A window whose area is `2m^2` opens on street where the street noise result in an intensity level at the window of 60 dB. How much acoustic power enters the window via sound waves. Now if an acoustic absorber is fitted at the window, how much energy from street will it collect in 5 h?

A window whose area is `2m^2` opens on street where the street noise r

1.	A window whose area is `2m^2` opens on street where the street noise result in an intensity level at the window of 60 dB. How much acoustic power enters the window via sound waves. Now if an acoustic absorber is fitted at the window, how much energy from street will it collect in 5 h?
Answer» Loudness level is given as `L=10log((I)/(I_0))` Thus, `10log((I)/(I_0))=60` `implies(I)/(I_0)=10^6` `impliesI=(10^-12xx10^6)=10^-6(W)/(m^2)=1mu(W)/(m^2)` and the power of sound is given by `P=IS` `=1xx10^-6xx2=2muW` Thus energy is given by `E=Pxxt` `=2xx10^-6xx5xx60xx60=36xx10^-3J`

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A window whose area is `2m^2` opens on street where the street noise result in an intensity level at the window of 60 dB. How much acoustic power enters the window via sound waves. Now if an acoustic absorber is fitted at the window, how much energy from street will it collect in 5 h?

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