Question

A concert loudspeaker suspended high off the ground emits 25.0 W of sound power. A small microphone with a 0.500 cm^2 area is 55.0 m from the speaker. A.What is the sound intensity at the position of the microphone? Express your answer with the appropriate units.

B.How much sound energy impinges on the microphone each second? Express your answer with the appropriate units.

Answer 1

The sound energy impinges on the microphone each second is
`3.285*10^(-8)\ J/s`.

Step-by-step explanation:

Given that,

Sound power = 25.0 W

Area = 0.500 cm²

Radius r = 55.0

(a). We need to calculate the sound intensity

Using formula of sound intensity

`I=(P)/(A)`

`I=(25.0)/(4\pi*(55)^2)`

`I=0.000657=0.657*10^(-3)\ W/m^2`

(b). We need to calculate the sound energy impinges on the microphone each second

Using formula of intensity

`P=IA`

Put the value into the formula

`P=0.657*10^(-3)*0.500*10^(-4)`

`P=3.285*10^(-8)\ J/s`

Hence, The sound energy impinges on the microphone each second is
`3.285*10^(-8)\ J/s`.