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yKpoI14uk [10]
3 years ago
10

The howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 8.9 km. 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/m2. A juvenile howler monkey has an coustic output of 63 µW. What is the ratio of the acoustic intensity produced by the juvenile howler to the reference intensity I0, at a distance of 210 m?
Physics
1 answer:
Airida [17]3 years ago
6 0

Answer:

\frac{I}{I_0}=113.68

Explanation:

P = Acoustic power = 63 µW

r = Distance to the sound source = 210 m

Acoustic power

P=IA\\\Rightarrow I=\frac{P}{A}\\\Rightarrow I=\frac{63\times 10^{-6}}{4\times \pi \times 210^2}

Threshold intensity = I_0=1\times 10^{-12}\ W/m^2

Ratio

\frac{I}{I_0}=\frac{\frac{63\times 10^{-6}}{4\times \pi \times 210^2}}{1\times 10^{-12}}\\\Rightarrow \frac{I}{I_0}=113.68

Ratio of the acoustic intensity produced by the juvenile howler to the reference intensity is 113.68

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