Physics: Principles and Problems
9th Edition
ISBN: 9780078458132
Textbook solutions
All Solutions
Page 455: Standardized Test Practice
Exercise 1
Step 1
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In 1987, a supernova was observed, so it was 34 years ago (2021). The scientist believed that galaxy was $d= 1.66 cdot 10^{21} hspace{0.5mm} mathrm{m}$ away from us. In 1987, they observed the light from that galaxy, and the speed of that light was $c = 3 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}$. To calculate the time for which the light was travel, we use the equation [ d= ct] Now, we can combine these information to calculate the time
begin{align*}
d&=ct\
t&= frac{d}{c}\
t&= frac{1.66 cdot 10^{21} hspace{0.5mm} mathrm{m}}{ 3 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}}\
t&=5.53 cdot 10^{12} hspace{0.5mm} mathrm{s}\
t&=5.53 cdot 10^{12} hspace{0.5mm} mathrm{s} cdot mathrm{frac{1 yr}{ 31.536 cdot 10^{6} s}} \
t&=1.75 cdot 10^{5} hspace{0.5mm} mathrm{yr}
end{align*}
To get precise value of time, we should add 34 years, but it is neglible to the value of $t$. So, we can say that te supernova explosion actually occur before [ framebox[1.1width]{$ therefore t=1.75 cdot 10^{5} hspace{0.5mm} mathrm{yr} $}]
begin{align*}
d&=ct\
t&= frac{d}{c}\
t&= frac{1.66 cdot 10^{21} hspace{0.5mm} mathrm{m}}{ 3 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}}\
t&=5.53 cdot 10^{12} hspace{0.5mm} mathrm{s}\
t&=5.53 cdot 10^{12} hspace{0.5mm} mathrm{s} cdot mathrm{frac{1 yr}{ 31.536 cdot 10^{6} s}} \
t&=1.75 cdot 10^{5} hspace{0.5mm} mathrm{yr}
end{align*}
To get precise value of time, we should add 34 years, but it is neglible to the value of $t$. So, we can say that te supernova explosion actually occur before [ framebox[1.1width]{$ therefore t=1.75 cdot 10^{5} hspace{0.5mm} mathrm{yr} $}]
Result
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$B.$ $1.75 cdot 10^{5} hspace{0.5mm} mathrm{yr}$
Exercise 2
Step 1
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The speed of a galaxy is $v=5.8 cdot 10^{6} hspace{0.5mm} mathrm{m/s}$, and it is moving away from us. An observer sees the light from that galaxy with a frequency of $f_{obs}=5.6 cdot 10^{14} hspace{0.5mm} mathrm{Hz}$. To find the frequency of emitted light $f$, we need to use equation [f_{obs}=f(1- frac{v}{c})] where $c= 3 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}}$. Now, we can calculate the frequency of emitted ligth
begin{align*}
f_{obs}&=f(1- frac{v}{c}) \
f&=f_{obs} (1- frac{v}{c})^{-1} \
f&= 5.6 cdot 10^{14} hspace{0.5mm} mathrm{Hz} (1- frac{5.8 cdot 10^{6} hspace{0.5mm} mathrm{m/s}}{3 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}}})^{-1} \
f&= 5.6 cdot 10^{14} hspace{0.5mm} mathrm{Hz} (1- 1.93 cdot 10^{-2})^{-1} \
f&= frac{5.6}{0.98} cdot 10^{14} hspace{0.5mm} mathrm{Hz} \
f&= 5.7 cdot 10^{14} hspace{0.5mm} mathrm{Hz}
end{align*}
The frequency of emitted light is [ framebox[1.1width]{$ therefore f= 5.7 cdot 10^{14} hspace{0.5mm} mathrm{Hz} $}]
begin{align*}
f_{obs}&=f(1- frac{v}{c}) \
f&=f_{obs} (1- frac{v}{c})^{-1} \
f&= 5.6 cdot 10^{14} hspace{0.5mm} mathrm{Hz} (1- frac{5.8 cdot 10^{6} hspace{0.5mm} mathrm{m/s}}{3 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}}})^{-1} \
f&= 5.6 cdot 10^{14} hspace{0.5mm} mathrm{Hz} (1- 1.93 cdot 10^{-2})^{-1} \
f&= frac{5.6}{0.98} cdot 10^{14} hspace{0.5mm} mathrm{Hz} \
f&= 5.7 cdot 10^{14} hspace{0.5mm} mathrm{Hz}
end{align*}
The frequency of emitted light is [ framebox[1.1width]{$ therefore f= 5.7 cdot 10^{14} hspace{0.5mm} mathrm{Hz} $}]
Result
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$C.$ $f= 5.7 cdot 10^{14} hspace{0.5mm} mathrm{Hz}$
Exercise 3
Solution 1
Solution 2
Step 1
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B) Red plus yellow produces magenta.
Result
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B) Red plus yellow produces magenta.
Step 1
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Let’s remember the light color combinations while solving this problem. We will check each one and with that find the one statement that is wrong.
a) Red plus green produces yellow. This one is true, we know that from the text.
b) Red plus yellow produces magenta. **This one is not correct**. Magenta is made by red and blue light, not red and yellow.
c) Blue plus green produces cyan. This one is also true.
d) Blue plus yellow produces white. White light is the light that is mixture of all colors. Because red and green make yellow light and then combining it with blue we get all three lights in one. That means this ones is true as well.
a) Red plus green produces yellow. This one is true, we know that from the text.
b) Red plus yellow produces magenta. **This one is not correct**. Magenta is made by red and blue light, not red and yellow.
c) Blue plus green produces cyan. This one is also true.
d) Blue plus yellow produces white. White light is the light that is mixture of all colors. Because red and green make yellow light and then combining it with blue we get all three lights in one. That means this ones is true as well.
Result
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b)
Exercise 4
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The illuminence of the sunlight is $E= 1 cdot 10^{5} hspace{0.5mm} mathrm{lx}$. A light on a stage has the luminous intensity of $5 cdot 10^{6} hspace{0.5mm} mathrm{cd}$, or if we write this as the luminous flux $P=4 pi 5 cdot 10^{6} hspace{0.5mm} mathrm{lm}$. To find the distance from a stage, we use the equation [E= frac{P}{4 pi d^{2}}] where $d$ is the distance from a stage. Now, we can calculate the distance as
begin{align*}
E&= frac{P}{4 pi d^{2}} \
d^{2}&= frac{P}{4 pi E} \
d^{2}&= frac{ 4 pi 5 cdot 10^{6} hspace{0.5mm} mathrm{lm}}{4 pi cdot 1 cdot 10^{5} hspace{0.5mm} mathrm{lx} } \
d^{2}&= 50 hspace{0.5mm} mathrm{m^{2}} \
d&= 7.1 hspace{0.5mm} mathrm{m}
end{align*}
A member of audience should be at the distance of [ framebox[1.1width]{$ therefore d= 7.1 hspace{0.5mm} mathrm{m} $}]
begin{align*}
E&= frac{P}{4 pi d^{2}} \
d^{2}&= frac{P}{4 pi E} \
d^{2}&= frac{ 4 pi 5 cdot 10^{6} hspace{0.5mm} mathrm{lm}}{4 pi cdot 1 cdot 10^{5} hspace{0.5mm} mathrm{lx} } \
d^{2}&= 50 hspace{0.5mm} mathrm{m^{2}} \
d&= 7.1 hspace{0.5mm} mathrm{m}
end{align*}
A member of audience should be at the distance of [ framebox[1.1width]{$ therefore d= 7.1 hspace{0.5mm} mathrm{m} $}]
Result
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$B.$ $d= 7.1 hspace{0.5mm} mathrm{m}$
Exercise 5
Solution 1
Solution 2
Result
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D) The color that an object appears to be is a result of the material absorbing specific light wavelengths and reflecting the rest.
Step 1
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We’ve already learned that the color we see is the wavelength of light which is reflected, while every other is absorbed by the material. That literally means the phrase **color** that was made **by subtraction of light**.
Result
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d)
Exercise 6
Step 1
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The illuminance of $60.0 hspace{0.5mm} mathrm{W}$ lightbulb is $E=9.35 hspace{0.5mm} mathrm{lx}$. Also, we know that lightbulb is at the distance of $d= 3.0 hspace{0.5mm} mathrm{m}$ from illuminated surface. To calculate the total luminous flux of the bulb we us [E= frac{P}{4 pi d^{2}}] where $P$ is the total luminous flux.Now, we can calculate the luminous flux as
begin{align*}
E&= frac{P}{4 pi d^{2}} \
P&= 4 pi E d^{2}\
P&= 4 pi 9.35 hspace{0.5mm} mathrm{lx} cdot (3.0 hspace{0.5mm} mathrm{m})^{2}\
P&=1057.46 hspace{0.5mm} mathrm{lm}\
P&=1.1 cdot 10^{3} hspace{0.5mm} mathrm{lm}
end{align*}
The total luminous flux of the bulb is [ framebox[1.1width]{$ therefore P=1.1 cdot 10^{3} hspace{0.5mm} mathrm{lm} $}]
begin{align*}
E&= frac{P}{4 pi d^{2}} \
P&= 4 pi E d^{2}\
P&= 4 pi 9.35 hspace{0.5mm} mathrm{lx} cdot (3.0 hspace{0.5mm} mathrm{m})^{2}\
P&=1057.46 hspace{0.5mm} mathrm{lm}\
P&=1.1 cdot 10^{3} hspace{0.5mm} mathrm{lm}
end{align*}
The total luminous flux of the bulb is [ framebox[1.1width]{$ therefore P=1.1 cdot 10^{3} hspace{0.5mm} mathrm{lm} $}]
Result
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$D.$ $P=1.1 cdot 10^{3} hspace{0.5mm} mathrm{lm}$
Exercise 7
Step 1
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The light from the Sun travels through space, so the light travels in a vacuum, with the speed $c = 3 cdot 10^{8} hspace{0.5mm} mathrm{m/s}$. The sunlight travels for [t= 8.0 hspace{0.5mm} mathrm{min} Rightarrow t= 480.0 hspace{0.5mm} mathrm{s}] so we can use the equation [d= ct] Now, we can calculate the distance
begin{align*}
d&= ct\
d&= 3 cdot 10^{8} hspace{0.5mm} mathrm{m/s} cdot 480.0 hspace{0.5mm} mathrm{s} \
d&= 1440 cdot 10^{8} hspace{0.5mm} mathrm{m} \
d&= 1.4 cdot 10^{11} hspace{0.5mm} mathrm{m} \
d&= 1.4 cdot 10^{8} hspace{0.5mm} mathrm{km}
end{align*}
The distance between the Sun and Earth is [ framebox[1.1width]{$ therefore d= 1.4 cdot 10^{8} hspace{0.5mm} mathrm{km} $}]
begin{align*}
d&= ct\
d&= 3 cdot 10^{8} hspace{0.5mm} mathrm{m/s} cdot 480.0 hspace{0.5mm} mathrm{s} \
d&= 1440 cdot 10^{8} hspace{0.5mm} mathrm{m} \
d&= 1.4 cdot 10^{11} hspace{0.5mm} mathrm{m} \
d&= 1.4 cdot 10^{8} hspace{0.5mm} mathrm{km}
end{align*}
The distance between the Sun and Earth is [ framebox[1.1width]{$ therefore d= 1.4 cdot 10^{8} hspace{0.5mm} mathrm{km} $}]
Result
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$C.$ $1.4 cdot 10^{8} hspace{0.5mm} mathrm{km}$
Exercise 8
Step 1
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The speed of light in a vacuum is $c= 3 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}$. The light has a wavelength of [ lambda=404 hspace{0.5mm} mathrm{nm} Rightarrow lambda=404 cdot 10^{-9} hspace{0.5mm} mathrm{m} ] Now, we can calculate the frequency
begin{align*}
f&= frac{c}{ lambda} \
f&= frac{ 3 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}}{ 404 cdot 10^{-9} hspace{0.5mm} mathrm{m}} \
f&= 7.43 cdot 10^{14} hspace{0.5mm} mathrm{Hz}
end{align*}
The frequency of the light is [ framebox[1.1width]{$ therefore f= 7.43 cdot 10^{14} hspace{0.5mm} mathrm{Hz} $}]
begin{align*}
f&= frac{c}{ lambda} \
f&= frac{ 3 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}}{ 404 cdot 10^{-9} hspace{0.5mm} mathrm{m}} \
f&= 7.43 cdot 10^{14} hspace{0.5mm} mathrm{Hz}
end{align*}
The frequency of the light is [ framebox[1.1width]{$ therefore f= 7.43 cdot 10^{14} hspace{0.5mm} mathrm{Hz} $}]
Result
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$D.$ $7.43 cdot 10^{14} hspace{0.5mm} mathrm{Hz}$
Exercise 9
Step 1
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The wavelength of emitted light is [ lambda = 525 hspace{0.5mm} mathrm{nm} ] The observed wavelength of light is [ lambda_{obs} = 473 hspace{0.5mm} mathrm{nm} ] Now, we can calculate the speed of an object
begin{align*}
lambda_{obs} – lambda &= pm frac{v}{c} lambda / : lambda \
frac{lambda_{obs}}{lambda} – 1 &=pm frac{v}{c}\
pm frac{v}{c}&= frac{ 473 hspace{0.5mm} mathrm{nm} }{ 525 hspace{0.5mm} mathrm{nm}} – 1 \
pm frac{v}{c}&= frac{473}{525} – 1 \
pm frac{v}{c}&= -0.099
end{align*}
Part of the equation $ frac{v}{c}$ has to be greater than zero, so the sign is “-“, which means, the celectial object is approaching.
begin{align*}
– frac{v}{c}&= -0.099\
frac{v}{c}&= 0.099\
v&= 0.099 c\
v&= 0.099 cdot 3 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}\
v&= 0.297 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}\
v&= 29.7 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}}\
end{align*}
The speed of a celectial object is [ framebox[1.1width]{$ therefore v= 29.7 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}} $}]
begin{align*}
lambda_{obs} – lambda &= pm frac{v}{c} lambda / : lambda \
frac{lambda_{obs}}{lambda} – 1 &=pm frac{v}{c}\
pm frac{v}{c}&= frac{ 473 hspace{0.5mm} mathrm{nm} }{ 525 hspace{0.5mm} mathrm{nm}} – 1 \
pm frac{v}{c}&= frac{473}{525} – 1 \
pm frac{v}{c}&= -0.099
end{align*}
Part of the equation $ frac{v}{c}$ has to be greater than zero, so the sign is “-“, which means, the celectial object is approaching.
begin{align*}
– frac{v}{c}&= -0.099\
frac{v}{c}&= 0.099\
v&= 0.099 c\
v&= 0.099 cdot 3 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}\
v&= 0.297 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}\
v&= 29.7 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}}\
end{align*}
The speed of a celectial object is [ framebox[1.1width]{$ therefore v= 29.7 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}} $}]
Result
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$$
v= 29.7 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}}
$$
v= 29.7 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}}
$$
Exercise 10
Step 1
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Intensity of nonpolarized light is $I_{0}$. That nonpolarized light goes through the polarizing filter, and the transmitted light is now polarized , with the intensity of [I_{1}= frac{I_{0}}{2} ] When polarized light goes through the polarizing filter, we can apply Malus’s Law [ I_{2} = I_{1} cos^{2} alpha ] In our case, the angle is $ alpha = 45 textdegree$, so we can calculate the intensity of transmitted light $I_{2}$
begin{align*}
I_{1}&= frac{I_{0}}{2}\
I_{2}&= I_{1} cos^{2} alpha \
I_{2}&= frac{I_{0}}{2} cos^{2} alpha \
I_{2}&= frac{I_{0}}{2} cos^{2} 45 textdegree \
I_{2}&= frac{I_{0}}{2} ( frac{ sqrt{2} }{2})^{2} \
I_{2}&= frac{I_{0}}{2} frac{1}{2}\
I_{2}&= frac{I_{0}}{4}
end{align*}
The light intensity that is emerging from the second polarizing filter [ framebox[1.1width]{$ therefore I_{2}= frac{I_{0}}{4} $}]
begin{align*}
I_{1}&= frac{I_{0}}{2}\
I_{2}&= I_{1} cos^{2} alpha \
I_{2}&= frac{I_{0}}{2} cos^{2} alpha \
I_{2}&= frac{I_{0}}{2} cos^{2} 45 textdegree \
I_{2}&= frac{I_{0}}{2} ( frac{ sqrt{2} }{2})^{2} \
I_{2}&= frac{I_{0}}{2} frac{1}{2}\
I_{2}&= frac{I_{0}}{4}
end{align*}
The light intensity that is emerging from the second polarizing filter [ framebox[1.1width]{$ therefore I_{2}= frac{I_{0}}{4} $}]
Result
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$$
I_{2}=frac{I_{0}}{4}
$$
I_{2}=frac{I_{0}}{4}
$$
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