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The volume control on a surround-sound amplifier is adjusted so the sound intensity level at the listening position increases from 23 to 61 dB. What is the ratio of the final sound intensity to the original sound intensity?

Respuesta :

Answer:

Ratio will be [tex]\frac{I_2}{I_1}=10^{3.8}=6309.5[/tex]

Explanation:

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

We know that level of intensity is given by

[tex]i=10log\frac{I}{I_0}[/tex]

So [tex]23=10log\frac{I_1}{I_0}[/tex]------eqn 1

And [tex]61=10log\frac{I_2}{I_0}[/tex]------eqn 2

Subtracting eqn 1 from eqn 2

[tex]61-23=10log\frac{I_2}{I_1}[/tex]

[tex]3.8=log\frac{I_2}{I_1}[/tex]

[tex]\frac{I_2}{I_1}=10^{3.8}=6309.5[/tex]

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