# 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?

Nov 3, 2015

$2.89 \times {10}^{- 5} {\text{W/m}}^{2}$

#### Explanation:

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

Iprop(1)/(d^2

$\therefore I = \frac{k}{{d}^{2}}$

For the 1st case we can write:

${I}_{1} = \frac{k}{{27}^{2}}$

$\therefore k = 3 \times {10}^{- 4} \times {27}^{2} = 0.2187 \text{W}$

For the 2nd case:

${I}_{2} = \frac{k}{{87}^{2}} = \frac{0.2187}{7569}$

${I}_{2} = 2.89 \times {10}^{- 5} {\text{W/m}}^{2}$