A bat flying in a cave emits a sound and receives its echo 0.1s later. Show that the distance to the wall of the cave is 17m.
Thanks!
distance = rate x time
speed of sound is 340 m/s (see your later question)
d = 340 x 0.1 = 34 meters.
BUT the sound must travel to the wall of the cave AND back so that makes it 17 m to the wall and 17 m back to the bat.
To determine the distance to the wall of the cave, we can use the speed of sound and the time it takes for the bat's sound to travel to the wall and back as an echo.
The speed of sound in air is approximately 343 meters per second.
Given that the sound emitted by the bat reaches the wall and returns as an echo in 0.1 seconds, we can infer that the total distance traveled by the sound is twice the distance to the wall.
Let's denote the distance to the wall as "d."
So, the total distance traveled by the sound is 2d.
We can use the formula:
Distance = Speed × Time
Using this formula, we have:
2d = (Speed of Sound) × (Time)
Plugging in the known values, we get:
2d = 343 m/s × 0.1 s
2d = 34.3 m
Dividing both sides of the equation by 2, we can isolate "d":
d = 34.3 m / 2
d = 17.15 m
Therefore, the distance to the wall of the cave is approximately 17 meters.