Question

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

Answer 1

Power emitted will be 0.314 mW

So option (A) will be correct option

Step-by-step explanation:

We have given threshold hearing
`I=10^(-12)W/m^2`

Distance is given r = 5 km =5000 m

We have to find the power emitted

Power emitted is equal to

`P=I* A`

`=10^(-12)* 4\pi r^2`

`=10^(-12)* 4* 3.14* (5000)^2`

=
`314* 10^(-6)watt=0.314mW`

So power emitted will be 0.314 mW

So option (A) will be correct option.