Explanation:
It is given that,
A bat flying in a cave emits a sound and receives its echo 0.1 s later, t = 0.1 s
The speed of sound, v = 340 m/s
We need to find the distance between the bat and the cave wall. Distance travelled is given by :
[tex]2d=v\times t[/tex]
[tex]2d=340\ m/s\times 0.1\ s[/tex]
d = 17 meters
It is the total distance between the bat and the cave wall is 17 meters. Hence, this is the required solution.
The distance of the cave wall will be d=17 m
It is given that
The echo sound comes back to the bat in tine t= 0.1s
Speed of the sound V=340 [tex]\frac{m}{s}[/tex]
So we have to find the distance traveled by the sound
Since the distance covered by the sound will be double
So
[tex]2d = v\times t[/tex]
By putting the values we get
[tex]2\times d=340\times 0.1[/tex]
[tex]d=17 m[/tex]
Thus the distance of the cave wall will be d=17 m
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