Respuesta :
Answer:
83dB
Explanation:
first we must convert decibels to intensity in [tex]W/m^2[/tex]
decibels: [tex]\beta =86dB[/tex]
we use the following equation for the conversion:
[tex]\beta =10log\frac{I}{I_{0}}[/tex]
Where [tex]I[/tex] is the intensity corresponding to the 86 decibels, and [tex]I_{0}[/tex] It is the minimum intensity that the human ear perceives: [tex]I_{0}=1x10^{-12}W/m^2[/tex]
So replacing [tex]\beta[/tex], and clearing for [tex]I[/tex]
[tex]86dB=10log\frac{I}{I_{0}}[/tex]
[tex]8.6dB=log\frac{I}{I_{0}}[/tex]
Now we use the exponential of 10 because this will cancel the logarithm.
[tex]10^{8.6}=10^{log\frac{I}{I_{0}} }[/tex]
[tex]10^{8.6}=\frac{I}{I_{0}}[/tex]
[tex]({I_{0})10^{8.6}=I[/tex]
Since [tex]I_{0}=1x10^{-12}W/m^2[/tex] , we have:
[tex](1x10^{-12}W/m^2)10^{8.6}=I[/tex]
[tex]I=3.98x10^{-4}W/m^2[/tex]
And because instead of two firecrackers now is only one, the intensity is reduced by half:
[tex]\frac{I}{2}=\frac{3.98x10^{-4}W/m^2}{2}=1.99x10^{-4}W/m^2[/tex]
And now returning to the decibel unit:
[tex]\beta =10log\frac{I}{I_{0}}[/tex]
we want to find the decibel value [tex]\beta[/tex] corresponding to the new intensity [tex]I=1.99x10^{-4}W/m^2[/tex]
[tex]\beta =10log\frac{1.99x10^{-4}W/m^2}{1x10^{-12}W/m^2}[/tex]
[tex]\beta =10log\frac{199.05x10^6}[/tex]
[tex]\beta =10(8.2989)[/tex]
[tex]\beta =82.989dB[/tex]≈ 83dB
the sound level if only one is exploted is 83dB
