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Sagot :
Answer:
required distance is 233.35 m
Explanation:
Given the data in the question;
Sound intensity [tex]I[/tex] = 1.62 × 10⁻⁶ W/m²
distance r = 165 m
at what distance from the explosion is the sound intensity half this value?
we know that;
Sound intensity [tex]I[/tex] is proportional to 1/(distance)²
i.e
[tex]I[/tex] ∝ 1/r²
Now, let r² be the distance where sound intensity is half, i.e [tex]I[/tex]₂ = [tex]I[/tex]₁/2
Hence,
[tex]I[/tex]₂/[tex]I[/tex]₁ = r₁²/r₂²
1/2 = (165)²/ r₂²
r₂² = 2 × (165)²
r₂² = 2 × 27225
r₂² = 54450
r₂ = √54450
r₂ = 233.35 m
Therefore, required distance is 233.35 m
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