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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 536: Practice Problems

Exercise 6

Step 1

1 of 2

$$
tt{ For an object moving towards the observer, the doppler formula is $f_{observed}=f_{source}(1+frac{v_{source}}{c})$}
$$

$$
begin{align*}
f_{observed}&=f_{source}(1+frac{v_{source}}{c})\
frac{f_{observed}}{f_{source}}&=(1+frac{v_{source}}{c})>1\
f_{observed}&>f_{source}
end{align*}
$$

Result

2 of 2

$$
tt{$$f_{observed}>f_{source}$$}
$$

Exercise 7

Step 1

1 of 2

Energy of a light wave is bottled in a particle like lumps.

Result

2 of 2

(d) energy.

Exercise 8

Step 1

1 of 1

Since the speed of the light in vacuum is a universal constant. The measurement of the speed of the light will not change.

Exercise 9

Step 1

1 of 2

Speed of distant of distant galaxies increases with increasing distance from the earth.

Result

2 of 2

The speed increases.

Exercise 10

Step 1

1 of 2

$$
tt{The basic defining characteristics of the photon are the energy that is carries and the no mass }
$$

Result

2 of 2

$$
tt{Energy,no mass}
$$

Exercise 11

Step 1

1 of 2

Suppose the frequency of source is $f_{source}$, and $v_{source}=c/2$.

Therefore when the source is directly moving towards me, the observed
frequency $f_{1}$ is

$$
f_{1}=f_{source}left(1+frac{v_{s}}{c}right)=f_{source}left(1+frac{c/2}{c}right)=frac{3}{2}f_{source}
$$

And when the source is moving directly away from me, the observed
frequency is

$$
f_{2}=f_{source}left(1-frac{v_{s}}{c}right)=f_{source}left(1-frac{c/2}{c}right)=frac{1}{2}f_{source}
$$

So the ratio is

$$
frac{f_{1}}{f_{2}}=3
$$

Result

2 of 2

Exercise 12

Step 1

1 of 2

One year is equal to $365times24times60times60=3.15times10^{7} {rm s}$.

Speed of light is $c=3times10^{8} {rm m/s}$

Therefore

one light year is

$$
1 {rm ly}=left(3times10^{8} {rm m/s}right)left(3.15times10^{7} {rm s}right)=9.45times10^{15} {rm m}=9.45times10^{12} {rm km}
$$

Result

2 of 2

$9.45times10^{12}$ km

Exercise 13

Step 1

1 of 2

Known:

$f_{s}=5.000times10^{14}$ Hz

$$
v_{s}=3325 {rm km/s}=3325times10^{3} {rm m/s}
$$

Unknown:

$$
f_{o}=?
$$

Solution:

$$
f_{o}=f_{s}left(1-frac{v_{s}}{c}right)=left(5.000times10^{14} {rm Hz}right)left(1-frac{left(3325times10^{3} {rm m/s}right)}{left(3times10^{8} {rm m/s}right)}right)=4.945times10^{14} {rm Hz}
$$

Result

2 of 2

$4.945times10^{14}$ Hz

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Page 536: Practice Problems

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