Answered

Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

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?

Sagot :

Answer:

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.

ttruong2023

Answer:

341

Explanation:

Just to confirm

Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.