70.1k views
20 votes
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?

User Robertos
by
3.7k points

2 Answers

7 votes

Answer:

341

Step-by-step explanation:

Just to confirm

User Kamara
by
4.0k points
7 votes

Answer:

Step-by-step 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.

User DrGeneral
by
4.3k points