A loudspeaker emits 31.6 W of acoustic power. A microphone with an effective cross-sectional area of...

Question

Question

A loudspeaker emits 31.6 W of acoustic power. A microphone with an effective cross-sectional area of 75.2 mm2 is located at a distance 194 m from the speaker. Calculate (a) the sound intensity at the location of the microphone, (b) the power incident on the microphone and (c) the total amount of energy that impinges on the microphone in 25.0 min.

science physics

Answer 1

Answer #1

a)

Intensity of microphone is

I=P/4pir²

I=31.6/4pi*194²

I=6.68*10^-5 W/m²

b)

Power incident on microphone

P_Mic=IA =(6.68*10^-5)(75.2*10^-6)

P_Mic=5.024*10^-9 W

c)

Total energy

E=Pt =(5.024*10^-9)(25*60)

E=7.537*10^-6 J

Answer 2

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