Question

A French submarine and a U.S. submarine move toward each other during maneuvers in motionless water in the North Atlantic. The French sub moves at 43.00 km/hkm/h , and the U.S. sub at 64.00 km/hkm/h . The French sub sends out a sonar signal (sound wave in water) at 1000 Hz. Sonar waves travel at 5470 km/hkm/h . What is the signal's frequency as detected by the U.S. sub

Answer 1

Answer:

$f_{U}= 1019.72hz$

Explanation:

From the equation we are told that:

Velocity of French sub $V_{U}= 43.00km/h$

Velocity of U.S. sub at $V_{F}|=64.00 km//h$

French Wave Frequency  $F_{F}=1000Hz$

Velocity of wave  $V_{s}=5470 km/h$

Generally the equation for Signal's frequency as detected by the U.S. is mathematically given by

Doppler effect

 $f_{U} = \frac{ f_F (vs + v_{U})}{(v_s - v_F) }$

 $f_{U}= \frac{ 1000 x ( 5470+64)}{(5470-43)}$

 $f_{U}= 1019.72hz$