Question

The speaker of a public address system emits 20 kW power considering it a point source. What is
the sound intensity level at a point 4.00 m away?



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Answer 1

Explanation:

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Answer 2

Hello!

We know that Intensity is power upon area

Power is given as 20 x 1000 W

Area ===> Surface area of a sphere is 4 x pi x (r^2) = 4 x 3.14 x 16 = 64 x 3.14

Intensity = 20000 / (64 x 3.14) = 99.58 dB or approximately 100.                       In the key they have given us 140, but answer is 100.

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