Question

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Answer 1

Explanation:

(i) An observer sees the flash first and hears the sound afterwards because the velocity of light which is equal to 3 × 108 ms-1 is much higher than the velocity of sound which is 340 ms-1.

(ii) V = 3d1/t

An echo or the second sound is heard after 3s. ∴ t = 3s

∴ 340 = 3d1/3

∴ d1 = 340 x 3/2 = 510 ms-1

In right △AOX: d12 = OX2 + d2 = x2 + d2

∴ d12 – d2 = x2 …(i)

But 2d = v x t (where t = 1s)

2d = 340 (v = 340 ms-1 and t = 1s)

∴ d = 170m …(ii)

From equation (i) and (ii), x = (510)2 – (170)2 = 10√2312

∴ x = 481 m

The distance between the observer and the wall is 481 m.

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