Sound intensity corresponding to 80 dB: [tex]1\cdot 10^{-4}W/m^2[/tex]
Step-by-step explanation:
The decibel of a normal conversation is 60 dB, which corresponds to an intensity of [tex]1\cdot 10^{-6} W/m^2[/tex]. (1)
The formula to use that relates the intensity of the sound in [tex]W/m^2[/tex] to the sound level in decibel is
[tex]\beta(dB)=10log_{10}(\frac{I}{I_0})[/tex]
Where
[tex]\beta[/tex] is the sound level in decibel
[tex]I[/tex] is the intensity
We can found [tex]I_0[/tex] by plugging in the numbers of (1):
[tex]\frac{\beta(dB)}{10}=log_{10}(\frac{I}{I_0})\\\frac{I}{I_0}=10^{\frac{\beta(dB)}{10}}\\I_0 = \frac{I}{10^{\frac{\beta(dB)}{10}}}=\frac{1\cdot 10^{-6} W/m^2}{10^{\frac{60}{10}}}=1\cdot 10^{-12} W/m^2[/tex]
Which corresponds to a sound level of 0 decibel.
Now we can find the intensity that corresponds to a sound level of
[tex]\beta=80 dB[/tex]
And from the equation we find:
[tex]I=I_0 10^{\frac{\beta(dB)}{10}}=(1\cdot 10^{-12})10^{\frac{80}{10}}=1\cdot 10^{-4}W/m^2[/tex]
Learn more about sound waves:
brainly.com/question/4899681
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