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A howler monkey is the loudest land animal and under some circumstances, can be heard up to a distance of 5.0km. Assume the acoustic output of a howler to be uniform in all directions and that the threshold of hearing is 1.0*10^-12 W/m^2. The acoustic power emitted by the howler is clostest to:

A) 0.31mW
B) 1.1mW
C) 3.2mW
D) 11mW

Respuesta :

Answer:

Power emitted will be 0.314 mW

So option (A) will be correct option

Explanation:

We have given threshold hearing [tex]I=10^{-12}W/m^2[/tex]

Distance is given r = 5 km =5000 m

We have to find the power emitted

Power emitted is equal to

[tex]P=I\times A[/tex]

[tex]=10^{-12}\times 4\pi r^2[/tex]

[tex]=10^{-12}\times 4\times 3.14\times (5000)^2[/tex]

=[tex]314\times 10^{-6}watt=0.314mW[/tex]

So power emitted will be 0.314 mW

So option (A) will be correct option.

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