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koban [17]
3 years ago
11

Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawate

r where the speed of sound is 1522 m/s. When the dolphin is swimming directly away at 8.5 m/s, the marine biologist measures the number of clicks occuring per second to be at a frequency of 2770 Hz.
1. What is the difference (in Hz) between this frequency and the number of clicks per second actually emitted by the dolphin?
Physics
1 answer:
Elenna [48]3 years ago
4 0

From Doppler effect we have that the frequency observed for the relation between the velocities is equivalent to the frequency observed. That is mathematically,

F_r = \frac{v}{v+v_s}F_s

Here,

Speed of sound in water v = 1522m/s

The Dolphin swims directly away from the observer with a velocity v_s = 8.5m/s

Observed frequency of the clicks produced by the dolphin F_r = 2770Hz

Replacing we have,

F_r = \frac{v}{v+v_s}F_s

F_s = \frac{v+v_s}{v}

F_s = 2770 (\frac{1522}{1522+8.5})

F_s = 2754.61Hz

Therefore the frequency emitted by the dolphin is 2754.61Hz

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4 years ago
A brick of mass 2.0kg is at rest. It falls to the ground through a
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I may not have the answer so i'll just give up some hints.

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Answer:

F_{Earth}= 15.57 N

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The weight of the sugar bag on Earth is:

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