SOLUTION: A surface receiving sound is moved from its original position to a position three times farther away from the source of the sound. The intensity of the received sound thus becomes

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Question 972982: A surface receiving sound is moved from its original position to a position three times farther away from the source of the sound. The intensity of the received sound thus becomes
A. three times as low.
B. three times as high.
C. nine times as high.
D. nine times as low.
I think it may be C nine times as high
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
.
Sound varies with the inverse of the square of the distance. A distance three times farther yields:
intensity=1%2F%283%5E2%29
intensity=1%2F9
The intensity at the new distance is 1/9 the original intensity.
The given answers make no sense. Once you decrease an amount 1 time, nothing is left. I think the intended answer is D, but this, too, is wrong. If we take a quantity, say 1, and take nine times it we get 9. Now if we decrease the original 1 by nine times 1 we have 1-0=-8. So even answer D would imply after making the distance three times greater the sound intensity is -8 (if the original intensity is considered as 1). In general, the phrases "three times as low", "three times smaller", or "three times less" have no meaning.