A man stands in front of a large wall. He claps two pieces of wood together regularly at 0.5 second intervals. The echo of the first clap coincides with the third clap. If the speed of sound in air is 340 m s-1, the approximate distance between the man and the wall is ?
The answer is 170m.
Explain briefly and show steps.
2007-02-25 10:45:37 · 2 個解答 · 發問者 SinG 1 in 科學 ➔ 其他：科學
The echo of the first clap coincides with the third clap. that means the sound comes back after 1 second. The sound travelled 340 m and the distance between the man and the wall is 340/2 = 170 m
2007-02-25 12:44:15 · answer #1 · answered by ? 7 · 0⤊ 0⤋
Assumes L be the distance between the man and the wall
The first and third clap of sound totally have travelled a distance of two L
As the echo of the first clap coincides with the third clap .
The first clap first travel to the wall , then return to the boy.
The clap has travelled for 0.5+0.5s=1s for the time between the first and the third clap
So: 2L=340*(0.5+0.5)
L=170m
2007-02-25 13:27:42 · answer #2 · answered by Joe 2 · 0⤊ 0⤋