# a person standing between two vertical cliffs and 640m away from nearest cliff shouted he heard first echo after 4 sec and second echo 3 sec later. calculate (i) velocity of sound in air (ii) distance between cliffs

1) velocity of sound in air = ( distance/ time) = (640/ 4) ms-1 = 160 ms-1

2) distance of second cliff = speed X time = {160 X (4 + 3)} m =1120 m

Distance between cliffs = (640 + 1120) m = 1760 m

• 17

but y r we adding 4 and 3 ???? @  debraj.

• -13
A radar signal is reflected by an aeroplane and is received 2* 10-5 after it was sent .If the speed of these wave is is 3*10-8 how far is the aeroplane

• 6
A sound of wavelength 0.332m has a time period of 10-3sec if the time period is decreased to 10-4sec.Calculate the wavelength and frequency
• -2
We know that 2d=vt
(2d as sound travels till the cliff and back)
Therefore v=2d/t=2*640/4 =320 m/s

Distance = vt/2 = 320*7/2 = 1120 m/s
• 5
Velocity =distance /time Velocity =640/4=160m/s Distance= velocity *time taken Distance=160*(4+3)m Distance =1140m
• 2
1760m
• 0
We're adding 4 and 3 because the first echo is heard after 4 seconds and the second is heard 3 seconds after the 1st one, which means that the second echo takes (4+3) sec or 7 sec.
• 0
What are you looking for?