Asked by Anonymous
The drilling of a jackhammer was measured at 131dB. The sound of whispering was measured at
28dB. (Note that log2=0.301). The ratio of the intensity of the drilling to that of the whispering would
be approximately:
(A) 2X10^10
(B) 3X10^103
(C) 3X10^10
(D) 2X10^103
(E) 3X10^9
(note: dB=10logI/I0) Ans:A
28dB. (Note that log2=0.301). The ratio of the intensity of the drilling to that of the whispering would
be approximately:
(A) 2X10^10
(B) 3X10^103
(C) 3X10^10
(D) 2X10^103
(E) 3X10^9
(note: dB=10logI/I0) Ans:A
Answers
Answered by
oobleck
The difference is 103 dB so the ratio is 10^(10.3) = 10^0.3 * 10^10 = 2*10^10
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