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What will be the distance for distinct echo on a planet share urinal velocity is 500ms-1 |
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Answer» Let the distance between the source of sound and the obstacle be x METRES .THUS the distance TRAVELLED by sound to reach BACK to the source d=2xAs the minimum time REQUIRED to hear an echo is equal to 0.1secThus d=vt2x=500(0.1)⟹x=25m |
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