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31 October, 01:01

A child sets off the firecracker at a distance of 100 m from the family house. what is the sound intensity β100 at the house?

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  1. 31 October, 01:02
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    To solve this problem, we use the formula:

    I100 / I1 = [P / 4π (100m) ^2] / [P / 4π (1m) ^2]

    I100 / I1 = 1 / 100^2

    I100 / I1 = 10^-4

    Therefore the change in intensity from 1m to 100m in decibels is:

    B100 - B1 = 10 log (10^-4) dB = - 40 dB

    So the intensity at 100m is calculated as:

    B100 = B1 - 40 dB = 140 dB - 40 dB = 100 dB

    Answer:

    100 dB
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