Question

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 1.67 10-6 W/m2 at a distance of 233 m from the explosion, at what distance from the explosion is the sound intensity half this value

Answer 1

the distance is 315.3696 m

Step-by-step explanation:

The computation of the distance is given below:

Given that

Sound intensity = 1.67 × 10^-6 W/m^2

And, the distance = 233 m

Now as we know that

Power = Intensity × surface area

1.67 × 10^-6 × 4π(233)^2 = 1.67 × 10^-6 ÷ 2× 4π × d^2

d^2 = 2 × (223)^2

= √2 × 223

= 315.3696 m

Hence, the distance is 315.3696 m