Question

A bat is using echolocation in a cave that is 15.5 degrees Celsius.

a. What is the speed of sound in the cave?

b. If one of the cave walls was 25m away from the bat, how long would it take for the sound’s echo to return to the bat?

Answer 1

a. v = 340.27 m/s

b. t = 0.15 s

Step-by-step explanation:

a.

The speed of sound at any given temperature can be found out by the use of the following formula:

`v = (331\ m/s)\sqrt{(T)/(273\ k)}`

where,

T = Absolute temperature = 15.5°C + 273 = 288.5 k

therefore,

`v = (331\ m/s)\sqrt{(288.5\ k)/(273\ k)}`

v = 340.27 m/s

b.

The time of echo to return can be found by the following formula:

`s = vt\\\\t = (s)/(v)`

where,

t = time = ?

s = distance = (2)(25 m) = 50 m

v = speed of sound = 340.27 m/s

Therefore,

`t = (50\ m)/(340.27\ m/s)`

t = 0.15 s