Question

A loudspeaker has a circular opening with a radius of 0.08514 m.The electrical power needed to operate the speaker is 24.3 W. Theaverage sound intensity at the opening is 15.8 W/m².What fraction of the electrical power is converted by the speakerinto sound power? (Give as a decimal number between zero and one.)

Answer 1

   Given :

Electric power ( P ) = 24.3 W

Radious ( r ) = 0.08514 m

Sound intensity ( I ) = 15.8 W/m².

    Sound power ( P_s ) = IA      [ A = Surface area = πr² ]

                                  = (15.8 W/m².) (0.08514 m)² (π)

                                  = ---- Watts

     So, required percentage = (P_s / 24.3 W ) * 100

Solve the above .

I hope it helps you