during takeoff, the sound intensity level of a jet engine is 110 db at a distance of 26 m . part a what is the sound intensity level at a distance of 1.0 km ?

Respuesta :

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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