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Sindrei [870]
3 years ago
5

The volume control on a surround-sound amplifier is adjusted so the sound intensity level at the listening position increases fr

om 23 to 61 dB. What is the ratio of the final sound intensity to the original sound intensity?
Physics
1 answer:
amid [387]3 years ago
5 0

Answer:

Ratio will be \frac{I_2}{I_1}=10^{3.8}=6309.5

Explanation:

We have given that intensity level is increases from 23 dB to 61 dB

We know that level of intensity is given by

i=10log\frac{I}{I_0}

So 23=10log\frac{I_1}{I_0}------eqn 1

And 61=10log\frac{I_2}{I_0}------eqn 2

Subtracting eqn 1 from eqn 2

61-23=10log\frac{I_2}{I_1}

3.8=log\frac{I_2}{I_1}

\frac{I_2}{I_1}=10^{3.8}=6309.5

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