After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. The rockets explode high in the air and the sound travels out uniformly in all directions. If the sound intensity is 1.62 10-6 W/m2 at a distance of 165 m from the explosion, at what distance from the explosion is the sound intensity half this value

required distance is 233.35 m

Explanation:

Given the data in the question;

Sound intensity = 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 is proportional to 1/(distance)²

i.e

∝ 1/r²

Now, let r² be the distance where sound intensity is half, i.e ₂ = ₁/2

Hence,

₂/₁ = 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

v 154.9= 0.9 length length s little over 154.9cm