Emma L.
asked 12/07/20How far away are the fish?
A sound pulse emitted underwater reflects off a school of fish and is detected at the same place 0.25 s
later (see the figure below). How far away are the fish? (Assume the water temperature is 20°C.)
1 Expert Answer

Johnny T. answered 12/07/20
Mathematics, Computer Science & Electrical Engineering Tutor
The speed of sound in water at a temperature of 200 C is:
- Fresh water ... ≈ 1,481 m./s.
- Sea water ... ≈ 1,522 m./s.
The problem states that it takes the sound wave to 0.25 s.
distance = rate * time
Since the problem did not state in which kind of water this was done, we shall solve the problem for both cases:
- Case I - Fresh Water:
- distance = rate * time
- distance = ( 1481 m./s. ) * ( 0.25 s. )
- distance = 370.25 m.
- distance ≈ 370 m.
- Case II - Sea Water:
- distance = rate * time
- distance = ( 1522 m./s. ) * ( 0.25 s. )
- distance = 380.5 m.
- distance ≈ 380 m.
CJ H.
This answer is incorrect. Because the pulse traveled to the fish and back to the transmitter you must divide the answer by 2. Fresh Water: d=370/2 = 185m. Salt Water: d=380/2 = 190m.10/01/22
Justina G.
The correct answer should be: 1,440 m/s * 0.25 s / 2 = 180m02/22/23
Brittany P.
Thank you Justina G so much!04/28/23
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CJ H.
This answer is incorrect. You must account for the distance traveled twice. The pulse goes to the fish and back to the transmitter, so you must divide by two (2). So, Fresh Water d=370/2 = 185m and salt water d=380/2 = 190 m.10/01/22