Since the Units presented are not in the International System we will proceed to convert them. We know that,
[tex]1 mi/h = 0.447 m/s[/tex]
So the speed in SI would be
[tex]V=95mi/h(\frac{0.447m/s}{1mi/h})[/tex]
[tex]V=42.465 m/s[/tex]
The change in frequency when the wave is reflected is
[tex]f'=f(1+\frac{V}{c})[/tex]
Or we can rearrange the equation as
[tex]f' = f + f\frac{V}{c}[/tex]
f' = Apparent frequency
f = Original Frequency
c = Speed of light
[tex]f'-f = f\frac{V}{c}[/tex]
[tex]\Delta f = f\frac{V}{c}[/tex]
Replacing,
[tex]\Delta f = (10.525*10^9)(\frac{42.465}{3*10^8})[/tex]
[tex]\Delta f =1489.8 Hz[/tex]
Since the waves are reflected, hence the change in frequency at the gun is equal to twice the change in frequency
[tex]\Delta f_T = 2 \Delta f[/tex]
[tex]\Delta f_T = 2(1489.8Hz)[/tex]
[tex]\Delta f_T = 2979.63Hz[/tex]
Therefore the increase in frequency is 2979.63Hz
The increase in frequency of these waves is equal to 2979.6 Hertz.
Given the following data:
Conversion:
1 mi/h = 0.447 m/s.
95.0 mi/h = [tex]95 \times 0.447[/tex] = 42.465 m/s.
To determine the increase in frequency:
Mathematically, the change in frequency of a wave is given by this formula:
[tex]F' = F(1+\frac{V}{c} )\\\\F' = F+F\frac{V}{c}\\\\F' - F=F\frac{V}{c}\\\\\Delta F = F\frac{V}{c}[/tex]
Where:
Substituting the given parameters into the formula, we have;
[tex]\Delta F = 10.525 \times 10^9 \times \frac{42.465}{3 \times 10^8}\\\\\Delta F = \frac{446.944 \times 10^9}{3 \times 10^8}\\\\\Delta F = 1489.8\;Hertz[/tex]
Since the waves were reflected, the increase in frequency toward the gun is double (twice) the change in frequency. Thus, we have;
[tex]F_2 = 2\Delta F\\\\F_2 = 2\times 1489.8\\\\F_2 = 2979.6\;Hertz[/tex]
Read more on frequency here: brainly.com/question/3841958
