Question

Suppose that a public address system emits sound uniformly in all directions and that there are no reflections. The intensity at a location 26 m away from the sound source is 3.0 × 10-4 W/m2. What is the intensity at a spot that is 71 m away?

Answer 1

Answer:

The intensity at a spot 71 m away is $4.02*10^{-5} Wm^{-2}$

Explanation:

Given:

Initial intensity,$I_{1}=3.0*10^{-4} Wm^{-2}$ at a distance, $d_{1} = 26 m$

Required:

New intensity, $I_{2} =?$ at a distance, $d_{2} = 71 m$

Using the inverse square law,

I ∝ $\frac{1}{d^{2} }$

⇒$I_{1}$$I_{1}d_{1}^{2} =I_{2}d_{2}^{2}$

$I_{2} =\frac{I_{1} d_{1}^{2} }{d_{2}^{2}} =\frac{3.0*10^{-4}*26^{2} }{71^{2} }$×

$I_{2}=4.02*10^{-5} Wm^{-2}$

Thus, the intensity at a spot that is 71 m away is $4.02*10^{-5} Wm^{-2}$