Suppose that a public address system emits sound uniformly in all directions and that there are no reflections. The intensity at a location 27 m away from the sound source is 3.0 × 10-4 #W/m^2#. What is the intensity at a spot that is 87 m away?

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Answer 1 · 2015-11-03T19:14:06

#2.89xx10^(-5)"W/m"^2#

I have assumed that the inverse square law is followed here:

#Iprop(1)/(d^2#

#:.I=(k)/(d^2)#

For the 1st case we can write:

#I_1=(k)/(27^2)#

#:.k=3xx10^(-4)xx27^2=0.2187"W"#

For the 2nd case:

#I_2=(k)/(87^2)=(0.2187)/(7569)#

#I_2=2.89xx10^(-5)"W/m"^2#