A Doppler radar sends a pulse at 6.00x109 Hz. It reflects off clouds

moving away at 8.52 m/s. What is the

change in frequency in the echo?

Explanation:

The problem is based on the concept of Doppler's effect of em wave .

Expression for apparent frequency can be given as follows

n = N x (V - v ) / ( V + v )

n is apparent frequency , N is real frequency , V is velocity of light  and v is velocity of cloud.

n = 6 x 10⁹ ( 3 x 10⁸ - 8.52 ) / ( 3 x 10⁸ + 8.52 )

= 6 x 10⁹ ( 3 x 10⁸ ) ( 3 x 10⁸ + 8.52 )⁻¹

= 6 x 10⁹ ( 3 x 10⁸ ) ( 3 x 10⁸)⁻¹ ( 1  + 8.52/3 x 10⁸ )⁻¹

= 6 x 10⁹ ( 1  - 8.52/3 x 10⁸ )

= 6 x 10⁹    - 6 x 10⁹x  8.52/ (3 x 10⁸ )

= 6 x 10⁹  1  - 170 .

So change in frequency = 170 approx.

lexy is right because she has the labeled s