Physics
A submarine produces an ultrasonic wave of velocity 1500 in water. The officer receives signal after 50 s of emission of ultrasonic waves. Find the distance of object which is present in the bottom of sea.
- 25.7 km
- 37.5 km
- 37.5 m
- 50 km
Sound
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Answer
37.5 km
Reason — As we know,
Total distance travelled by the sound in going and then coming back = 2d
Given,
t = 50 s
V = 1500 ms-1
Substituting the values in the formula above, we get,
Converting m to km
1000 m = 1 km
∴ 37500 m = km = 37.5 km
Hence, distance of the submarine from the object = 37.5 km
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