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?

a. v = 340.27 m/s

b. t = 0.15 s

Explanation:

a.

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

where,

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

therefore,

v = 340.27 m/s

b.

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

where,

t = time = ?

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

v = speed of sound = 340.27 m/s

Therefore,

t = 0.15 s

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.