Respuesta :
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
The sound intensity varies inversely as the square of the distance from the
explosion source.
- The distance from the explosion at which the sound intensity is half of 1.52 × 10⁻⁶ W/m², is approximately 173.95 meters
Reasons:
The sound intensity at 123 m = 1.52 × 10⁻⁶ W/m²
Required:
The distance at which the sound intensity is half the given value.
Solution:
Sound intensity is given by the formula;
[tex]I \propto \mathbf{ \dfrac{1}{d^2}}[/tex]
Which gives;
I × d² = Constant
I₁ × d₁² = I₂ × d₂²
Where;
I₁ = The sound intensity at d₁
I₂ = The sound intensity at d₂
[tex]d_2 = \mathbf{ \sqrt{\dfrac{I_1 \times d_1^2}{I_2} }}[/tex]
When the sound intensity is half the given value, we have;
I₂ = 0.5 × I₁
I₂ = 0.5 × 1.52 × 10⁻⁶ = 7.6 × 10⁻⁷
Therefore;
[tex]d_2 = \sqrt{\dfrac{1.52 \times 10^{-6} \times 123^2}{7.6 \times 10^{-7}} } \approx 173.95[/tex]
The distance from the explosion at which the sound intensity is half of the
sound intensity at 123 meters from the explosion, d₂ ≈ 173.95 m.
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