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Calculate how many times more intense is 90 dB sound compared to 40 dB sound? |
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Answer» Solution :Given DATA: `L=log(I)/(I_(0))=logI-logI_(0)` We get`90 dB = 9B=logI_(1)-logI_(0)`...(1) `40dB=4B=logI_(2)-logI_(0)`...(2) Subtract (2) from (1) `-50dB=5B=logI_(1)-logI_(2)` `5=log_(10)((I_(1))/(I_(2)))` `(I_(1))/(I_(2))=10^(5)` |
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