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crimeas [40]
4 years ago
11

If you install special sound-reflecting windows that reduce the sound intensity level by 30.0 db , by what factor have you reduc

ed the sound intensity?
Physics
1 answer:
shusha [124]4 years ago
3 0

Answer: The sound intensity was reduced by -14.77 dB.

Explanation:

The level of the intensity of the sound with respect to the reference value is called Sound Intensity level.

L_I=10\log\frac{I}{I_o} dB

I = Intensity of the sound

I_o = Reference sound intensity

According to question, sound-reflecting windows that reduce the sound intensity level by 30.0 db .

Let the intensity of the sound be I_o and reduced intensity be I.

I=\frac{I_o}{30} dB

L_I=10\log\frac{I}{I_o} dB

L_I=10\log\frac{\frac{I_o}{30}}{I_o}=10\log\frac{1}{30} dB=-14.77 dB

The sound intensity was reduced by -14.77 dB.

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