Answer:
The distance when the intensity is halved is 173.95 m
Explanation:
Given;
initial intensity of the sound, I₁ = 1.52 x 10⁻⁶ W/m²
initial distance from the explosion, d₁ = 123 m
final intensity of the sound, I₂ = ¹/₂ (1.52 x 10⁻⁶ W/m²) = 0.76 x 10⁻⁶ W/m²
Intensity of sound is inversely proportional to the square of distance between the source and the receiver.
I ∝ ¹/d²
I₁d₁² = I₂d²
(1.52 x 10⁻⁶)(123)² = (0.76 x 10⁻⁶)d₂²
d₂² = (1.52 x 10⁻⁶ x 123²) / (0.76 x 10⁻⁶)
d₂² = 30258
d₂ = √30258
d₂ = 173.95 m
Therefore, the distance when the intensity is halved is 173.95 m