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A Person Standing Between Two Vertical Cliffs and is 640 M Away from the Nearest Cliff. He Shouted and Heard the First Echo After 45s and the Second Echo After Further 3s. Calculate the Speed - Physics

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Question

A person standing between two vertical cliffs and is 640 m away from the nearest cliff. He shouted and heard the first echo after 45s and the second echo after further 3s. Calculate the speed of sound in air and the distance between the cliffs.

Numerical

Solution

Given: d1 = 640m; t1 = 4s
t2 = 4 + 3 = 7
First echo is heard from the nearest cliff.
Total distance travelled = 2d = 2 × 640 = 1280 m
v = `(2"d"_1)/"t"_1=1280/4` = 320 m/s
Second echo is heard from the first cliff
v = `(2"d"_2)/"t"_2`
⇒ 320 ms−1 = `("2d"_2)/7`
d2 = `(320xx7)/2` = 1120 m
hence, the distance between two cliff
d1 + d2 = 640 + 1120 = 1760 m.

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Chapter 6: Echoes and Vibrations of Sound - Long Numerical

APPEARS IN

ICSE Physics [English] Class 10
Chapter 6 Echoes and Vibrations of Sound
Long Numerical | Q 3

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