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Lelechka [254]
2 years ago
5

A small source of sound waves emits uniformly in all directions. The total power output of the source is P. By what factor must

P increase if the sound intensity level at a distance of 20.0 m from the source is to increase 5.00 dB?
Physics
1 answer:
Paladinen [302]2 years ago
4 0

Answer:

3.16

Explanation:

We know that sound level in dB = 10log(I'/I) where I = initial intensity at 20.0m from source = P/A where P = total power and A = area at 20.0 m and I' = New intensity when the sound level increases by 5.00 dB at 20.0 m from source = P'/A where P' = Power at 5.00 dB and A = area at 20.0 m

dB = 5.00 dB. So,

5.00 dB = 10log(P'/A ÷ P/A)

5.00 dB = 10log(P'/P)

5/10 = log(P'/P)

0.5 = log(P'/P)

taking anti-logarithm of both sides,

10⁰°⁵ = P'/P

P'/P = 3.16

P' = 3.16P

So, P must be increased by a factor of 3.16 for the sound intensity level at a distance of 20.0 m from the source is to increase 5.00 dB.

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