\left[\\frac{{100 \\mathrm{{~m}}}}{{6 \\mathrm{{~s}}}}\\right] \\frac{{1}}{{0.35 \\mathrm{{~m}}}}=\\frac{{47.6 \\mathrm{{rad}}}}{{\\mathrm{{s}}}}$
        Adding radians in the units makes the meaning clearer.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_10_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $1.2 \\mathrm{{~kg}}$ croquet ball moving at $2.0 \\mathrm{{~m}} / \\mathrm{{s}}$ struck from behind by the impulse force shown in Fig. 10-8. What is the final velocity of the croquet ball?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The initial momentum is
        $p_i=1.2 \\mathrm{{~kg}}(2.0 \\mathrm{{~m}} / \\mathrm{{s}})=2.4 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-10-8.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        The impulse integral, the difference in momentum. can be calculated by inspection. The area under the curve (one-half the base times the height) is
        $\\Delta(m v)=\\int F(t) d t=(1 / 2)(0.20 \\mathrm{{~s}}) 100 \\mathrm{{~N}}=10 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}}$
        The final momentum then is
        $p_f=12.4 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}}$
        and the final is
        $\\displaystyle v_f=\\frac{{12.4 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}}}}{{1.2 \\mathrm{{~kg}}}}=10.3 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_10_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $1.0 \\mathrm{{~kg}}$ steel ball is attached to a light-weight $1.0 \\mathrm{{~m}}$ long rod pivoted at the other end. The ball is released at the horizontal and strikes a $3.0 \\mathrm{{~kg}}$ steel block resting on a surface with coefficient of friction 0.25 . How far does the block travel?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This problem is similar to the ballistic pendulum problem in that conservation of energy and conservation of momentum and then work-encrgy have to be applied correctly. From the description of the collision assume that it is elastic.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-10-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        First calculate the velocity of the ball as it hits the block. Potential energy Goesinto kinetic energy. $m g h=(1 / 2) m v^2 \\quad$ or $\\quad v=\\sqrt{{2 g h}}=\\sqrt{{2\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 1.0 \\mathrm{{~m}}}}=4.4 \\mathrm{{~m}} / \\mathrm{{s}}$
        The collision is elastic and $v_{{\\mathrm{{BI}}}}=0$, so
        $\\displaystyle v_{{\\mathrm{{B}} 2}}=\\frac{{2 m_{{\\mathrm{{A}}}}}}{{m_{{\\mathrm{{A}}}}+m_{{\\mathrm{{B}}}}}} v_{{\\mathrm{{Al}}}}=\\frac{{2 m_{{\\mathrm{{A}}}}}}{{m_{{\\mathrm{{A}}}}+m_{{\\mathrm{{B}}}}}} \\sqrt{{2 g h}}=\\frac{{1}}{{2}} \\sqrt{{2 g h}}=2.2 \\mathrm{{~m}} / \\mathrm{{s}}$
        The gives the initial velocity of the struck block. The kinetic energy Goesinto work against friction.
        $\\displaystyle \\frac{{1}}{{2}} m_{{\\mathrm{{B}}}} v_{{\\mathrm{{B}} 2}}^2=\\mu m_{{\\mathrm{{B}}}} g L . \\quad \\text {{ or }} \\quad L=\\frac{{v_{{\\mathrm{{B}} 2}}^2}}{{2 \\mu g}}=\\frac{{1}}{{2 \\mu g}} \\frac{{2 g h}}{{4}}=\\frac{{h}}{{4 \\mu}}=\\frac{{1.0 \\mathrm{{~m}}}}{{4 \\cdot 0.25}}=1.0 \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_10_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $1.5 \\mathrm{{~kg}}$ block traveling at $6.0 \\mathrm{{~m}} / \\mathrm{{s}}$ strikes a $2.5 \\mathrm{{~kg}}$ block at rest. After an elastic collision, what are the velocities of the hlocks?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This is the special case where the struck block is at rest so
        $\\displaystyle v_{{\\mathrm{{A}} 2}}=\\frac{{m_{{\\mathrm{{A}}}}-m_{{\\mathrm{{B}}}}}}{{m_{{\\mathrm{{A}}}}+m_{{\\mathrm{{B}}}}}} v_{{\\mathrm{{A}} 1}}=\\frac{{1.5-2.5}}{{4.0}} 6.0 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}=-1.5 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}$
        $\\displaystyle v_{{\\mathrm{{B}} 2}}=\\frac{{2 m_{{\\mathrm{{A}}}}}}{{m_{{\\mathrm{{A}}}}+m_{{\\mathrm{{B}}}}}} v_{{A 1}}=\\frac{{2 \\cdot 1.5}}{{4.0}} 6.0 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}=4.5 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}$
        the striking block rebounds at (-) $1.5 \\mathrm{{~m}} / \\mathrm{{s}}$, and the struck block moves off at $4.5 \\mathrm{{~m}} / \\mathrm{{s}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_10_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $6.0 \\mathrm{{~g}}$ bullet is fired horizontally into a $2.8 \\mathrm{{~kg}}$ block resting on a horizontal surface with coefficient of friction 0.30 . The bullet comes to rest in the block, and the block slides $0.65 \\mathrm{{~m}}$ before coming to a stop. What is the velocity of the bullet?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-10-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        Assume the bullet comes to rest in the block before the block moves appreciably and that all the momentum in the bullet is transferred to the bullet-block combination.
        $m v=(m+M) V$
        Once the bullet-block combination is moving at $V$, the kinetic energy, $(1 / 2)(m+M) V^2$, Goesinto work to overcome friction $\\mu(m+M) g x$
        $(1 / 2)(m+M) V^2=\\mu(m+M) g x \\quad$ or $\\quad V^2 / 2=\\mu g x$
        Substituting
        $\\displaystyle \\frac{{1}}{{2}}\\left(\\frac{{m v}}{{m+M}}\\right)^2=\\mu g x$
        or
        $\\displaystyle v=\\frac{{m+M}}{{m}} \\sqrt{{2 \\mu g x}}=\\frac{{2.806}}{{0.006}} \\sqrt{{2(0.30)\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 0.65 \\mathrm{{~m}}}}=914 \\mathrm{{~m}} / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_10_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A ballistic pendulum, a device for measuring the speed of a bullet, consists of a block of wood suspended by cord. When the bullet is fired into the block, the block is free to rise. How high does a $5.0 \\mathrm{{~kg}}$ block rise when a $12 \\mathrm{{~g}}$ bullet traveling at $350 \\mathrm{{~m}} / \\mathrm{{s}}$ is fired into it?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-10-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        soln_s = f"""
        This is a most interesting and instructive problem. The collision between the bullet and the block is clearly inelastic (the bullet comes to rest in the block). Part of the kinetic energy of the bullet Goesinto friction as the bullet burrows its way into the block. Therefore mechanical energy is not conserved.

        Because the collision is inelastic, apply conservation of momentum of the collision. Before the collision, all the momentum is in the $m v$ of the bullet. After the collision, the momentum is in the $(m+M) V$ of the block and bullet. We assume that the bullet comes to rest (transfer all its momentum) before there is appreciable motion of the bullet-block combination.
        $m v=(m+M) V$
        After the collision, the rise of the block is determined by energy analysis. The kinetic energy of the block Goesinto potential energy.
        $(1 / 2)(m+M) V^2=(m+M) g h \\text {{ or }} V^2 / 2=g h$
        Substituting for $V$ from $m v=(m+M) V$
        $\\displaystyle \\frac{{1}}{{2}}\\left(\\frac{{m}}{{m+M}}\\right)^2 v^2=g h \\quad v=\\frac{{m+M}}{{m}} \\sqrt{{2 g h}} \\quad \\text {{ or }} \\quad h=\\frac{{v^2}}{{2 g}}\\left(\\frac{{m}}{{m+M}}\\right)^2$
        giving the relation hetween the velocity of the bullet and the height the block and bullet rise. For this problem
        $\\displaystyle h=\\frac{{(350 \\mathrm{{~m}} / \\mathrm{{s}})^2}}{{2 \\cdot 9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2}}\\left(\\frac{{0.012}}{{5.012}}\\right)^2=3.6 \\mathrm{{~cm}}$
        In this problem the 0.012 can be neglected in comparison to 5.0 . This is not always the case so we write $m+M$ as 5.012 as a reminder to include both $m+M$ in the calculation.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_9_11(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Rabbit food in the form of pellets is poured onto a scale pan at the rate of 400 pellets per second. Each pellet has a mass of $20 \\mathrm{{~g}}$ and falls a distance of $2.0 \\mathrm{{~m}}$. Assuming the pellets do not bounce, calculate the scale reading at $7.0 \\mathrm{{~s}}$.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        There are two components of the force. Momentum is transferred to the scale pan giving a constant force. The pellets are accumulating in the scale pan giving a time dependent part of the force.
        First calculate the velocity of the pellets as they hit the scale pan.
        $(1 / 2) m v^2=m g h \\quad$ or $\\quad v=\\sqrt{{2 g h}}=\\sqrt{{2\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 2.0 \\mathrm{{~m}}}}=6.26 \\mathrm{{~m}} / \\mathrm{{s}}$
        The momentum for each pellet is $m v=0.020 \\mathrm{{~kg}}(6.26 \\mathrm{{~m}} / \\mathrm{{s}})=0.125 \\mathrm{{~kg}} . \\mathrm{{m}} / \\mathrm{{s}}$
        The total force due to this momentum transfer is $400 \\mathrm{{mv}}$ 's per second
        $\\displaystyle F_{{\\text {{momentum }}}}=\\frac{{400(0.125 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}})}}{{1.0 \\mathrm{{~s}}}}=50 \\mathrm{{~N}}$
        The force due to accumulation is the weight of each pellet times the number per second ( $\\mu$ ). times the number of seconds.
        $\\displaystyle F_{{\\text {{accumulation }}}}=m g \\mu \\Delta t=20 \\times 10^{{-3}} \\mathrm{{~kg}}\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)(400 / \\mathrm{{s}}) 7.0 \\mathrm{{~s}}=549 \\mathrm{{~N}}$
        The scale reading at $7.0 \\mathrm{{~s}}$ is the sum of these two components or $599 \\mathrm{{~N}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_9_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $3.0 \\mathrm{{~kg}}$ cat is in a $24 \\mathrm{{~kg}}$ boat. The cat is $10 \\mathrm{{~m}}$ from the shore. The cat walks $3.0 \\mathrm{{~m}}$ toward the shore. How far is the cat from the shore? Assume no friction between boat and water.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        There are no external forces so the center of mass of the cat-boat system is constant. Knowing that the center of mass doesn't move is all that is necessary to do this problem.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-9-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        Referring to Fig. 9-5, write the center of mass of the cat-boat system before the cat walks. ( $M$ is the mass of the boat and $m$ the cat.) Then write the center of mass of the cat-boat system after the cat walks.
        $\\displaystyle x_{{c m}}=\\frac{{M x_b+m x_c}}{{M+m}} \\quad x_{{c m}}=\\frac{{M x_b^{{\\prime}}+m x_c^{{\\prime}}}}{{M+m}}$
        Because there are no external forces the centers of mass are the same so
        $M x_b+m x_c=M x_b^{{\\prime}}+m x_c^{{\\prime}}$
        Watch the algebra and solve for $x_{{\\mathrm{{c}}}}$. Use a word statement before writing the algebra statement. "Ihe final distance of the cat from the shore is equal to the original distance of the cat from the shore plus the displacement of the cat relative to the boat and the displacement of the boat relative to the shore." Read the sentence, look at the diagram, and write
        $x_c^{{\\prime}}=10 \\mathrm{{~m}}-3.0 \\mathrm{{~m}}+\\left(x_b^{{\\prime}}-x_b\\right)$
        Now substitute from $M\\left(x_b^{{\\prime}}-x_b\\right)=m\\left(x_c-x_c^{{\\prime}}\\right)$ or $x_b-x_b=(m / \\mathrm{{M}})\\left(x_c-x_{{\\mathrm{{c}}}}\\right)$
        $M\\left(x_b^{{\\prime}}-x_b\\right)=m\\left(x_c-x_c^{{\\prime}}\\right) \\text {{ or }} x_b^{{\\prime}}-x_b=(m / M)\\left(x_c-x_c^{{\\prime}}\\right),$
        $x_c^{{\\prime}}=7+(1 / 8)\\left(10-x_c^{{\\prime}}\\right)$ and $8 x_c^{{\\prime}}=56+\\left(10-x_c^{{\\prime}}\\right)$ so $x_c^{{\\prime}}=7.33 \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_9_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation described in problems 9-7 and 9-8, what is the rate of change in velocity for the boxcars?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This is a very interesting calculus problem that involves taking the total derivative. Since there are no external forces, the total change in $m v$ must equal zero so
        $d(m v)=m d v+v d m=0 \\quad \\text {{ or }} \\quad m d v=-v d m$
        Now write $m$ as a function of time $m=m_{{\\circ}}+r t=4000 \\mathrm{{~kg}}+(1.0 \\mathrm{{~kg}} / \\mathrm{{s}}) t$
        The derivative of $m$ is $d m=r d t$.
        $\\displaystyle \\frac{{d v}}{{v}}=-\\frac{{d m}}{{m}}=-\\frac{{r}}{{m_o+r t}} d t$
        Using the previous two equations and rearranging
        Introduce a change of variable
        $\\displaystyle u=m_o+r t \\text {{ with }} d u=r d t \\text {{ so }} \\frac{{d v}}{{v}}=-\\frac{{d u}}{{u}}$
        Integrating, 1n $v=-\\ln u+\\ln K$. Use $\\ln K$ because it is a convenient form for the constant.
        Now rearrange: $1 \\mathrm{{n}} v+\\ln u=\\ln K$ and $\\ln u v=\\ln K$ or $u v=K$
        change the variable back to $t$ so $\\left(m_0+r t\\right) v=\\mathrm{{K}}$.
        Evaluate the constant from the condition that at
        $t=0, m_n=14000 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}} \\text {{ so that }} K=14000 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}}$
        The relation between
        $\\displaystyle v=\\frac{{K}}{{m_o+r t}}=\\frac{{14000 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}}}}{{(4000 \\mathrm{{~kg}}+1.0 \\mathrm{{~kg}} / \\mathrm{{s}} \\cdot t)}}$
        The velocity at
        $\\displaystyle t=500 \\mathrm{{~s}} \\text {{ is }}\\left.v\\right|_{{t=500}}=\\frac{{14000 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}}}}{{4500 \\mathrm{{~kg}}}}=3.11 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}$
        Conservation of momentum and a little calculus produce the $v$ versus $t$ relation.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_9_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Continuing problem $9-7$, rain falls into the open boxcar so that the mass increases at $1.0 \\mathrm{{~kg}} / \\mathrm{{s}}$. What is the velocity of the boxcars at 500 s?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The total momentum of the boxcars is $4000 \\mathrm{{~kg}}(3.5 \\mathrm{{~m}} / \\mathrm{{s}}=14000 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}}$. Assume there is no horizontal component of the rain to change the momentum in the direction of motion of the boxcars. The mass increases by $(1.10 \\mathrm{{~kg}} / \\mathrm{{s}}) 3.5 \\mathrm{{~m}} / \\mathrm{{s}}=500 \\mathrm{{~kg}}$. The momentum is a constant, so the new velocity is
        $14000 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}}=(4500 \\mathrm{{~kg}}) v_R \\quad$ or $\\quad v_R=3.11 \\mathrm{{~m}} / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_9_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $3000 \\mathrm{{~kg}}$ closed boxcar traveling at $3.0 \\mathrm{{~m}} / \\mathrm{{s}}$ is overtaken by a $1000 \\mathrm{{~kg}}$ open boxcar traveling at $5.0 \\mathrm{{~m}} / \\mathrm{{s}}$. The cars couple together. Find the resultant speed of the combination.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The momentum before coupling is the same as the momentum after coupling (no external forces).
        $3000 \\mathrm{{~kg}}(3.0 \\mathrm{{~m}} / \\mathrm{{s}})+1000 \\mathrm{{~kg}}(5.0 \\mathrm{{~m}} / \\mathrm{{s}})=(4000 \\mathrm{{~kg}}) v \\quad \\text {{ or }} \\quad v=3.5 \\mathrm{{~m}} / \\mathrm{{s}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_9_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        James Bond is skiing along being pursued by Goldfinger, also on skis. Assume no friction. Mr. Bond, at $100 \\mathrm{{~kg}}$, fires backward a $40 \\mathrm{{~g}}$ bullet at $800 \\mathrm{{~m}} / \\mathrm{{s}}$. Goldfinger, at $120 \\mathrm{{~kg}}$, fires forward at Bond with a similar weapon. What is the relative velocity change after the exchange of 6 shots each? No bullets hit Bond or Goldfinger.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The problem is analyzed with conservation of momentum. The $m_b v_b$ of the bullet fired by Bond increases his momentum by $m_B \\delta v_B$. Remember that each bullet Bond fires increase his velocity. Set $m_b v_b=m_B \\delta v_B$ and solve for $\\delta v_B$
        $40 \\times 10^{{-3}} \\mathrm{{~kg}}(800 \\mathrm{{~m}} / \\mathrm{{s}})=(100 \\mathrm{{~kg}}) \\Delta v_B \\quad \\text {{ or }} \\quad \\Delta v_B=0.32 \\mathrm{{~m}} / \\mathrm{{s}}$
        The $40 \\times 10^{{-3}} \\mathrm{{~kg}}$ bullet is small compared to the $100 \\mathrm{{~kg}}$ of Bond. and it would not affect the calculation. The $\\delta v$ notation is used to indicate that each bullet fired by Bond causes a change in his velocity.
        Goldfinger, on the other hand, has his momentum decreased. In his case, $m_b v_b=m_G \\delta v_G$.
        Putting in the numbers
        $32 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}}=(120 \\mathrm{{~kg}}) \\Delta v_G \\quad \\text {{ or }} \\quad \\Delta v_G=0.26 \\mathrm{{~m}} / \\mathrm{{s}}$
        Bond goes faster and Goldfinger goes slower with the total change in velocity $0.58 \\mathrm{{~m}} / \\mathrm{{s}}$ for each pair of shots fired. For six shots this amounts to a difference of $3.48 \\mathrm{{~m}} / \\mathrm{{s}}$. If Bond and Goldfinger had been traveling at the same speeds, then after this exchange Bond would have a relative speed advantage of $3.48 \\mathrm{{~m}} / \\mathrm{{s}}$.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_9_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $75 \\mathrm{{~kg}}$ hockey player traveling at $12 \\mathrm{{~m}} / \\mathrm{{s}}$ collides with a $90 \\mathrm{{~kg}}$ player traveling, at right angles to the first, at $15 \\mathrm{{~m}} / \\mathrm{{s}}$. The players stick together. Find their resultant velocity and direction. Assume the ice surface to be frictionless.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This problem can be analyzed by conservation of momentum. Calculate the momenta and draw a vector diagram.
        $p_1=75 \\mathrm{{~kg}}(12 \\mathrm{{~m}} / \\mathrm{{s}})=900 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}} \\text {{ and }} p_2=90 \\mathrm{{~kg}}(15 \\mathrm{{~m}} / \\mathrm{{s}})=1350 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-9-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        The angle of the two hockey players is from $\\tan \\theta=\\frac{{1350}}{{900}}=1.5 \\quad$ or $\\quad \\theta=56^{{\\circ}}$ or $\\theta-56^{{\\circ}}$,
        and the resulting momentum is $p=\\sqrt{{1350^2+900^2}} \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}}=1620 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}}$
        The players move off with velocity
        $\\displaystyle v=\\frac{{p}}{{m_1+m_2}}=9.83 \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}}}$
        at an angle of $56^{{\\circ}}$ to the original direction of the $75 \\mathrm{{~kg}}$ player.

        Second Solution: A more formal approach is to write a conservation of momentum statement equating the total (vector) momentum before the collision to the total (vector) momentum after the collision. Take the plus $i$ direction as the initial direction of the first player and the plus $j$ direction as the original direction of the second player. Using the numbers already calculated
        $[900 i+1350 j] \\mathrm{{kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}}=(165 \\mathrm{{~kg}}) v_f \\quad$ or $\\quad v_f=[5.45 i+8.18 j] \\mathrm{{m}} / \\mathrm{{s}}$
        As an exercise verify the final velocity of $9.83 \\mathrm{{~m}} / \\mathrm{{s}}$ at the $56^{{\\circ}}$ angle.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_9_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Make the pellet gun of problem 9-3 fully automatic and capable of firing 10 pellets per second. Calculate the force these pellets make on a target where the pellets do not bounce.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This problem is solved by calculating the average momentum transferred to the target per unit of time. The momentum of each pellet is $2.4 \\mathrm{{~kg}} . \\mathrm{{m}} / \\mathrm{{s}}$. The force on the target is calculated from the simple expression
        $F=\\frac{{\\Delta p}}{{\\Delta l}}=\\frac{{10(2.4 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}})}}{{1.0 \\mathrm{{~s}}}}=24 \\mathrm{{~N}}$
        The total momentum transferred each second is 10 individual momenta of each pellet.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_9_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A 5.0g pellet is compressed against a spring in a gun of mass $300 \\mathrm{{~g}}$. The spring is released and the gun allowed to recoil with no friction as the pellet leaves the gun. If the speed of the recoiling gun is $8.0 \\mathrm{{~m}} / \\mathrm{{s}}$, what is the speed of the pellet?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This problem is solved by application of the law of conservation of momentum. This law can be applied because there is a no external force. Since there is no external force, all the $m v$'s must add to zero.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-9-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        $m_g v_g=m_p v_p \\quad$ or $\\quad m_g v_g-m_p v_p=0$
        The conservation of momentum statement on the left is hased on the simple observation "bullet gn one way, gun go the other," while the formal statement that the $m v$'s add to zero is on the right. With a well-labeled diagram of the situation, the statement on the left is probably easier to visualize, that is, the momenta are equal and opposite. Putting in the numbers
        $\\displaystyle 300 \\mathrm{{~g}}(8.0 \\mathrm{{~m}} / \\mathrm{{s}})=5.0 \\mathrm{{~g}} \\cdot v_p \\quad \\text {{ or }} \\quad v_p=480 \\mathrm{{~m}} / \\mathrm{{s}}$
        As a check, not that the momentum of the gun $p_s=m_s v_s=2.4 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}}$ and the momentum of the pellet $p_p=m_p v_p=$ $2.4 \\mathrm{{~kg}} . \\mathrm{{m}} / \\mathrm{{s}}$ are numerically equal and since they are in opposite directions add to zero.
        The energy of each is $m v^2 / 2$ or $p^2 / 2 m$. So for the gun
        $\\displaystyle K E_{{\\mathrm{{gun}}}}-\\frac{{(2.4 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}})^2}}{{2 \\cdot 0.30 \\mathrm{{~kg}}}}=9.6 \\mathrm{{~J}}$
        Performing the same calculation for the pellet
        $\\displaystyle K E_{{\\text {{pellet }}}}=\\frac{{(2.4 \\mathrm{{~kg}} \\cdot \\mathrm{{m}} / \\mathrm{{s}})^2}}{{2 \\cdot 0.010 \\mathrm{{~kg}}}}=288 \\mathrm{{~J}}$
        The total energy stored in the spring is the sum of these energies or 298J.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_9_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the same collection of masses at the same points, as in problem 9-1, add forces as shown in Fig. 9-2 to each mass and find the resultant acceleration of the center of mass.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-9-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        soln_s = f"""
        Remember: the acceleration of the center of mass is the total mass times the vector sum of all these forces. Applying equation $9-3$
        $(50 \\mathrm{{~kg}}) a_{{c m}}=\\left[30 \\cos 45^{{\\circ}} i+30 \\sin 45^{{\\circ}} j-24 i-40 j\\right] \\mathrm{{N}}=[-3 i-19 j] \\mathrm{{N}}$
        so
        $\\displaystyle a_{{c m}}=\\left[-\\frac{{3}}{{50}} i-\\frac{{19}}{{50}} j\\right] \\frac{{\\mathrm{{m}}}}{{\\mathrm{{s}}^2}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_9_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the center of mass for a distribution of mass points as shown in Fig. 9-1.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-9-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        The $r_{{\\mathrm{{cm}}}}$ form indicates that the calculation is to be done in vector notation, so (The units are often left our of calculations involving unit vectors and added at the end.)
        $\\displaystyle r_{{c m}}=\\frac{{1}}{{50}}[20(3 i+3 j)+18(-2 i-2 j)+12(i-j)]$
        $\\displaystyle \\left.r_{{c m}}-\\frac{{1}}{{50}}[36 i+12 j)\\right]-\\frac{{36}}{{50}} i+\\frac{{12}}{{50}} j=0.72 i+0.24 j$
        These mass points act as if all of their mass $(50 \\mathrm{{~kg}})$ were at the point $(0.72,0.24)$. If these three masses were placed on a plate of negligible mass, the balance point would be at $(0.72,0.24)$.

        The law of conservation of momentum can be viewed as a consequence of the statement: The total mass of a collection of particles times the acceleration of the center of mass equals the applied, or external, force, that is, the sum of the forces on all the individual component masses. This is of course a vector equation.
        $M a_{{c m}}=F_{{e x t}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_8_11(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider an elevator weighing $4000 \\mathrm{{~N}}$ held $5.0 \\mathrm{{~m}}$ above a spring with a force constant of $8000 \\mathrm{{~N}} / \\mathrm{{m}}$. The elevator falls onto the spring while subject to a frictional retarding force (brake) of $1000 \\mathrm{{~N}}$. Describe the motion of the elevator.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-8-8.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        The potential energy of the elevator with respect to the top of the spring is
        $m g h=4000 \\mathrm{{~N}}(5.0 \\mathrm{{~m}})=20000 \\mathrm{{~J}}$
        When the elevator falls, this energy, less the energy lost due to the frictional brake, is available to compress the spring.
        $20000 \\mathrm{{~J}}-1000 \\mathrm{{~N}}(5.0 \\mathrm{{~m}})=15000 \\mathrm{{~J}}=\\text {{ Energy to compress spring }}$
        The friction brake stays on while the elevator is in contact with the spring so this energy Goesinto $k x^2 / 2+f x$. In cquation form this is
        $15000 \\mathrm{{~J}}=k\\left(x^2 / 2\\right)+f x$
        This expression is only approximately correct since if the spring is compressed a distance $x$, then this $m g x$ provides an additional amount of potential energy. The correct equation that determines the compression of the spring is
        $\\displaystyle m g(h+x)=\\frac{{k x^2}}{{2}}+f(h+x) \\quad \\text {{ or }} \\quad \\frac{{k x^2}}{{2}}+(f-m g) x+(f-m g) h=0$
        Putting numbers into the equation
        $(4000 \\mathrm{{~N}} / \\mathrm{{m}}) x^2+(1000 \\mathrm{{~N}}-4000 \\mathrm{{~N}}) x+(1000 \\mathrm{{~N}}-4000 \\mathrm{{~N}}) 5.0 \\mathrm{{~m}}=0$
        The units are correct so $4000 x^2-3000 x-15000=0$ or $4 x^2-3 x-15=0$ which has solution
        $\\displaystyle x=\\frac{{3 \\pm \\sqrt{{9-4(4)(-15)}}}}{{2(4)}}=2.3 \\mathrm{{~m}}$
        The energy stored in the spring at this (compressed) point is
        $\\displaystyle \\frac{{k x^2}}{{2}}=8000 \\mathrm{{~N}} / \\mathrm{{m}} \\frac{{(2.3 \\mathrm{{~m}})^2}}{{2}}=22040 \\mathrm{{~J}}$
        This amount of energy will raise the elevator and do work against frietion according to the equation
        $22040 \\mathrm{{~J}}=m g x^{{\\prime}}+f x^{{\\prime}} \\quad \\text {{ or }} \\quad 22040 \\mathrm{{~J}}=(5000 \\mathrm{{~N}}) x^{{\\prime}} \\quad \\text {{ or }} \\quad x^{{\\prime}}=4.4 \\mathrm{{~m}}$
        This places the elevator $2.1 \\mathrm{{~m}}$ above the spring.
        This analysis can be repeated until the elevator comes to a stop. As an exercise perform one more "bounce" of the elevator starting at $2.1 \\mathrm{{~m}}$ above the spring.

        The elevator will continue to bounce up and down off the spring until the weight of the elevator is equal to the force of the spring plus the frictional force
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_8_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        If $30 \\%$ of the energy lost in problem 8-9 is available to move the block along a surface with coefficient of friction 0.80 , how far will it move?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Equate the energy available to the work performed
        $0.30 \\Delta K E=\\mu \\mathrm{{mg}} L \\quad \\text {{ so }} \\quad 0.30(488 \\mathrm{{~J}})=0.80(3.0 \\mathrm{{~kg}})\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) L \\quad \\text {{ or }} \\quad L=0.2 \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_8_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider a 10gm bullet passing through a $3.0 \\mathrm{{~kg}}$ block resting on a table. The velocity of the bullet is 400 $\\mathrm{{m}} / \\mathrm{{s}}$ on entering the block and $250 \\mathrm{{~m}} / \\mathrm{{s}}$ on exit. Calculate the energy lost by the bullet in passing through the block.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The energy is all kinetic, so calculating before and after
        $K E_{{\\mathrm{{i}}}}=(1 / 2)\\left(10 \\times 10^{{-3}} \\mathrm{{~kg}}\\right)(400 \\mathrm{{~m}} / \\mathrm{{s}})^2=800 \\mathrm{{~J}}$
        $K E_{{\\mathrm{{f}}}}=(1 / 2)\\left(10 \\times 10^{{-3}} \\mathrm{{~kg}}\\right)(250 \\mathrm{{~m}} / \\mathrm{{s}})^2=312 \\mathrm{{~J}}$
        The kinetic energy lost or $\\Delta K E=488 \\mathrm{{~J}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_8_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the track shown in problem 8-7 take $\\mu=0.10, R=1.0 \\mathrm{{~m}}$, and $L=2.0 \\mathrm{{~m}}$ and find where the mass stops.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The only way the mass loses energy is by sliding along the flat "friction" part of the track. The initial potential energy Goesinto frictional work.
        $(m g) 1.0 \\mathrm{{~m}}=0.10(m g x) \\quad$ or $\\quad x=10 \\mathrm{{~m}}$
        The mass crosses the friction area 5 times and ends up at the edge of the friction surface opposite from where it started.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_8_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Place a mass on a track made up of a flat section, $L$, with coefficient of friction, $\\mu$, and two frictionless semicircular surfaces of radius $R$. Let the mass start from the top of one of the semicircles and calculate where it comes to rest.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-8-7.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        The initial potential energy is $m g R$. When the mass encounters the friction surface, this (potential) energy is dissipated in doing work to overcome friction. Assuming the energy lost due to friction in one traverse is less than the initial potential energy, the mass will rise to a height (on the opposite semicircle) $R^{{\\prime}}$ dictated by the energy statement
        $m g R^{{\\prime}}=m g R \\quad \\mu m g L$
        After another traverse of the flat portion of the track the height will be dictated by
        $m g R^{{\\prime \\prime}}=m g R-2 \\mu m g L$
        and so on until all the original potential energy is dissipated.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_8_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Go back to an incline plane problem (similar to problem 8-3) but with the addition of a spring with a force constant of $300 \\mathrm{{~N}} / \\mathrm{{m}}$ placed at the bottom of the incline, as shown in Fig. 8-6. Calculate the compression of the spring when the block slides down the plane.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        An approximate answer can be obtained by setting $m g h$, where $h$ is the vertical distance from the block to the spring, equal to $k x^2 / 2$, the compression of the spring.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-8-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        $\\displaystyle 2.5 \\mathrm{{~kg}}\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 4.0 \\mathrm{{~m}}=\\frac{{300 \\mathrm{{~N}}}}{{\\mathrm{{~m}}}} \\frac{{x^2}}{{2}} \\quad \\text {{ or }} \\quad x=0.81 \\mathrm{{~m}}$
        The compression of the spring gives the block a little more potential energy so a more correct statement would take into account this "additional height," $h$.
        $\\displaystyle m g h+(m g x) \\sin 40^{{\\circ}}=\\frac{{k x^2}}{{2}} \\quad h^{{\\prime}}=x\\left(\\sin 40^{{\\circ}}\\right)$
        $\\displaystyle 2.5 \\mathrm{{~kg}}\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 4.0 \\mathrm{{~m}}+2.5 \\mathrm{{~kg}}\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) x\\left(\\sin 40^{{\\circ}}\\right)=\\frac{{300 \\mathrm{{~N}} x^2}}{{\\mathrm{{~m}}}}$
        Eliminating the units and cleaning up the numbers yields a quadratic
        $150 x^2-15.7 x-98=0$
        with solution
        $\\displaystyle x=\\frac{{15.7 \\pm \\sqrt{{15.7^2-4(150)(-98)}}}}{{2(150)}}=\\frac{{15.7 \\pm 243}}{{300}}=0.86 \\mathrm{{~m}}$
        This value is somewhat larger than calculated above as expected, due to the "additional height."
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_8_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation described in problem 8-4, find the velocity with which the block leaves the spring.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Start this problem with the spring compressed the $0.91 \\mathrm{{~m}}$. At this point the potential energy stored in the spring is used to push the block $0.91 \\mathrm{{~m}}$ back across the friction surface. The energy remaining as the block leaves the spring is the original kinetic energy less the energy expended in two $0.91 \\mathrm{{~m}}$ trips across this friction surface. In word-equation form this is
        $K E_{{\\text {{initial }}}}-\\mathrm{{Two}}\\left(\\mu m g x^{{\\prime}} s\\right)=K E_{{\\text {{final }}}}$
        This is a Goesinto problem. Initial $K E$ Goesinto two trips across the friction surface plus the final $K E$. The total work performed in the block moving across this surface is
        $2\\left[0.25(3.0 \\mathrm{{~kg}})\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 0.91 \\mathrm{{~m}}\\right]=13.4 \\mathrm{{~J}}$
        The velocity of the block as it flies off the spring is
        $\\displaystyle \\frac{{(3.0 \\mathrm{{~kg}}) 144 \\mathrm{{~m}}^2 / \\mathrm{{s}}^2}}{{2}}-13.4 \\mathrm{{~J}}-\\frac{{(3.0 \\mathrm{{~kg}}) v_f^2}}{{2}} \\quad \\text {{ or }} \\quad v_f=11.6 \\mathrm{{~m}} / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_8_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $5.0 \\mathrm{{~kg}}$ block sliding at $12 \\mathrm{{~m}} / \\mathrm{{s}}$ along a frictionless surface encounters a spring of force constant of 500 $\\mathrm{{N}} / \\mathrm{{m}}$ resting on a surface with coefficient of friction 0.25 . The friction surface is only under the spring. What is the maximum compression of the spring?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Before looking at Fig. 8-5 try drawing the diagram yourself. On exams you sometimes have to work a problem from only a word description. As you are working through these problems practice drawing the diagrams, and do not be hesitant in using several diagrams. Notice the previous problem where the system was depicted with the vector diagram beside it. Sketching the system and all the vectors on a single diagram is unnecessary confusion. Use multiple diagrams and reduce the clutter.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-8-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        When the block encounters the spring and the friction surface all the energy is kinetic, $m v^2 / 2$. This energy Goesinto compressing the spring and sliding along the friction surface. In equation form
        $\\displaystyle \\frac{{m v^2}}{{2}}=\\frac{{k x^2}}{{2}}+\\mu m g x$
        Re careful in writing this statement. Many people write the kinetic energy correctly, the energy to compress the spring correctly, and then write the force to slide along the friction surface. Be aware of this mistake and check the units to be sure the units on one side of the equation are the same as the units on the other side of the equation. You may have noticed that in all problems the first equation is written with units just as a reminder that checking units will reduce the number of errors. It is very frustrating to make a simple mistake that could have been avvided had units been cliecked in the problen before proceeding with the numerical calculations.
        Putting in numbers
        $\\displaystyle \\frac{{3.0 \\mathrm{{~kg}}(12 \\mathrm{{~m}} / \\mathrm{{s}})^2}}{{2}}=\\frac{{500 \\mathrm{{~N}}}}{{\\mathrm{{~m}}}} \\frac{{x^2}}{{2}}+0.25(3.0 \\mathrm{{~kg}})\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) x$
        The units are correct so without units the equation is
        $250 x^2+7.35 x-216=0$
        and
        $\\displaystyle x=\\frac{{-7.35 \\pm \\sqrt{{7.35^2-4(250)(-216)}}}}{{2 \\cdot 250}}=\\frac{{-7.4 \\pm 464.8}}{{500}}=0.91 \\mathrm{{~m}}$
        This is the amount the spring is compressed as the block comes to a stop.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_8_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider the incline plane with friction shown in Fig. 8-4 and calculate the velocity of the block at the bottom of the plane using force analysis, then work-energy analysis.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The vector diagram is also shown in Fig. 8-4. The unbalanced force of $20 \\mathrm{{~N}}$ acts on the 5.0kg block causing it to accelerate at $4.0 \\mathrm{{~m}} / \\mathrm{{s}}^2$ down the plane.
        The slant height of the plane is $6.0 \\mathrm{{~m}} / \\sin 35^{{\\circ}}=10.5 \\mathrm{{~m}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-8-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        This acceleration over the $10.5 \\mathrm{{~m}}$ results in a velocity
        $v^2=2 a\\left(x-x_o\\right)=2\\left(4.0 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 10.5 \\mathrm{{~m}} \\quad \\text {{ or }} \\quad v=9.2 \\mathrm{{~m}} / \\mathrm{{s}}$
        Second Solution: Now do the same problem using work-energy analysis. When the block is at the top of the plane, the energy is
        $m g h=5.0 \\mathrm{{~kg}}\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 6.0 \\mathrm{{~m}}=294 \\mathrm{{~J}}$
        At the bottom of the plane all of this (potential) energy has gone into kinetic energy (velocity of the block) except for the amount used to do work against the frictional force. The work (against the frictional force) is
        $\\mu N L=0.20(40 \\mathrm{{~N}}) 10.5 \\mathrm{{~m}}=84 \\mathrm{{~J}}$
        The energy "left" after the block has slid down the plane is the $294 \\mathrm{{~J}}$ at the top of the plane minus the $84 \\mathrm{{~J}}$ lost due to friction, and this energy (=210J) is manifest in the velocity of the block (kinetic energy). This "energy" is equal to the $m v^2 / 2$, the kinetic energy of the block at the bottom of the plane.
        $\\displaystyle \\frac{{(5.0 \\mathrm{{~kg}}) v^2}}{{2}}=210 \\mathrm{{~J}} \\quad \\text {{ or }} \\quad v=9.2 \\mathrm{{~m}} / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_8_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Now complicate problem 8-1 by adding a coefficient of friction of 0.20 for the horizontal surface. How far does the block slide while compressing the spring?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Now the potential energy Goesinto compressing the spring and overcoming friction. This statement is very helpful in writing the equation. In words, the potential energy, $m g h$, equals the energy to compress the spring, $k x^2 / 2$ plus the work to overcome friction, $\\mu m g x$, or
        $\\displaystyle m g h=\\frac{{k x^2}}{{2}}+\\mu m g x$
        or
        $100 \\mathrm{{~J}}=(200 \\mathrm{{~N}} / \\mathrm{{m}}) x^2+0.20(3.0 \\mathrm{{~kg}})\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) x$
        Eliminating the units, which are correct, the equation becomes
        $200 x^2+5.9 x-100=0$
        with solutions
        $\\displaystyle x=\\frac{{-5.9 \\pm \\sqrt{{5.9^2-4(200)(-100)}}}}{{2(200)}}=\\frac{{-5.9 \\pm 283}}{{400}}=0.70 \\mathrm{{~m}}$
        The negative root is inappropriate for this problem. The bluck compresses the spring 0.70 m while sliding on this frictional surface.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_8_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Place a $3.0 \\mathrm{{~kg}}$ block at the top of a $3.4 \\mathrm{{~m}}$ high frictionless incline. At the bottom of the incline the block encounters a spring with a constant of $400 \\mathrm{{~N}} / \\mathrm{{m}}$ on a horizontal surface. No energy is lost to friction. How far is the spring compressed?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-9-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        The energy at the top of the plane, which is the same as the energy at the bottom of the plane, Goesinto compressing the spring. The energy at the top of the plane is, from the example problem, 100J so
        $\\displaystyle 100 \\mathrm{{~J}}=\\frac{{k x^2}}{{2}} \\quad \\text {{ or }} \\quad x=\\left[\\frac{{200 \\mathrm{{~J}} \\cdot \\mathrm{{m}}}}{{400 \\mathrm{{~N}}}}\\right]^{{1 / 2}}=0.71 \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_7_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the work performed in compressing the spring of problem 7-4 the next $3.0 \\mathrm{{~cm}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The general expression for the work performed is $k x^2 / 2$. The work to compress the spring from $3.0 \\mathrm{{~cm}}$ to $6.0 \\mathrm{{~cm}}$ is the work to compress it $6.0 \\mathrm{{~cm}}$ less the work necessary to compress it $3.0 \\mathrm{{~cm}}$.
        $\\displaystyle W_{{3 \\rightarrow 6}}=\\frac{{200 \\mathrm{{~N}}}}{{\\mathrm{{~m}}}} \\cdot \\frac{{1}}{{2}}\\left[\\left(6.0 \\times 10^{{-2}} \\mathrm{{~m}}\\right)^2-\\left(3.0 \\times 10^{{-2}} \\mathrm{{~m}}\\right)^2\\right]=0.27 \\mathrm{{~J}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_7_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the work performed in compressing a spring with a force constant of $200 \\mathrm{{~N}} / \\mathrm{{m}}$ the first $3.0 \\mathrm{{~cm}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The work performed is calculated with equation 7-5 as
        $\\displaystyle W=\\frac{{k x^2}}{{2}}=\\frac{{200 \\mathrm{{~N}}}}{{\\mathrm{{~m}}}} \\cdot \\frac{{\\left(3.0 \\times 10^{{-2}} \\mathrm{{~m}}\\right)^2}}{{2}}=9.0 \\times 10^{{-2}} \\mathrm{{~N}} \\cdot \\mathrm{{m}}=9.0 \\times 10^{{-2}} \\mathrm{{~J}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_7_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the work done by the force of gravity acting on a $70 \\mathrm{{~kg}}$ student sliding down a $30^{{\\circ}}$ inclined slide a slant distance of $10 \\mathrm{{~m}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The first thing to calculate is the force due to gravity, $m \\mathrm{{~g}}$, which is $70 \\mathrm{{~kg}} .9 .8 \\mathrm{{~m}} / \\mathrm{{s}}_2=686 \\mathrm{{~N}}$. Place this vector on Fig. 7-4 pointing down.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-7-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        Calculate the force in the direction of the incline $F=686 \\mathrm{{~N}} \\cos 60^{{\\circ}}=343 \\mathrm{{~N}}$
        The work performed by the gravitational force is $W=F\\cdot s=343 \\mathrm{{~N}} \\cdot 10 \\mathrm{{~m}}=3430 \\mathrm{{~J}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_7_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the work performed by a constant force of $40 \\mathrm{{~N}}$ pushing a block a distance of $3.0 \\mathrm{{~m}}$ along a horizontal surface at an angle of $20^{{\\circ}}$ with the horizontal as shown in Fig. 7-3.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-7-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        soln_s = f"""
        The work performed is calculated with equation 7-2.
        $W=F \\cdot s=40 \\mathrm{{~N}} \\cos 20^{{\\circ}} \\cdot 3.0 \\mathrm{{~m}}=113 \\mathrm{{~J}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_7_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Calculate the work performed by a constant force of $60 \\mathrm{{~N}}$ acting in the direction of a displacement of $3.0 \\mathrm{{~m}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-9-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        The force is acting in the direction of the displacement, so the work is the product of force times displacement.
        $W=F \\cdot x=60 \\mathrm{{~N}}\\cdot3.0 \\mathrm{{~m}}=180 \\mathrm{{~J}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_6_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $16 \\mathrm{{~kg}}$ monkey wishes to raise a $20 \\mathrm{{~kg}}$ mass by climbing (accelerating) up a rope that passes over a pulley attached to the mass. How much must the monkey accelerate up the rope in order to raise the mass?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-6-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')

        soln_s = f"""
        The mass produces a force or tension in the rope of
        $T_{{\\text {{mass }}}}=20 \\mathrm{{~kg}}\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)=196 \\mathrm{{~N}}$
        The mass of the monkey hanging on the rope produces a force of
        $T_{{\\text {{monkey }}}}=16 \\mathrm{{~kg}}\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)=157 \\mathrm{{~N}}$
        To just balance the mass the monkey must accelerate up the rope to produce a force of (196-157) $\\mathrm{{N}}=39 \\mathrm{{~N}}$.
        $39 \\mathrm{{~N}}=(16 \\mathrm{{~kg}}) a \\text {{ or }} \\quad a=2.4 \\mathrm{{~m}} / \\mathrm{{s}}^2$
        At this acceleration, the tension in the rope is $196 \\mathrm{{~N}}$.
        To raise the mass, the monkey must accelerate at a rate greater than $2.4 \\mathrm{{~m}} / \\mathrm{{s}}^2$.
        Any acceleration greater than $2.4 \\mathrm{{~m}} / \\mathrm{{s}}^2$ will increase the tension in the rope by an amount equal to the "additional" acceleration times the mass of the monkey.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_6_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        $100 \\mathrm{{~kg}}$ astronaut produces a force (weight) on the surface of the earth of $980 \\mathrm{{~N}}$. What force (weight) would the astronaut produce on the surface of the moon where the " $\\mathrm{{g}}$ " is about one-sixth of the $g$ on earth?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The force or weight would be
        $W_{{\\text {{moon }}}}=100 \\mathrm{{~kg}}\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) \\frac{{1}}{{6}}=163 \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_6_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the tension in a rope with the $70 \\mathrm{{~kg}}$ mountain climber of problem 6-3 accelerating (climbing) up the rope at a constant acceleration of $1.0 \\mathrm{{~m}} / \\mathrm{{s}}^2$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        In this case the tension in the rope is increased by ma.
        $T=(9.8+1.0) \\mathrm{{m}} / \\mathrm{{s}}^2(70 \\mathrm{{~kg}})=756 \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_6_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A rope (fastened at the top) is hanging over a cliff. What is the tension in the rope with a $70 \\mathrm{{~kg}}$ mountain climber sliding down the rope at a constant acceleration of $6.0 \\mathrm{{~m}} / \\mathrm{{s}}^2$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Performing a little "thought experiment," the tension in the rope is less than if he were at zero velocity or sliding at constant velocity. In this instance the $6.0 \\mathrm{{~m}} / \\mathrm{{s}}^2$ must be subtracted from the $9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2$.
        $T=(9.8-6.0) \\mathrm{{m}} / \\mathrm{{s}}^2(70 \\mathrm{{~kg}})=266 \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_6_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $12 \\mathrm{{~kg}}$ flower pot is hanging by a cord from the roof of an elevator. What is the tension in the cord when the elevator is stationary and when it is accelerating upward at $3.0 \\mathrm{{~m}} / \\mathrm{{s}}^2$ ?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First Solution: When the elevator is stationary the tension in the cord is
        $T=12 \\mathrm{{~kg}}\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)=118 \\mathrm{{~N}}$
        When the elevator is accelerating upward the tension in the cord is increased by an amount equal to ma
        $T=118 \\mathrm{{~N}}+12 \\mathrm{{~kg}} \\cdot 3.0 \\mathrm{{~m}} / \\mathrm{{s}}^2=154 \\mathrm{{~N}}$
        
        Second Solution: Now take another, more analytical, look at the problem. When the elevator is at rest the tension in the cord equals $m g$. so writing a force statement, $T-m g=m a$ where $a$ is zero, produces $T=m g$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-6-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')

        soln_s = f"""
        When the elevator is accelerating upward at $3.0 \\mathrm{{~m}} / \\mathrm{{s}}^2$ the $T-m g=m a$ equation has an acceleration. The hard part of doing the problem this way is keeping the algebraic sign of the acceleration correct. Looking at Fig. 6-1, $T$ and $a$ have the same sign so write
        $T-12 \\mathrm{{~kg}}\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)=12 \\mathrm{{~kg}}\\left(3.0 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) \\quad \\text {{ or }} \\quad T=154 \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_18(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the speed, at which no side force due to friction is required, for a car traveling at a radius of $240 \\mathrm{{~m}}$ on a $20^{{\\circ}}$ banked road?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Figure 5-20 shows a profile of the car on the banking and the vector diagram.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-20.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        Start with $m g$ acting down. The normal force (between the car and road) is normal to the surface with the vertical component equal to $m g$ and the horizontal component equal to $m v^2 / r$.
        $N \\cos 20^{{\\circ}}=m g \\quad N \\sin 20^{{\\circ}}=m v^2 / r$
        And again,
        $\\displaystyle \\frac{{N \\sin 20^{{\\circ}}}}{{N \\cos 20^{{\\circ}}}}=\\tan 20^{{\\circ}}=\\frac{{v^2}}{{r g}}$
        so,
        $v=\\sqrt{{r g \\tan 20^{{\\circ}}}}=\\sqrt{{240 \\mathrm{{~m}}\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) \\tan 20^{{\\circ}}}}=29 \\mathrm{{~m}} / \\mathrm{{s}}$
        Notice that in this problem the mass of the car does not enter into the calculation.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_17(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A conical pendulum is a mass on the end of a cord, where the mass moves at constant speed in a circle with the cord tracing out a cone. A conical pendulum of length $1.2 \\mathrm{{~m}}$ moves in a circle of radius $0.20 \\mathrm{{~m}}$. What is the period of the pendulum?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Note that the mass is not given. Start with a vector diagram of the forces on the mass as shown in Fig. $5-19$.
        The $m g$ must equal the vertical component of the tension in the string $T \\cos \\theta=m g$.
        The horizontal component is due to the centripetal force so $T \\sin \\theta=m v^2 / r$.
        Dividing the second equation by the first eliminates $T$ and $m$ or $\\tan \\theta=v^2 / r g$.
        The angle is from $\\sin \\theta=r / L$ making $\\tan \\theta=0.17$ and
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-19.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        
        soln_s = f"""
        $v=\\sqrt{{r g \\tan \\theta}}=\\sqrt{{0.20 \\mathrm{{~m}}\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 0.17}} =0.57 \\mathrm{{~m}} / \\mathrm{{s}}$
        The period is from $2 \\pi r / T=v$ or $\\displaystyle T =\\frac{{2 \\pi r}}{{v}}=\\frac{{2 \\pi(0.20 \\mathrm{{~m}})}}{{0.57 \\mathrm{{~m}} / \\mathrm{{s}}}}=2.2 \\mathrm{{~s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_16(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A 1-oz gold coin is placed on a turntable turning at 33-1/3 RPM (revolutions per minute). What is the coefficient of friction between the coin and the turntable if the maximum radius. before the coin slips. is $0.14 \\mathrm{{~m}}$ ?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The frictional force between the coin and turntable provides the center-directed force to keep the coin on the turntable (see Fig. 5-18). This center-directed force must equal $m y^2 / r$.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-18.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        This is a force-balance problem. In equation form: frictional force $=$ centripetal force, or $\\mu m g=m v^2 / r \\quad$ or $\\quad \\mu=v^2 / r g$
        The velocity is $2 \\pi \\mathrm{{r}}$ times the number of $2 \\pi r$ 's per minute, $100 / 3$, or
        $\\displaystyle v=2 \\pi \\cdot 0.14 \\mathrm{{~m}} \\frac{{100}}{{3}} \\frac{{1}}{{\\min }} \\frac{{1 \\mathrm{{~min}}}}{{60 \\mathrm{{~s}}}}=0.49 \\mathrm{{~m}} / \\mathrm{{s}}$
        Therefore,
        $\\displaystyle \\mu=\\frac{{v^2}}{{r g}}=\\frac{{(0.49 \\mathrm{{~m}} / \\mathrm{{s}})^2}}{{0.14 \\mathrm{{~m}}\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)}}=0.18$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_15(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An $0.60 \\mathrm{{~kg}}$ rubber stopper is whirled in a horizontal circle of $0.80 \\mathrm{{~m}}$ radius at a rate of 3.0 revolutions per second. What is the tension in the string?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        
        soln_s = f"""
        Three revolutions per second means three $2 \\pi \\mathrm{{r}}$ 's (circumferences) per second so
        $\\displaystyle v=\\frac{{3 \\cdot 2 \\pi(0.80 \\mathrm{{~m}})}}{{1.0 \\mathrm{{~s}}}}=15 \\mathrm{{~m}} / \\mathrm{{s}}$
        The tension in the string, which is providing the centripetal force, is
        $\\displaystyle F=m \\frac{{v^2}}{{r}}=0.60 \\mathrm{{~kg}} \\frac{{(15 \\mathrm{{~m}} / \\mathrm{{s}})^2}}{{0.80 \\mathrm{{~m}}}}=170 \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_14(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        What is the (centripetal) force produced hy the tires acting on the pavement for the car of problem 5-13 if the car is $1200 \\mathrm{{~kg}}$ of mass?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $F=m a_{{r a d}}=1200 \\mathrm{{~kg}}\\left(9.3 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)=11200 \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_13(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A certain car driven in a circle can exert a maximum side force of $0.95 \\mathrm{{~g}}$. What is the maximum speed for this car driven in a circle of $160 \\mathrm{{~m}}$ radius?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        An $0.95 \\mathrm{{~g}}$ means a side force or acceleration directed at right angles to the direction of travel of $0.95\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)=9.3 \\mathrm{{~m}} / \\mathrm{{s}}^2$. Using $a_{{r a d}}=v^2 / r$
        $v=\\sqrt{{a_{{\\text {{rad }}}} r}}=\\sqrt{{9.3 \\mathrm{{~m}} / \\mathrm{{s}}^2(160 \\mathrm{{~m}})}}=38.6 \\mathrm{{~m}} / \\mathrm{{s}}$
        At a speed greater than $38.6 \\mathrm{{~m}} / \\mathrm{{s}}$ the car will "slide out of the circle.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_12(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider a flat surface with a coefficient of friction of 0.20 with a $2.0 \\mathrm{{~kg}}$ mass and a $3.0 \\mathrm{{~kg}}$ mass connected together with a $6.0 \\mathrm{{~kg}}$ mass along a slant as shown in Fig. 5-15. The coefficient of friction on the slant is zero. Calculate the acceleration of the system.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-15.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        In this problem note that the tensions are different leading to three unknowns, $T_1, T_2$, and $a$. The solution will require three equations from three vector diagrams as shown in Fig. 5-16. Assume the acceleration of the system is clockwise.
        The frictional retarding force on the $2.0 \\mathrm{{~kg}}$ block is $f=\\mu \\mathrm{{N}}=0.20(19.6 \\mathrm{{~N}})=3.92 \\mathrm{{~N}}$.
        The frictional retarding force on the $3.0 \\mathrm{{~kg}}$ block is $f=\\mu N=0.20(29.4 \\mathrm{{~N}})=5.88 \\mathrm{{~N}}$.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-16.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        
        The equations for the masses are: 
        $T_2=3.92 \\mathrm{{~N}}=2.0 \\mathrm{{~kg}} \\cdot a$
        $T_1-T_2-5.88 \\mathrm{{~N}}=3.0 \\mathrm{{~kg}} . a$
        $24.8 \\mathrm{{~N}}-T_1=6.0 \\mathrm{{~kg}} . a$
        The last two equations added together are: $18.9 \\mathrm{{~N}}-T_2=9.0 \\mathrm{{~kg}} \\cdot a$ and adding this to the first equation yields $15.0 \\mathrm{{~N}}=11 \\mathrm{{~kg}} . a$ or $a=1.36 \\mathrm{{~m}} / \\mathrm{{s}}_2$ The tensions come from the first and third equations for the masses.
        $T_2=3.92 \\mathrm{{~N}}+2.0 \\mathrm{{~kg}}\\left(1.36 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)=6.64 \\mathrm{{~N}}$
        $T_1=24.8 \\mathrm{{~N}}-6.0 \\mathrm{{~kg}}\\left(1.36 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)=16.6 \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_11(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider a double-incline plane with angles, masses, and coefficients of friction as shown in Fig. 5-12. Calculate the acceleration of the system.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-12.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        
        soln_s = f"""
        $\\Lambda$ ssume the system moves counterclockwise and set up the vector diagram.
        The equation for $m_1$ is $(30-15-T) \\mathrm{{N}}=6.0 \\mathrm{{~kg}} \\text {{ a }}$
        The equation for $m_2$ is $(T-62-10) \\mathrm{{N}}=9.0 \\mathrm{{~kg}} \\cdot \\mathrm{{a}}$
        Adding the equations $-57 \\mathrm{{~N}}=15 \\mathrm{{~kg}} . \\mathrm{{a}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-13.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        
        soln_s = f"""
        This is a negative acceleration. We cannot simply reverse the sign of $a$ to find the correct answer since if the system were moving the other way the frictional forces would have to be reversed. Further analysis requires us to consider another vector diagram assuming the system moves in a clockwise direction.
        The numbers have all been calculated so the vector diagram can be written down easily and is shown in Fig. 5-14
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-14.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        
        soln_s = f"""
        The new equations are:
        $(T-30-15) \\mathrm{{N}}=6.0 \\mathrm{{~kg}} \\cdot a$
        $(62-T-10) \\mathrm{{N}}=9.0 \\mathrm{{~kg}} \\cdot a$
        Adding the equations yields $7.0 \\mathrm{{~N}}=15 \\mathrm{{~kg}} . a$ or $a=0.54 \\mathrm{{~m}} / \\mathrm{{s}}^2$
        The tension in the cord is from the first equation: $T=45 \\mathrm{{~N}}+6.0 \\mathrm{{~kg}}\\left(0.54 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)=48 \\mathrm{{~N}}$
        If the acceleration of the system were negative for both the clockwise and counterclockwise calculations then we would conclude that the system would not move.

        The final complication in these incline plane problems is the introduction of three masses and two connecting cords.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $65^{{\\circ}}$-incline plane has a mass of $20 \\mathrm{{~kg}}$ on the plane where the coefficient of friction is 0.40 and a mass of $50 \\mathrm{{~kg}}$ hanging free. Calculate the acceleration of the system.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        
        soln_s = f"""
        Set up the vector diagram assuming motion counterclockwise. In the diagram the frictional force is calculated as a maximum of $33 \\mathrm{{~N}}$.
        The equation for $m_2$ is $490 \\mathrm{{~N}}-T=50 \\mathrm{{~kg}} \\cdot a$
        The equation for $m_1$ is $\\quad T-178 \\mathrm{{~N}}-33 \\mathrm{{~N}}=20 \\mathrm{{~kg}} . a$
        $\\Lambda$ dding the equations $279 \\mathrm{{~N}}=70 \\mathrm{{~kg}} \\cdot a$ or $a=4.0 \\mathrm{{~m}} / \\mathrm{{s}}^2$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-11.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A $50 \\mathrm{{~kg}}$ sled is pulled along a level surface at constant velocity by a constant force of $200 \\mathrm{{~N}}$ at an angle of $30^{{\\circ}}$. What is the coefficient of friction between the sled and the surface?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The first step in this problem is to find the components of the applied force and the mg of the sled. These are shown on the vector diagram. Notice that the normal force, the actual force between the sled and the friction surface is $m g$ minus the component of the applied force lifting up on the sled. Sometimes it is easy to miss this vertical component of the applied force. This is the major mistake in this type of problem. To help visualize this vertical component of force try placing the origin for the vectors in the middle of the sled rather than at one end where a rope would be attached.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-10.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The normal force is $m g(=490 \\mathrm{{~N}})$ minus the vertical component of the applied force, $100 \\mathrm{{~N}}$, or $390 \\mathrm{{~N}}$. Since the sled is moving at constant velocity the forces must be in equilibrium. The horizontal component of the applied force must equal the frictional retarding force, the coefficient of friction times the normal force.
        $\\mu \\cdot 390 \\mathrm{{~N}}=173 \\mathrm{{~N}}$
        so $\\mu=0.44$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation of problem 5-7 increase the coefficient of friction to 0.80 . What is the acceleration of the system?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This makes the vector diagram look like Fig. 5-9.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-9.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The equation for $m_2$ is $T-392 \\mathrm{{~N}}=50 \\mathrm{{~kg}} \\cdot a$
        The quation for $m_2$ is $294-T=30 \\mathrm{{~kg}} . a$
        Adding the tow equations $-98 \\mathrm{{~N}}=80 \\mathrm{{~kg}} \\cdot a \\quad \\text {{ or }} \\quad a=-1.2 \\mathrm{{~m}} / \\mathrm{{s}}^2$
        Clearly the system does not accelerate counterclockwise! This is an illustration of the second property of frictional forces: the frictional force is less than or equal to $\\mu N$. In this case the frictional force just balances the tension in the cord. The frictional force, as calculated, can be up to $392 \\mathrm{{~N}}$. In order for $m_2$ not to move, the tension in the cord must be $294 \\mathrm{{~N}}$. (If $m_2$ is not moving, the fores must be in balanec.) Therefore the forecs on $m_1$ must be $294 \\mathrm{{~N}}$ due to the tension in the cord and $294 \\mathrm{{~N}}$ due to friction.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation of problem 5-6 add a coefficient of friction $\\mu=0.20$ between the block and table.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This makes the vector diagram look like Fig. 5-8. Notice that the only addition is the frictional retarding force $\\mu N=98 \\mathrm{{~N}}$.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-8.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The equation for $m_1$ is $T-98 \\mathrm{{~N}}=50 \\mathrm{{~kg}} \\cdot a$
        The equation for $m_2$ is $\\quad 294 \\mathrm{{~N}}-T=30 \\mathrm{{~kg}} . a$
        Add the equations to obtain $196 \\mathrm{{~N}}=80 \\mathrm{{~kg}} . a \\quad$ or $a=2.4 \\mathrm{{~m}} / \\mathrm{{s}}^2$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Two masses are arranged with one, of $50 \\mathrm{{~kg}}$, on a frictionless table and the other, of $30 \\mathrm{{~kg}}$, attached by a cord and hanging, over a frictionless pulley, off the edge of the table. Find the acceleration of the system.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Assume the system moves clock wise.
        The equation for $m_1$ is
        $T=50 \\mathrm{{~kg}} \\cdot a$
        The equation for $m_2$ is
        $294 \\mathrm{{~N}}-T=30 \\mathrm{{~kg}} \\cdot a$
        Adding the two equations
        $294 \\mathrm{{~N}}=80 \\mathrm{{~kg}} \\cdot a \\text {{ or }} a=3.7 \\mathrm{{~m}} / \\mathrm{{s}}^2$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-7.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Another popular problem is the Atwood machine. A simple example is shown in Fig. 5-6. Find the acceleration of the system.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Assume the system moves counterclockwise.
        The equation for $m_{{\\mathrm{{l}}}}$ is $\\quad 98 \\mathrm{{~N}}-T-10 \\mathrm{{~kg}} . a$
        The equation for $m_2$ is $\\quad T-147 \\mathrm{{~N}}=15 \\mathrm{{~kg}} . a$
        Add the equations $-49 \\mathrm{{~N}}=25 \\mathrm{{~kg}} . a \\quad$ or $a=-2.0 \\mathrm{{~m}} / \\mathrm{{s}}_2$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        file_name = 'how-to-solve-physics-problems-fig-5-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The system accelerates clockwise! The tension in the cord is from the equation for $m_1$.
        $T=98 \\mathrm{{~N}}+(10 \\mathrm{{~kg}}) 2.0 \\mathrm{{~m}} / \\mathrm{{s}}^2=118 \\mathrm{{~N}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The next complication in force problems with incline planes is a double-incline plane as shown in Fig. 5-4.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The vector diagram for these masses is shown in Fig. 5-5. Assume the system moves clockwise.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The equation for $m_1$ is $\\quad T-76 \\mathrm{{~N}}=12 \\mathrm{{~kg}} . a$
        The equation for $m_2$ is $83 \\mathrm{{~N}}-T=20 \\mathrm{{~kg}} . a$
        Adding the two equations $7.0 \\mathrm{{~N}}=32 \\mathrm{{~kg}} . a \\quad$ or $\\quad a=0.22 \\mathrm{{~m}} / \\mathrm{{s}}^2$
        The tension in the cord is $T=76 \\mathrm{{~N}}+12 \\mathrm{{~kg}} \\cdot 0.22 \\mathrm{{~m}} / \\mathrm{{s}}_2=79 \\mathrm{{~N}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Place a $10 \\mathrm{{~kg}}$ mass on a frictionless $35^{{\\circ}}$ incline plane, and attach a second $20 \\mathrm{{~kg}}$ mass via a cord to hang vertically as shown in Fig. 5-3. Calculate the acceleration of the system.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        This problem introduces the concept of the tension in the connecting cord. The most convenient way to visualize this tension is that if the cord were cut and a force meter inserted it would read a certain tension (force). This tension acts, as shown, on each mass. Notice the $35^{{\\circ}}$ angle of the plane and the $35^{{\\circ}}$ angle between the normal and the direction of $m g$.
        We'll do the problem first with the tension and later without the tension.
        To write equations relating the unbalanced force to the acceleration, we need to make an assumption as to which way the masses move. Assume that the system (of masses) moves to the right or clockwise.
        Start by writing equations relating the unbalanced force on each mass. On $m_1$ the unbalanced force is $T$ - $56 \\mathrm{{~N}}$, and this force makes the $10 \\mathrm{{~kg}}$ accelerate.
        $T-56 \\mathrm{{~N}}=10 \\mathrm{{~kg}} \\cdot a_1$
        Likewise on $m_2$, the unbalanced force to make the mass accelerate is
        $196 \\mathrm{{~N}}-T=20 \\mathrm{{~kg}} \\cdot a_2$
        The assumption that the system accelerates clockwise dictates the directions of the forces in these two equations. Look again at these equations and notice that if we had assumed the acceleration were counterclockwise then the signs would be different. In frictionless problems it is all right to have a negative acceleration. The negative sign means that you guessed wrong in assuming the acceleration direction. This is not true with friction problems.

        Add the two equations and take the accelerations the same. The accelerations are the same because the masses are linked together with the cord.
        $140 \\mathrm{{~N}}=30 \\mathrm{{~kg}} \\cdot a \\quad$ or $\\quad a=4.7 \\mathrm{{~m}} / \\mathrm{{s}}^2$

        Now the tension in the cord is easily calculated from the equation for $m_1$.
        $T=56 \\mathrm{{~N}}+10 \\mathrm{{~kg}} \\cdot 4.7 \\mathrm{{~m}} / 2=103 \\mathrm{{~N}}$
        Another way to view this problem is to say that the unbalanced force on $m_1$ and $m_2$ is $(196-56) \\mathrm{{N}}=140 \\mathrm{{~N}}$. This force makes the total mass of $30 \\mathrm{{~kg}}$ accelerate, so
        $140 \\mathrm{{~N}}=30 \\mathrm{{~kg}} \\cdot a \\text {{ or }} a=4.7 \\mathrm{{~m}} / \\mathrm{{s}}^2$
        This is a very quick way to find the acceleration of the system, but it is inconvenient to find the tension in the cord.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Now place the $20 \\mathrm{{~kg}}$ mass (with no external force applied) on a frictionless $50^{{\\circ}}$ incline plane. What is the acceleration of the mass?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The force acting on the mass is due to gravity so set up a vector diagram (some authors call this a foree diagram or free body diagram) starting with this $196 \\mathrm{{~N}}$ force acting down. The mass is constrained to move down the plane, so the force we want is the component of this gravitational force acting down the plane. In Fig. 5-2 the gravitational force is shown with components down the plane, $F_p$, and normal, $F_w$ to the plane. Notice the geometry of the situation where the $50^{{\\circ}}$ angle of the plane is the same as the angle between the gravitational force and the normal (force).
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        file_name = 'how-to-solve-physics-problems-fig-5-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The $150 \\mathrm{{~N}}$ force acting down the plane causes the $20 \\mathrm{{~kg}}$ mass to accelerate at
        $a=\\frac{{150 \\mathrm{{~N}}}}{{20 \\mathrm{{~kg}}}}=7.5 \\mathrm{{~m}} / \\mathrm{{s}}^2$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_5_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A 40-N force is applied to a $20 \\mathrm{{~kg}}$ block resting on a horizontal frictionless table. Find the acceleration?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The acceleration is
        $\\displaystyle a=\\frac{{F}}{{m}}=\\frac{{40 \\mathrm{{~N}}}}{{20 \\mathrm{{~kg}}}}=2.0 \\mathrm{{~m}} / \\mathrm{{s}}^2$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_4_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A physicist turned motorcycle stunt rider will jump a $20 \\mathrm{{~m}}$ wide row of cars. The launch ramp is $30^{{\\circ}}$ and $9.0 \\mathrm{{~m}}$ high. The land ramp is also $30^{{\\circ}}$ and is $6.0 \\mathrm{{~m}}$ high. Find the minimum speed for the launch.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-4-7.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        First write down the six equations of motion.
        $a_x=0 \\qquad a_y=-9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2$
        $v_x=v_o \\cos 30^{{\\circ}} \\qquad v_y=v_o \\sin 30^{{\\circ}}-\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t$
        $x=v_o \\cos 30^{{\\circ}} t \\qquad y=v_o \\sin 30^{{\\circ}} t-\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2$
        The minimum $v_0$ is dictated by the condition that the rider be at $x=20 \\mathrm{{~m}}$ and $y=-3.0 \\mathrm{{~m}}$. So put these conditions into the equations for $x$ and $y$.
        $20 \\mathrm{{~m}}=v_o \\cos 30^{{\\circ}} t \\qquad-3.0 \\mathrm{{~m}}=v_o \\sin 30^{{\\circ}} t-\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2$
        At this point we can solve for $v_0$ by substituting $\\displaystyle t=\\frac{{20 \\mathrm{{~m}}}}{{v_o \\cos 30^{{\\circ}}}}$ from the first equation into the second equation.
        $\\displaystyle -3.0 \\mathrm{{~m}}=(20 \\mathrm{{~m}}) \\tan 30^{{\\circ}}-\\frac{{4.9 \\times 400 \\mathrm{{~m}}^3}}{{v_o^2 \\cos ^2 30^{{\\circ}} \\mathrm{{s}}^2}} \\quad \\text {{ yields }} \\quad v_o=13.4 \\mathrm{{~m}} / \\mathrm{{s}}$
        This is the minimum velocity to make the jump.
        As a follow on to this problem calculate the velocity to just hit the end of the ramp. This velocity gives an upper limit to the jump velocity.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_4_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A baseball is hit at a $45^{{\\circ}}$ angle and a height of $0.90 \\mathrm{{~m}}$. The ball travels a total distance of $120 \\mathrm{{~m}}$. What is the initial velocity of the ball? What is the height of the ball above a $3.0 \\mathrm{{~m}}$ fence $100 \\mathrm{{~m}}$ from where the ball is hit?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        This problem is unique in that it requests the initial velocity, a number usually given in the problem. Also note that the angle is $45^{{\\circ}}$, the angle for maximum range. The other interesting feature of the problem is the question concerning the height at some specific point down range. Set up the problem in the conventional way assigning $v_0$ to the initial velocity.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-4-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_problem(file_path, 'image/jpg')
        
        soln_s = f"""
        The next question is how to handle the ball being hit at $0.90 \\mathrm{{~m}}$ above the ground. Since $v_0$ is specified at the point where the ball is hit, and this is where we start counting time, then the origin for the coordinate system should be put here. We just need to keep in mind that this origin is $0.90 \\mathrm{{~m}}$ above the ground level.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-4-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        Now write the six equations.
        $\\displaystyle \\begin{{array}}{{ll}} a_x=0 & a_y=-9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2 \\\\ v_x=v_o \\cos 45^{{\\circ}} & v_y=v_o \\sin 45^{{\\circ}}-9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2 t \\\\ x=v_o \\cos 45^{{\\circ}} t & y=v_o \\sin 45^{{\\circ}} t-4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2 t^2 \\end{{array}}$
        The first question reduces to finding $v_o$ for the 120 m hit. Algebraically this means that when $y=-0.9 \\mathrm{{~m}}$, $x= 120 \\mathrm{{~m}}$, so write the two equations (for $x$ and $y$ ) with these conditions
        $120 \\mathrm{{~m}}=v_o \\cos 45^{{\\circ}} t \\qquad-0.9 \\mathrm{{~m}}=v_o \\sin 45^{{\\circ}} t-\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2$
        Note that so write
        $0=0.90 \\mathrm{{~m}}+120 \\mathrm{{~m}}-\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2 \\qquad$ or $\\displaystyle \\qquad t^2=\\frac{{120.9}}{{4.9}} \\mathrm{{~s}}^2 \\quad$ and $\\quad t=4.97 \\mathrm{{~s}}$
        This is the time of flight, so put this time into the equation for $x$.
        $120 \\mathrm{{~m}}=v_o \\cos 45^{{\\circ}}(4.97 \\mathrm{{~s}})$, to find $v_o$ as $\\displaystyle v_o=\\frac{{120 \\mathrm{{~m}}}}{{\\cos 45^{{\\circ}} \\cdot 4.97 \\mathrm{{~s}}}}=34.1 \\mathrm{{~m}} / \\mathrm{{s}}$
        The next part of the problem asks for the height above the fence at $100 \\mathrm{{~m}}$. Again we need to find the time for $x$ to be $100 \\mathrm{{~m}}$ down range and substitute this value of $t$ into the $y$ equation
        $\\displaystyle 100 \\mathrm{{~m}}=(34.1 \\mathrm{{~m}} / \\mathrm{{s}}) \\cos 45^{{\\circ}} t \\quad \\text {{ or }} \\quad t=\\frac{{100 \\mathrm{{~s}}}}{{34.1 \\cos 45^{{\\circ}}}}=4.15 \\mathrm{{~s}}$
        This is the time for the ball to go $100 \\mathrm{{~m}}$. The height of the ball at this time is
        $\\left.y\\right|_{{t=4.15}}=(34.1 \\mathrm{{~m}} / \\mathrm{{s}}) \\cos 45^o(4.15 \\mathrm{{~s}})-\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)(4.15 \\mathrm{{~s}})^2=15.6 \\mathrm{{~m}}$
        Remember that this is $15.6 \\mathrm{{~m}}$ above the zero of the coordinate system which is $0.90 \\mathrm{{~m}}$ above the ground. So the ball is $16.5 \\mathrm{{~m}}$ above the ground at this point and for a $3.0 \\mathrm{{~m}}$ high fence is $13.5 \\mathrm{{~m}}$ above the fence.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_4_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        An airplane traveling at $100 \\mathrm{{~m}} / \\mathrm{{s}}$ drops a bomb from a height of $1500 \\mathrm{{~m}}$. Find the time of flight, distance traveled, and the velocity components as the bomb strikes the ground.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Start by placing the origin of the coordinate system at the point where the bomb is released. And take the direction the bomb falls due to gravity as positive $x$ and the horizontal position as $y$. This is different from the conventional orientation, but it is convenient in this problem because the six equations all come out positive!
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-4-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        Now write down the six equations governing the motion.
        $\\displaystyle \\begin{{array}}{{ll}} a_x=9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2 & a_y=0 \\\\ v_x=\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t & v_y=v_o=100 \\mathrm{{~m}} / \\mathrm{{s}} \\\\ x=\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t & y=(100 \\mathrm{{~m}} / \\mathrm{{s}}) t \\end{{array}}$
        With these six equations we can answer all the questions in the problem. We need only to "translate" the word questions into algebra questions.
        a) The time for the bomb to reach the earth means: Find the time when $x=1500 \\mathrm{{~m}}$ ?
        $1500 \\mathrm{{~m}}=\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2 \\quad \\text {{ yields }} \\quad \\tau=17 \\mathrm{{~s}}$
        b) Ilow far does the bomb travel horizontally means: Find the value of $y$ when $t=17 \\mathrm{{~s}}$ ?
        $\\left.y\\right|_{{t=17}}=(100 \\mathrm{{~m}} / \\mathrm{{s}}) 17 \\mathrm{{~s}}=1700 \\mathrm{{~m}}$
        c) Find the velocity components at impact means: Find $v_x$ and $v_y$ at $t=17 \\mathrm{{~s}}$ ?
        $\\left.v_x\\right|_{{t-17}}=\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 17 \\mathrm{{~s}}=167 \\mathrm{{~m}} / \\mathrm{{s}}$
        $\\left.v_y\\right|_{{t=17}}-100 \\mathrm{{~m}} / \\mathrm{{s}}$
        d) Where is the airplane when the bomb strikes the earth means: What is $y$ and $t=17 \\mathrm{{~s}}$ ?
        Remember that the plane and the bomb have the same velocity in the horizontal direction.
        $\\left.y\\right|_{{\\text {{plane }}}}=(100 \\mathrm{{~m}} / \\mathrm{{s}}) 17 \\mathrm{{~s}}=1700 \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_4_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation of problem 4-1 suppose the field is covered with fog down to $3.0 \\mathrm{{~m}}$ above the ground. What are the times for the ball's entry into and exit out of the fog?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The times are found from the equation for $y$ as a function of time with $y$ set equal to $3.0 \\mathrm{{~m}}$.
        $3.0 \\mathrm{{~m}}=(10.6 \\mathrm{{~m}} / \\mathrm{{s}}) t-\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2$
        This is a quadratic and removing the units it reads $4.9 t_2-10.6 t+3.0=0$
        and is solved with the quadratic formula.
        $\\displaystyle t=\\frac{{10.6 \\pm \\sqrt{{10.6^2-4(4.9) 3.0}}}}{{2(4.9)}}=\\frac{{10.6 \\pm 7.3}}{{9.8}}=0.34,1.8$
        As an exercise find the $x$ positions for these times.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_4_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A soccer player kicks a ball at an initial velocity of $18 \\mathrm{{~m}} / \\mathrm{{s}}$ at an angle of $36^{{\\circ}}$ to the horizontal. Find the time of flight, range, maximum height, and velocity components at $t=0$, mid-range, and at impact.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The acceleration is down so set up the coordinate system with $x$ horizontal (along the ground ) and $y$ vertical, and place $v_0$ on the graph. Write the velocity in component form, and calculate $v_{{x 0}}$ and $v_{{00}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-4-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        Now, and this and this is the key to getting the problem right, write down the six equations governing the motion starting with the accelerations (the accelerations are the easiest to write).
        $\\begin{{array}}{{ll}} a_x=0 & a_y=-9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2 \\\\ v_x=14.6 \\mathrm{{~m}} / \\mathrm{{s}} & v_y=10.6 \\mathrm{{~m}} / \\mathrm{{s}}-\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t \\\\ x=(14.6 \\mathrm{{~m}} / \\mathrm{{s}}) t & y=(106 \\mathrm{{~m}} / \\mathrm{{s}}) t-\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2 \\end{{array}}$
        Now look at the motion, which is parabolic in $x-y$, keeping in mind the properties of parabolas.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-4-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The ball is on the ground at $t=0$ and $t=\\tau$, the time of flight. To find these times, set $y=0$ and obtain (without units)
        $10.6 t-4.9 t^2=0 \\quad$ or $\\quad t(10.6-4.9 t)=0$
        which gives $t=0$ and $t=10.6 / 4.9=2.16 \\mathrm{{~s}}$, the time of flight.
        The range is the value of $x$ at $2.16 \\mathrm{{~s}}:\\left.x\\right|_{{t=2.16}}=(14.6 \\mathrm{{~m}} / \\mathrm{{s}}) 2.16 \\mathrm{{~s}}=31.5 \\mathrm{{~m}}$
        The maximum height, and $v_y=0$, occur at $1.08 \\mathrm{{~s}}$ (one-half the time of flight). Therefore the maximum height is $y$ at $t=1.08 \\mathrm{{~s}}$
        $y_{{\\max }}=\\left.y\\right|_{{i=1.08}}=(10.6 \\mathrm{{~m}} / \\mathrm{{s}}) 1.08 \\mathrm{{~s}}-\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)(1.08 \\mathrm{{~s}})^2=5.73 \\mathrm{{~m}}$
        The velocity components are:
        at $t=0: \\quad v_x=14.6 \\mathrm{{~m}} / \\mathrm{{s}}, \\quad v_y=10.6 \\mathrm{{~m}} / \\mathrm{{s}}$
        at $t=1.08 \\mathrm{{~s}} \\quad v_x=14.6 \\mathrm{{~m}} / \\mathrm{{s}}, \\quad v_y=0$
        at $t=2.16 \\mathrm{{~s}} \\quad v_x=14.6 \\mathrm{{~m}} / \\mathrm{{s}}, \\quad v_y=-10.6 \\mathrm{{~m}} / \\mathrm{{s}}$
        The procedure for doing the problem is to write the initial velocity in component form, write the six equations for acceleration, velocity, and position, and then perform the mathematical operations answering questions about the problem.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_3_10(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        In a situation similar to problem 3-9, the tank is filled with a different liquid causing the acceleration in the tank to be $6.0 \\mathrm{{~m}} / \\mathrm{{s}}^2$ and the time to traverse the window, $0.10 \\mathrm{{~s}}$. Calculate the height of the liquid above the window, the time to reach the bottom of the tank, and the velocity of the ball when it reaches the bottom of the tank.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        
        soln_s = f"""
        Diagram the problem as in Fig. 3-8.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-3-8.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        From the data about the window, calculate the velocity of the ball at the top of the window.
        $x-x_o=v_t t+(1 / 2) a t^2 \\qquad 1.0 \\mathrm{{~m}}=v_t 0.40 \\mathrm{{~s}}+\\left(3.0 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)(0.40 \\mathrm{{~s}})^2 \\qquad \\text {{ or }} \\qquad v_t=1.3 \\mathrm{{~m}} / \\mathrm{{s}}$
        The velocity at the bottom of the window is from
        $v_b^2=v_l^2+2 a\\left(x-x_o\\right)=(1.3 \\mathrm{{~m}} / \\mathrm{{s}})^2+2\\left(6.0 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 1.0 \\mathrm{{~m}} \\quad \\text {{ or }} \\quad v_b=3.7 \\mathrm{{~m}} / \\mathrm{{s}}$
        The time to reach the bottom of the tank is from $x-x_0=v_b t+(1 / 2) a t_2$
        $12 \\mathrm{{~m}}=(3.7 \\mathrm{{~m}} / \\mathrm{{s}}) t+\\left(3.0 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2$
        and eliminating the units $3.0 t^2+3.7 t-12=0$ so
        so
        $t=\\frac{{-3.7 \\pm \\sqrt{{(3.7)^2-4(3.0)(-12)}}}}{{2(3)}}=\\frac{{-3.7 \\pm 12.6}}{{6}}=1.5,-2.7$
        This (positive) time is for the ball to travel from the bottom of the window to the bottom of the tank. Assuming the ball started at zero velocity the distance from the top of the liquid to the top of the window, $x$. will come from $v_t^2=v_o^2+2 a x_t$


        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_3_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        You are observing steel balls falling at a constant velocity in a liquid-filled tank. The window you are using is one meter high and the bottom of the window is twelve meters from the bottom of the tank. You observe a ball falling past the window taking 3.0 s to pass the window. Calculate the time required to reach the bottom of the tank, after the ball has reached the bottom of the window.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The situation is diagrammed in Fig. 3-7.
        The observed velocity is $1.0 \\mathrm{{~m}} / 3.0 \\mathrm{{~s}}$ so $12 \\mathrm{{~m}}=(1.0 \\mathrm{{~m}} / 3.0 \\mathrm{{~s}}) t$ or $t=36 \\mathrm{{~s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-3-7.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_3_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A boat is passing under a bridge. The deck of the boat is $15 \\mathrm{{~m}}$ below the bridge. A small package is to be dropped from the bridge onto the deck of the boat when the boat is $25 \\mathrm{{~m}}$ from just below the drop point. What (boat) speed is necessary to have the package land in the boat?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Calculate the time for the package to fall the $15.0 \\mathrm{{~m}}$ using $x-x_o=v_o t+(1 / 2) a t^2$ $15 \\mathrm{{~m}}=\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2 \\quad$ or $\\quad t=1.7 \\mathrm{{~s}}$
        The boat must move at $25 \\mathrm{{~m}} / 1.7 \\mathrm{{~s}}=14 \\mathrm{{~m}} / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_3_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A coconut is dropped from a height of $60 \\mathrm{{~m}}$. One second later a second coconut is thrown down with an initial velocity. Both coconuts reach the ground at the same time. What was the initial velocity of the second coconut?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        In problems where there is a time delay it is usually best to calculate the position, velocity, and acceleration of the first particle at the time when the second particle starts to move. In the train problem in the previous chapter there was a position difference between the two trains at $t=0$ that was easily translated into the equations. It is possible to do time delay problems with a time differential in one set of equations. The difficulty with this approach is that it is easy to get an algebraic sign wrong. If you say that the time for the second particle is the time for the first plus the difference between them, then it is essential that the algebraic sign of the difference be correct. It is much easier, especially when you are learning how to do problems with a time delay, to do them in this slower, but inherently more accurate, method. First, calculate the state of the first particle when the second one begins moving. Then, write the two sets of equations describing the motion with this instant as $t=0$.

        Calculate the position and velocity of the first coconut at the end of one second, the time when the second one starts.
        $x=(1 / 2) a t^2=(1 / 2)\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)(1.0 \\mathrm{{~s}})^2=4.9 \\mathrm{{~m}} \\quad v=a t=9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2 1.0 \\mathrm{{~s}}=9.8 \\mathrm{{~m}} / \\mathrm{{s}}$
        Since everything (position, velocity, and acceleration) is positive down, orient the coordinate system for positive down with the origin at the top. Now diagram the problem as in Fig. 3-6. At the instant the second coconut is thrown down, the first coconut has position $4.9 \\mathrm{{~m}}$, velocity $9.8 \\mathrm{{~m}} / \\mathrm{{s}}$ and acceleration $9.8 \\mathrm{{~m}} \\mathrm{{~s}}^2$.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-3-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        Since both coconuts strike the ground at the same time use the conditions of the first coconut to find the total time.
        First calculate the velocity at impact of the first coconut.
        $v_1^2=v_{{1 o}}^2+2 a\\left(x_1-x_{{1 o}}\\right)=(9.8 \\mathrm{{~m}} / \\mathrm{{s}})^2+2\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)(60-4.9) \\mathrm{{m}}=1174 \\mathrm{{~m}}^2 / \\mathrm{{s}}^2$
        $v_1=34.3 \\mathrm{{~m}} / \\mathrm{{s}}$


        The time for the second coconut to reach the ground is the same as the time for the first coconut to go from $4.9 \\mathrm{{~m}}$ to $60 \\mathrm{{~m}}$ or the time for the first coconut to go from $9.8 \\mathrm{{~m}} / \\mathrm{{s}}$ to $34.3 \\mathrm{{~m}} / \\mathrm{{s}}$.

        This comes from $v=v_v+a t$. where $\\mathrm{{v}}_v$ is $\\mathrm{{v}}_{{l v}}$, the velocity of the first coconut when the second one is thrown down, and $v$ is the velocity of the first coconut at the ground.
        $34.3 \\mathrm{{~m}} / \\mathrm{{s}}=9.8 \\mathrm{{~m}} / \\mathrm{{s}}+\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t \\quad$ or $\\quad t=2.5 \\mathrm{{~s}}$
        Now that we have the time for the second coconut to travel the $60 \\mathrm{{~m}}$ we can find its initial velocity from $x$ - $x_0=$ $v_{{20}} t+(1 / 2) a t_2$ where $t$ is the total time for the second coconut.
        $60 \\mathrm{{~m}}=v_{{20}}(2.5 \\mathrm{{~s}})+\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)(2.5 \\mathrm{{~s}})^2 \\quad \\text {{ or }} \\quad v_{{20}}=11.5 \\mathrm{{~m}} / \\mathrm{{s}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_3_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A parachutist descending at a constant rate of $2.0 \\mathrm{{~m}} / \\mathrm{{s}}$ drops a smoke canister at a height of $300 \\mathrm{{~m}}$. Find the time for the smoke canister to reach the ground and its velocity when it strikes the ground. Then find the time for the parachutist to reach the ground, the position of the parachutist when the smoke canister strikes the ground, and an expression tor the distance between the smoke canister and the parachutist.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Diagram the system as shown in Fig. 3-5 taking displacement, velocity, and acceleration as positive down with the origin at the point where the canister is dropped.
        The time for the smoke canister to reach the ground is from equation 3-4.
        $v_x^2=v_o^2+2 a\\left(x-x_o\\right)=9.0 \\mathrm{{~m}}^2 / \\mathrm{{s}}^2+\\left(19.6 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 290 \\mathrm{{~m}} \\quad \\text {{ or }} \\quad v=75 \\mathrm{{~m}} / \\mathrm{{s}}$ 
        $x=v_o t+(1 / 2) a t^2 \\quad \\text {{ or }} \\quad 300 \\mathrm{{~m}}=(2.0 \\mathrm{{~m}} / \\mathrm{{s}}) t+\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-3-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        Without units the equation is $4.9 t^2+2.0 t-300=0$ with solutions
        $\\displaystyle t=\\frac{{-2 \\pm \\sqrt{{4-4(4.9)(-300)}}}}{{2 \\cdot 4.9}}=\\frac{{-2 \\pm 76.7}}{{9.8}}=7.6,-8.0$
        The time for the canister to reach the ground is 7.6s.
        The velocity when it strikes the ground is
        $v^2=(2.0 \\mathrm{{~m}} / \\mathrm{{s}})^2+2\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 300 \\mathrm{{~m}}=5884 \\mathrm{{~m}}^2 / \\mathrm{{s}}^2 \\quad \\text {{ or }} \\quad v=77 \\mathrm{{~m}} / \\mathrm{{s}}$
        The time for the parachutist to reach the ground is from equation 3-2.
        $300 \\mathrm{{~m}}=(2.0 \\mathrm{{~m}} / \\mathrm{{s}}) t \\quad \\text {{ or }} \\quad t=150 \\mathrm{{~s}}$
        When the canister strikes the ground the parachutist has dropped $(2.0 \\mathrm{{~m}} / \\mathrm{{s}}) 7.6 \\mathrm{{~s}}=15 \\mathrm{{~m}}$ and is $285 \\mathrm{{~m}}$ above the ground.
        The expression for the distance between the canister and the parachutist is
        $x_c-x_p=(2.0 \\mathrm{{~m}} / \\mathrm{{s}}) t+\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2-(2.0 \\mathrm{{~m}} / \\mathrm{{s}}) t=\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_3_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A bottle of champagne is dropped by a balloonist. The balloon is rising at a constant velocity of $3.0 \\mathrm{{~m}} / \\mathrm{{s}}$. It takes 8.0s for the bottle of champagne to reach the ground. Find the height of the balloon when the bottle was dropped, the height of the balloon when the bottle reached the ground, and the velocity with which the bottle strikes the ground.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Diagram the situation as shown in Fig. 3-4 being especially careful about the relative orientation (algebraic signs) of displacement, velocity, and acceleration. There are several possibilities as regards the origin and direction of the coordinate system. Take the origin at the height of the balloon when the bottle is dropped; the position of the balloon at $t=0$. Take the displacement as positive down. The main reason for taking the displacement as positive down is that the acceleration is down and the initial velocity up making two positives and one negative. As time goes on, however, displacement, velocity, and acceleration will be positive. This choice seems to make for fewer minus signs and less chance for error with an algebraic sign.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        file_name = 'how-to-solve-physics-problems-fig-3-4.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        Write the equation for the height of the balloon (when the bottle was dropped) counting time from when the bottle was dropped
        $x=(-3.0 \\mathrm{{~m}} / \\mathrm{{s}}) t+\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2=(-3.0 \\mathrm{{~m}} / \\mathrm{{s}}) 8.0 \\mathrm{{~s}}+\\left(4.9 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 64 \\mathrm{{~s}}^2=290 \\mathrm{{~m}}$
        The height of the balloon when the bottle reached the ground would be the height when the bottle was dropped plus the amount the balloon rose in the $8.0 \\mathrm{{~s}}$ it took the bottle to reach the ground, or
        $290 \\mathrm{{~m}}+(3.0 \\mathrm{{~m}} / \\mathrm{{s}}) 8.0 \\mathrm{{~s}}=314 \\mathrm{{~m}}$
        The velocity on impact $v=v_0+a t=-3.0 \\mathrm{{~m}} / \\mathrm{{s}}+\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 8.0 \\mathrm{{~s}}=75 \\mathrm{{~m}} / \\mathrm{{s}}$.
        This velocity could also be calculated using
        $290 \\mathrm{{~m}}+(3.0 \\mathrm{{~m}} / \\mathrm{{s}}) 8.0 \\mathrm{{~s}}=314 \\mathrm{{~m}}$

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_3_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation of problem 3-3 calculate the maximum height above the top of the building and the time for the ball to reach maximum height.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The time for the ball to reach maximum height is from equation 3-2. Note that at maximum height the velocity must be zero. Again watch the signs closely. It doesn't matter how you choose the signs, but the acceleration has to be opposite the velocity. The equations
        $0^2=(8.0 \\mathrm{{~m}} / \\mathrm{{s}})^2-2\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)\\left(x-x_o\\right) \\quad \\text {{ or }} \\quad x=3.3 \\mathrm{{~m}}$
        yield the same result, $t=0.82 \\mathrm{{~s}}$. Because of symmetry, it takes the ball the same amount of time to reach maximum height as it does for the ball to return to the original level.
        Calculate the height above the top of the building from equation 3-5.
        $0^2=(8.0 \\mathrm{{~m}} / \\mathrm{{s}})^2-2\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)\\left(x-x_o\\right) \\quad \\text {{ or }} \\quad x=3.3 \\mathrm{{~m}}$
        A thorough understanding of these four problems will keep you from making sign mistakes in problems like these.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_3_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Add a different complication to problem 3-1. Throw the ball up from the top of the building with a velocity of $8.0 \\mathrm{{~m}} / \\mathrm{{s}}$. Find the time for the ball to reach the ground and the velocity on impact
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Diagram the problem as in Fig. 3 3. Take $x$ the displacement as positive down and the velocity as negative up. It is important to remember that the sign of the velocity is opposite that of the displacement. It doesn't matter whether the velocity is negative and the distance down positive or vice versa. It does matter that the signs be opposite! Getting a sign wrong is the source of many, possibly most, errors in falling body problems.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-5-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        Calculate the time of flight and note the signs:
        Rearranging and without units, this equation is $4.9 t^2-8.0 t-40=0$,
        with solution
        $\\displaystyle t=\\frac{{8.0 \\pm \\sqrt{{64-4(4.9)(-40)}}}}{{9.8}}=3.8,-2.2$
        The positive time is the obvious choice. Note the numbers used for the solutions to the quadratics in time for this and problem $3-2$.
        The velocity when the ball strikes the ground is
        $v^2=(-8.0 \\mathrm{{~m}} / \\mathrm{{s}})^2+2\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 40 \\mathrm{{~m}}=848 \\mathrm{{~m}}^2 / \\mathrm{{s}}^2 \\quad \\text {{ or }} \\quad v=29 \\mathrm{{~m}} / \\mathrm{{s}}$
        Notice that whether the $v$ term is a positive or negative number, the result is the same. If the ball is thrown up with a certain velocity or down with the same velocity, the velocity at impact is the same. This is to be expected from the symmetry of the equations. If the ball is thrown up with a eertain velocity, then on the way down it passes the same level (from which it was thrown) with (numerically) that same velocity.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_3_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Now add a complication to problem $3-1$ by throwing the ball down with an initial velocity of $8.0 \\mathrm{{~m}} / \\mathrm{{s}}$. Find the time for the ball to reach the ground and the velocity on impact.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Again diagram the problem as in Fig. 3-2. Note that in this problem the displacement and velocity are positive down.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        file_name = 'how-to-solve-physics-problems-fig-3-2.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The time of flight is from $40 \\mathrm{{~m}}=(8.0 \\mathrm{{~m}} / \\mathrm{{s}}) t \\mathrm{{n}}+(1 / 2)\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2$
        which rearranged and without units is $4.9 t^2+8.0 t-40=0$.
        with solution by formula of
        $\\displaystyle t=\\frac{{-8.0 \\pm \\sqrt{{64-4(4.9)(-40)}}}}{{2(4.9)}}=2.2,-3.8 .$
        The positive time is the obvious choice.
        The velocity at the ground level is from
        $v^2=\\left(8.0 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)^2+2\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 40 \\mathrm{{~m}}=848 \\mathrm{{~m}}^2 / \\mathrm{{s}}^2 \\quad v=29 \\mathrm{{~m}} / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_3_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Consider a ball dropped from the top of a $40 \\mathrm{{~m}}$ tall building. Calculate everything possible.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        "Calculate everything possible" is an unusual request. Usually early in your study of falling body problems, there is a problem that asks for something that does not at all seem like it has anything to do with the information given. It's almost like asking "What color is the building?" When this happens, and it probably will, the way to do the problem is to calculate what you can and let the information you generate lead you through the problem. Let's apply this approach to this problem. First, diagram the problem as shown in Fig. 3-1. Place the origin at the top of the building with displacement, velocity, and acceleration (g) all positive down.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-3-1.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        Since most of the kinematic equations contain the time, this is usually one of the first things to calculate. Use equation 3-4 to find the time for the ball to strike the ground.
        $x-x_o=v_o t+(1 / 2) a t^2 \\quad$ or $\\quad 40 \\mathrm{{~m}}=(1 / 2)\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2 \\quad$ and $\\quad t=2.9 \\mathrm{{~s}}$
        Knowing the time we can calculate the velocity from equation 3-2.
        $v=v_o+g t=\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 2.9 \\mathrm{{~s}}=28 \\mathrm{{~m}} / \\mathrm{{s}}$
        Alternatively we can use equation 3-4.
        $v^2=v_o^2+2 a\\left(x-x_o\\right)=2\\left(9.8 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 40 \\mathrm{{~m}}=784 \\mathrm{{~m}}^2 / \\mathrm{{s}}^2 \\quad \\text {{ or }} \\quad v=28 \\mathrm{{~m}} / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_2_9(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Change problem 2-8 by giving the second train an initial velocity of $25 \\mathrm{{~m}} / \\mathrm{{s}}$. This will give a real time for the collision. Find the collision time.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The situation is now as shown in the Fig. 2-8.
        Again setting the expressions for $x_1$ and $x_2$ equal and dropping the units produces $200+12 t=25 t-0.10 t^2$ or $0.10 t^2-13 t+200=0$ with solutions
        $t=\\frac{{+13 \\pm \\sqrt{{169-4(0.10) 200}}}}{{2(0.10)}}=\\frac{{13 \\pm \\sqrt{{89}}}}{{0.20}}=\\frac{{13 \\pm 8.3}}{{0.20}}=17.8,112$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-2-8.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The two times correspond to when $x_1=x_2$. The earliest time is the first coincidence and the end of the (physical) problem. The position at this time can be obtained from either expression for $x$.
        $\\left.x_1\\right|_{{t=17.8}}=200 \\mathrm{{~m}}+(12 \\mathrm{{~m}} / \\mathrm{{s}}) 17.8 \\mathrm{{~s}}=414 \\mathrm{{~m}}$
        Verify this distance by using $x_2$.
        The velocity of the second train at collision is
        $\\left.v_2\\right|_{{t=17.8}}=25 \\mathrm{{~m}} / \\mathrm{{s}}-\\left(0.20 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 17.8 \\mathrm{{~s}}=21.4 \\mathrm{{~m}} / \\mathrm{{s}}$
        The relative velocity (between the two trains) is $v=1.4 \\mathrm{{~m}} / \\mathrm{{s}}$.
        The two times are the result of the quadratic in $t$. The two solutions occur when the curves cross. The equation for $x_1=200+12 t$ is a straight line of slope 12 starting at 200 . The equation for $x_2=25 t-0.10 t^2$ is a parabola that opens down. Figure 2-9 (not to scale) shows the two curves.

        While the "mathematics" produces two times, the reality of the problem dictates the earlier time as the one for the "collision."
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-2-9.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_2_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Two trains are traveling along a straight track, one behind the other. The first train is traveling at $12 \\mathrm{{~m}} / \\mathrm{{s}}$. The second train, approaching from the rear, is traveling at $20 \\mathrm{{~m}} / \\mathrm{{s}}$. When the second train is $200 \\mathrm{{~m}}$ behind the first, the operator applies the brakes producing a constant deceleration of $0.20 \\mathrm{{~m}} / \\mathrm{{s}}^2$. Will the trains collide, and if so where and when?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First, diagram the situation. Our strategy will be to write down the equations for each train using equations (2-1) through (2-6) and the information provided in the problem. Take $t=0$ when the brakes are applied and the first train is $200 \\mathrm{{~m}}$ ahead of the second. This makes the position of the first train $200 \\mathrm{{~m}}$ at $t=0$ $\\left(x_{{10}}\\right.$ in equation 2-2$)$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-2-7.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The question as to whether the trains collide means: "Is there a real (time) solution to the equation resulting from setting $x_1=x_2$.
        $200 \\mathrm{{~m}}+(12 \\mathrm{{~m}} / \\mathrm{{s}}) t=(20 \\mathrm{{~m}} / \\mathrm{{s}}) t-\\left(0.10 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2$
        If there are no real solutions to this equation then the trains do not collide. Drop the units and write $0.10 t^2-8 t+$ $200=0$ which is solved by quadratic formula
        $\\displaystyle t=\\frac{{+8 \\pm \\sqrt{{64-4(0.10) 200}}}}{{2(0.10)}}=\\frac{{8 \\pm \\sqrt{{-16}}}}{{0.20}}$
        Since there are no real solutions to this equation there are no times when the trains collide.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_2_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation of problem 2-5. what is the position, velocity, and acceleration of each vehicle when vehicle 2 has traveled twice the distance of vehicle 1 ?

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The time when this occurs is when $x_2=2 x_1: 5 t^2=60 t$ or $5 t(t-12)=0$
        This gives times of $t=0$ and $t=12$. The time $t=0$ is correct, though uninteresting. The time $t=12$ is the physically interesting answer.
        At $t=12:\\left.\\quad x_1\\right|_{{t=12}}=(30 \\mathrm{{~m}} / \\mathrm{{s}}) 12 \\mathrm{{~s}}=360 \\mathrm{{~m}}$
        From the original statement of the problem, $v_1=v_{{10}}=30 \\mathrm{{~m}} / \\mathrm{{s}}$ and $a_1=0$.
        Now solve for the remaining variables for the second vehicle by substitution
        $\\left.x_2\\right|_{{t=12}}=\\left(5 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)(12 \\mathrm{{~s}})^2=720 \\mathrm{{~m}}$
        $\\left.v_2\\right|_{{t=12}}=10 \\mathrm{{~m}} / \\mathrm{{s}}^2 12 \\mathrm{{~s}}=120 \\mathrm{{~m}} / \\mathrm{{s}}$
        From the original statement of the problem $a_2=10 \\mathrm{{~m}} / \\mathrm{{s}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_2_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation in problem 2-5, when do the vehicles have the same velocity?
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        In algebra this means to set the equations for velocity equal $\\left(v_1=v_2\\right)$ and solve for the time. Remember that three of the four equations of motion are functions of time so most question are answered by first calculating the time for a certain condition to occur.
        $30 \\mathrm{{~m}} / \\mathrm{{s}}=\\left(10 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t \\quad\\text {{ or }} \\quad t=3.0 \\mathrm{{~s}}$
        Now that we know when, we can calculate where they have the same velocity. Use either equation for position and $t=3.0 \\mathrm{{~s}}$
        $\\left.x_1\\right|_{{t=3.0}}=(30 \\mathrm{{~m}} / \\mathrm{{s}}) 3.0 \\mathrm{{~s}}=90 \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_2_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Two vehicles are at position $x=0$ at $t=0$. Vehicle 1 is moving at constant velocity of $30 \\mathrm{{~m}} / \\mathrm{{s}}$. Vehicle 2, starting from rest, has acceleration of $10 \\mathrm{{~m}} / \\mathrm{{s}}^2$. A typical question of this situation is "Where do they pass?"
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        First diagram the situation.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        file_name = 'how-to-solve-physics-problems-fig-2-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The question "Where do they pass?" translated into algebra means "What is the value of $x$ when they pass?" This can be determined by writing equations for the position of each vehicle and equating $x_1-v_{{10}} t=(30 \\mathrm{{~m}} / \\mathrm{{s}}) t$ and $x_2=(1 / 2) a_2 t^2=\\left(5 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2$
        Setting $x_1=x_2$ gives the time when they pass as $30 t=5 t(t-6)=0$; so the vehicles pass at $t=0$ and $t=6$. Putting $t=6 \\mathrm{{~s}}$ in either equation for $x\\left(x_1\\right.$ or $\\left.x_2\\right)$ gives $180 \\mathrm{{~m}}$ as the distance.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_2_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation of problem 2-1 find the velocity of the train at $120 \\mathrm{{~m}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $v^2=v_o^2+2 a x=(20 \\mathrm{{~m}} / \\mathrm{{s}})^2+2\\left(0.50 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 120 \\mathrm{{~m}}=400 \\mathrm{{~m}}^2 / \\mathrm{{s}}^2+120 \\mathrm{{~m}}^2 / \\mathrm{{s}}^2=520 \\mathrm{{~m}}^2 / \\mathrm{{s}}^2$
        $v=23 \\mathrm{{~m}} / \\mathrm{{s}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_2_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation of problem 2-1 find the time required for the train to reach $100 \\mathrm{{~m}}$
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $x-x_o=v_o t+(1 / 2) a t^2$ so $100 \\mathrm{{~m}}=(20 \\mathrm{{~m}} / \\mathrm{{s}}) t+(1 / 2)\\left(0.50 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) t^2$
        This is a quadratic equation in $t$. Without units the equation is $t^2+80 t-400=0$
        and has solutions
        $\\displaystyle t=\\frac{{-80 \\pm \\sqrt{{6400-4(1)(400)}}}}{{2(1)}}=\\frac{{-80 \\pm 89}}{{2}}=4.5,-85$
        The negative answer is inappropriate for this problem so take $t=4.5 \\mathrm{{~s}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_2_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        For the situation of problem 2-1, calculate the position of the train at 20 s.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $x=v_o t+(1 / 2) a t^2=(20 \\mathrm{{~m}} / \\mathrm{{s}}) 20 \\mathrm{{~s}}+(1 / 2)\\left(0.50 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right)(20 \\mathrm{{~s}})^2=500 \\mathrm{{~m}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_2_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        A train starts from rest (at position zero) and moves with constant acceleration. On first observation the velocity is $20 \\mathrm{{~m}} / \\mathrm{{s}}$ and $80 \\mathrm{{~s}}$ later the velocity is $60 \\mathrm{{~m}} / \\mathrm{{s}}$. At $80 \\mathrm{{~s}}$, calculate the position, average velocity, and the constant acceleration over the interval.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Diagram the problem.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-2-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        Calculate the acceleration:
        $\\displaystyle a=\\frac{{v-v_o}}{{t}}=\\frac{{60 \\mathrm{{~m}} / \\mathrm{{s}}-20 \\mathrm{{~m}} / \\mathrm{{s}}}}{{80 \\mathrm{{~s}}}}=0.50 \\mathrm{{~m}} / \\mathrm{{s}}^2$
        Calculate the distance traveled over this $80 \\mathrm{{~s}}$ :
        $\\displaystyle x=v_o t+(1 / 2) a t^2=(20 \\mathrm{{~m}} / \\mathrm{{s}}) 80 \\mathrm{{~s}}+(1 / 2)\\left(0.50 \\mathrm{{~m}} / \\mathrm{{s}}^2\\right) 6400 \\mathrm{{~s}}^2=3200 \\mathrm{{~m}}$
        The average velocity is
        $\\displaystyle v_{{\\text {{avg }}}}=\\frac{{v+v_0}}{{2}}=\\frac{{20 \\mathrm{{~m}} / \\mathrm{{s}}+60 \\mathrm{{~m}} / \\mathrm{{s}}}}{{2}}=40 \\mathrm{{~m}} / \\mathrm{{s}}$
        If the acceleration is constant then the average velocity is the average of $20 \\mathrm{{~m}} / \\mathrm{{s}}$ and $60 \\mathrm{{~m}} / \\mathrm{{s}}$, or $40 \\mathrm{{~m}} / \\mathrm{{s}}$, and at an average velocity of $40 \\mathrm{{~m}} / \\mathrm{{s}}$ and $80 \\mathrm{{~s}}$, the distance traveled is $3200 \\mathrm{{~m}}$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_1_8(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Form the cross product of $\\boldsymbol{{A}}=23<37^{{\\circ}}$ and $\\boldsymbol{{C}}=47<193^{{\\circ}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The vectors and their components are shown in Fig. 1-14. Crossing $\\boldsymbol{{A}}$ into $\\boldsymbol{{C}}$ defines the angle as $156^{{\\circ}}$ so that $A C \\sin 156^{{\\circ}}-440$ with the direction out of the paper as give by the right-hand rule.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-1-14.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        Doing the same problem with determinants
        $\displaystyle A \\times C=\\left|\\begin{{array}}{{ccc}} i & j & k \\\\ 18.4 & 13.8 & 0 \\\\ -45,8 & -10.6 & 0 \\end{{array}}\\right|$
        $A \\times C=k[18.4(-10.6)-13.8(-45.8)]=437 k$
        The numeric difference, 440 versus 437 , is due to rounding the components to three significant figures.
        In simple problems it is easier to find the cross product from the geometric definition. In more complicated problems, or where it is difficult to be sure of the direction of the vector product, the determinant form is more coonvenient.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_1_7(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Form the dot product of $\\boldsymbol{{A}}=23<37^{{\\circ}}$ and $\\boldsymbol{{B}}=14<-35^{{\\circ}}$.

        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Using the first definition $\\boldsymbol{{A}} \\cdot \\boldsymbol{{B}}=A \\cdot B \\cos \\theta=23.14 \\cos 72^{{\\circ}}=100$.
        The second definition of the dot product requires the components of the vectors. The components of $\\boldsymbol{{A}}$ (from Fig. 1-1) are $a_1=18.4$ and $a_y=13.8$. The components of $\\boldsymbol{{B}}$ are in Fig. 1-11.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-1-11.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The dot product is
        $A \\cdot B=a_x b_x+a_y b_y=18.4 \\cdot 11.5+13.8(-8.0)=100$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_1_6(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Find the resultant of the two forces $F_1=800 \\mathrm{{~N}}<47^{{\\circ}}$ and $F_2=600 \\mathrm{{~N}}<140^{{\\circ}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Place the force vectors with components on the same diagram (Fig. 1-8).
        $\\begin{{aligned}} & F_{{1 x}}=800 \\cos 47^{{\\circ}}=546 \\\\ & F_{{1 y}}=800 \\sin 47^{{\\circ}}=585 \\\\ & F_{{2 x}}=600 \\cos 40^{{\\circ}}=460 \\\\ & F_{{2 y}}=600 \\sin 40^{{\\circ}}=386\\end{{aligned}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-1-8.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_1_5(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Add the vectors $\\boldsymbol{{A}}=13<50^{{\\circ}}, \\boldsymbol{{B}}=15<-60^{{\\circ}}$, and $\\boldsymbol{{C}}=17<20^{{\\circ}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        Diagram each vector along with the components.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        file_name = 'how-to-solve-physics-problems-fig-1-6.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The sum of these vectors $\\boldsymbol{{R}}=\\boldsymbol{{A}}+\\boldsymbol{{B}}+\\boldsymbol{{C}}$ is $\\boldsymbol{{R}}=1.2 \\boldsymbol{{i}}-11.5 \\boldsymbol{{j}}$. This resultant vector is diagrammed in Fig. 1-7 along with the magnitude and angle.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        file_name = 'how-to-solve-physics-problems-fig-1-7.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        The vector is $\\boldsymbol{{R}}=11.6<-84^{{\\circ}}$
        The key to getting vector addition problems correct is to use multiple diagrams. Most mistakes in vector problems are sign mistakes. And the way to avoid sign mistakes is to use diagrams extensively.

        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')

        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_1_4(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Diagram $T=A-C$ and write in number plus angle form.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The diagram is started by drawing the components on the coordinate system. With the components the magnitude and angle can be calculated $T=68.7<20.8^{{\\circ}}$.
        $T=\\sqrt{{64.2^2+24.4^2}}=68.7$
        $\\theta=\\tan ^{{-1}}(24.4 / 64.2)=20.8^{{\\circ}}$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-1-5.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_1_3(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The vector $\\boldsymbol{{A}}$ (Fig. 1-1) would be written as $\\boldsymbol{{A}}=18.4 \\boldsymbol{{i}}+13.8 \\boldsymbol{{j}}$ and $\\boldsymbol{{C}}$ (problem 1-1) would be written as $\\boldsymbol{{C}}=-45.8 i-10.6 j$
        Subtract $\\mathbf{{C}}$ from $\\mathbf{{A}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        $T=A-C=64.2 i+24.4 j$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_1_2(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        The vector $\\boldsymbol{{A}}$ (Fig. 1-1) would be written as $\\boldsymbol{{A}}=18.4 \\boldsymbol{{i}}+13.8 \\boldsymbol{{j}}$ and $\\boldsymbol{{C}}$ (problem 1-1) would be written as $\\boldsymbol{{C}}=-45.8 i-10.6 j$
        Add the vectors $\\boldsymbol{{A}}$ and $\\boldsymbol{{C}}$.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The addition of $\\boldsymbol{{A}}$ and $\\boldsymbol{{C}}$ is now accomplished by adding the components
        $S=A+C=-27.4 i+3.2 j$
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
            
            
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _problem_1_1(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        Diagram the vector $C=47<193^{{\\circ}}$ and write the components.
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        The components are $c_{{\\mathrm{{x}}}}=47 \\cos 193^{{\\circ}}=-45.8$ and $c_y=47 \\sin 193^{{\\circ}}=-10.6$.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        file_name = 'how-to-solve-physics-problems-fig-1-3.png'
        file_path = self._init_file_path(file_name)        
        problemTemplate.add_solution(file_path, 'image/jpg')
        
        soln_s = f"""
        Taking the sine or cosine of the $193^{{\\circ}}$ angle will produce the appropriate negative numbers, but drawing a figure and using the principle angle is a better procedure. Less mistakes are made from figures than from the readout of calculators. When the vector and its components are drawn on the coordinate system there can be no mistake that 45.8 and 10.6 are both negative.
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _sample(self, *args, **kwargs):
        problemTemplate = KProblemTemplate()
        problemTemplate.problem_type = kwargs.get('problem_type')

        problem_s = f"""
        """
        problemTemplate.add_problem(problem_s.strip(), 'text/html')
        
        soln_s = f"""
        """
        problemTemplate.add_solution(soln_s.strip(), 'text/html')
        
        answer_s = f"""
        """
        problemTemplate.add_answer(answer_s.strip(), 'text/html')
        
        return problemTemplate

    def _init_file_path(self, file_name):
        import pathlib
        from xv.util import locatePath
        rootFolder = pathlib.Path(__file__).parent.absolute()
        file_path = locatePath(file_name, 
                               parent_depth = 6,
                               rootFolder = rootFolder,
                               part_match = False,
                               search_type = 'file',
                               verbose = self.verbose,
                               )
        return os.path.join(rootFolder, file_path)