1.

A man is standing between two vertical walls 680m apart. He claps his hands and hears two distinct echoes after 0.9 seconds and 1.1 second respectively. What is the speed of sound in the air?

Answer»

Solution :VELOCITY =`("DISTANCE TRAVELLED by sound")/("Time taken")`
`d_(1)=(vt_(1))/(2)&d_(2)=(vt_(2))/(2)`
`d=(V)/(2)(t_(1)+t_(2))`
`v=(2d)/((t_(1)+t_(2)))`
`=(2xx680)/((0.9+1.1))`
`=(2xx680)/(2)`
`v=680ms^(-2)`
Given data.s
d=680m
`t_(1)=0.9s`
`t_(2)=1.1s`
v=?


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