Baseball scouts use a radar gun to measure the speed of pitches. One particular model of radar gun emits a microwave signal at a frequency of 10.525 GHz (GHz = x10^9 Hz). What is the increase in frequency of these waves as seen by a 40.2 m/s fastball headed straight toward the radar gun? (Express your answer in the following format: #.# x 10^# Hz)

Respuesta :

Answer:

[tex]2.8\times 10^3 Hz[/tex]

Explanation:

We are given that

Frequency,f=[tex]10.525GHz=10.525\times 10^9 Hz[/tex]

Speed,v=40.2 m/s

We have to find the increase in frequency.

We know that

Speed of light,c=[tex]3\times 10^8m/s[/tex]

[tex]\Delta f=f\times \frac{v}{c}[/tex]

[tex]\Delta f=10.525\times 10^9\times \frac{40.2}{3\times 10^8}[/tex]

[tex]\Delta f=1.4\times 10^3Hz[/tex]

The waves are reflected therefore the change in frequency  at the gun is equal to twice the change in frequency.

Therefore, increase in frequency=[tex]2\Delta f=2\times 1.4\times 10^3=2.8\times 10^3 Hz[/tex]