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Sagot :
The sound intensity level at a distance of 1.0 km is 94.149 dB
The expression of the intensity of the sound is,
B=10log([tex]\frac{I}{Io}[/tex])
I=Io([tex]10^{\frac{B}{100} }[/tex])
= ([tex]10^{-12}[/tex])(10[tex]\frac{110}{10}[/tex])
= 0.1 W/[tex]m^{2}[/tex]
The new intensity of the sound is,
I' = [tex](\frac{r1}{r2}) ^{2} (I)[/tex]
= [tex](\frac{26m}{1000m} )^{2}(0.1W/m^{2} )[/tex]
= 0.00026W/[tex]m^{2}[/tex]
The intensity level of the sound is ,
β = 10log([tex]\frac{I}{Io}[/tex])
= 10log([tex]\frac{0.0026W/m^{2} }{10^{-12 W/m^{2} } }[/tex])
= 94.149 dB
Sound intensity, also called sound intensity, is defined as the power carried by a sound wave per unit area in a direction perpendicular to that area. The SI unit of intensity, which includes sound intensity, is watts per square meter (W/m2). One application is the noise measurement of sound intensity in air at the listener's location as the amount of sound energy.
Learn more about sound intensity here :
https://brainly.ph/question/5547134
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