a lady produces sound in front of a cliff and hears and echo in 5 seco

1.	a lady produces sound in front of a cliff and hears and echo in 5 seconds on moving closer to the Cliff by 165 m she hears the echo in 4 seconds what is the distance of the lady from the Cliff now
Answer» ong>Answer: Lets assume that the distance of the lady from the front of the CLIF= x. From x distance from the cliff, echo is HEARD in 5 seconds. THEREFORE, It TAKES 4 secods for the echo to be heard from a distance of x-165m. Since, T is DIRECTLY proportional to the distance. Therefore, D/T=k(constant) Therefore, D1/T1=D2/T2 =>x/5=(x-165)/4 =>4x= 5x-825 =>825=5x-4x=x Therefore, The distance between the cliff now = x-165 = 825-165= 660m *Hope it helps. Like and follow if it helps* ^_^

a lady produces sound in front of a cliff and hears and echo in 5 seconds on moving closer to the Cliff by 165 m she hears the echo in 4 seconds what is the distance of the lady from the Cliff now

Answer»

ong>Answer:

Lets assume that the distance of the lady from the front of the CLIF= x.

From x distance from the cliff, echo is HEARD in 5 seconds.

THEREFORE, It TAKES 4 secods for the echo to be heard from a distance of x-165m.

Since, T is DIRECTLY proportional to the distance.

Therefore, D/T=k(constant)

Therefore, D1/T1=D2/T2

=>x/5=(x-165)/4

=>4x= 5x-825

=>825=5x-4x=x

Therefore, The distance between the cliff now = x-165 = 825-165= 660m

Hope it helps. Like and follow if it helps ^_^

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a lady produces sound in front of a cliff and hears and echo in 5 seconds on moving closer to the Cliff by 165 m she hears the echo in 4 seconds what is the distance of the lady from the Cliff now

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