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 1.77 10-6 W/m2 at a distance of 261 m from the explosion, at what distance from the explosion is the sound intensity half this value?

Answer 1

d2 = 369.11 m

Step-by-step explanation:

`i_(1)` intensity of the sound at 261 m.

d1 distance of 261 m from te explosion

`i_(2)` intensity of the sound at d2 (
`i_(1)`)/2

d2 new distance

from te question,

we understand that I2 =(
`i_(1)`)/2 , so replacing in (1) we have:

`i_(1)` * d1² =(
`i_(1)`)/2* d2²

d2:

2 *
`i_(1)` * d1² / I1 = d2² ---> From here,
`i_(1)` can be striked out

d2² = 2di²

substituting the values of d1:

d2² = 2 * 261²

d2 = √136,242

d2 = 369.11 m

Answer 2

369.11 m

Step-by-step explanation:

To solve this, we need first to write the expression to calculate the distance:

I1 * d1² = I2 * d2² (1)

Where:

I1 intensity of the sound at 261 m.

d1 distance of 261 m.

I2 intensity of the sound at d2 (Half of I1)

d2 distance required.

Now we know that I2 = 1/2I1, so replacing in (1) we have:

I1 * d1² = 1/2I1 * d2²

Solving now for d2:

2 * I1 * d1² / I1 = d2² ---> From here, I1 gets canceled so:

d2² = 2di²

Replacing the values of d1:

d2² = 2 * 261²

d2 = √136,242

d2 = 369.11 m