a doppler radar sends a pulse at 6.00×109 hz. it reflects off clouds moving away at 8.52 m/s. what is the…

Explanation:

Step1: Recall Doppler - shift formula for moving observer

The Doppler - shift formula when the source is stationary and the observer is moving away is $\Delta f=-\frac{v}{c}f_0$, where $v$ is the velocity of the observer, $c = 3\times10^{8}\ m/s$ is the speed of light, and $f_0$ is the original frequency of the wave.

Step2: Identify the given values

We are given that $v = 8.52\ m/s$, $c=3\times10^{8}\ m/s$, and $f_0 = 6.00\times10^{9}\ Hz$.

Step3: Calculate the change in frequency

Substitute the values into the formula: $\Delta f=-\frac{8.52}{3\times 10^{8}}\times6.00\times 10^{9}$. First, calculate $\frac{8.52\times6.00\times 10^{9}}{3\times 10^{8}}=\frac{8.52\times60}{3}= 170.4$. Since the clouds are moving away, the change in frequency is negative. So, $\Delta f=- 170.4\ Hz$. The magnitude of the change in frequency is $170.4\ Hz$.

Answer:

$170.4$