1 answer

A car traveling 40.0 m/s overtakes another car going only 25.0 m/s. When the faster car...

Question:

A car traveling 40.0 m/s overtakes another car going only 25.0 m/s. When the faster car is still behind the slower one, it sounds a horn of frequency 1600.0Hz. What is the frequency heard by the driver of the slower car?


Answers

f' = ((1 + - Uo/V) / (1 -+ Us/V)) f

Uo would be the slower car
Us would be the faster car
V is the speed of sound

f = ((1 + -25/340)/(1 - +40/340))*1600 = 1680 Hz

.

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