Respuesta :

danielmaduroh

Explanation:

Here we have that the sound level at a point is 115dB so we want to know the intensity of sound at that point in [tex]W/m^2[/tex], that is, we need to convert from Intensity expressed in dB to intensity expressed in [tex]W/m^2[/tex]. So:

[tex]I(dB)=10log(\frac{I}{I_{0}}) \\ \\ \\ But: \\ \\ I_{0}=1\times 10^{-12}W/m^2 \\ \\ I(dB)=115dB \\ \\ \\ Substituting: \\ \\ 115=10log(\frac{I}{1\times 10^{-12}}) \\ \\ \\[/tex]

[tex]Isolating \ I: \\ \\ log(\frac{I}{1\times 10^{-12}})=\frac{115}{10} \\ \\ log(\frac{I}{1\times 10^{-12}})=11.5 \\ \\ \frac{I}{1\times 10^{-12}}=10^{11.5} \\ \\ \frac{I}{1\times 10^{-12}}=3.16\times 10^{11} \\ \\ I=(1\times 10^{-12})(3.16\times 10^{11}) \\ \\ I=3.16\times 10^{-12}\times 10^{11} \\ \\ I=3.16\times 10^{-12+11} \\ \\ \boxed{I=3.16\times 10^{-1}W/m^2}[/tex]

facundo3141592

For a sound level of 115 dB, the sound intensity is 0.316 w/m^2

How to get the sound intensity?

We know that the relation between sound level and sound intensity is given by:

[tex]B = 10*Log_{10}(\frac{I}{I_0} )[/tex]

Where I₀ = 10^{-12} W/m^2

B = 115 dB

Replacing that, we can solve for I.

[tex]115 = 10*log_{10}(\frac{I}{10^{-12}} ) = 10*\frac{ln(\frac{I}{10^{-12}})}{ln(10)} \\\\11.5*ln(10) = ln(\frac{I}{10^{-12}} )[/tex]

Now we apply the exponential function in both sides to get:

[tex]exp(11.5*ln(10)) = exp(ln(\frac{I}{10^{-12}} ))\\\\exp(11.5*ln(10)) = \frac{I}{10^{-12}} \\\\exp(11.5*ln(10))*10^{-12} = I = 0.316[/tex]

So the sound intensity is 0.316 w/m^2

If you want to learn more about sound intensity, you can read:

https://brainly.com/question/17062836

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