Sound intensity at a microphone

[vertabatt]

Homework Statement

A concert loudspeaker suspended high off the ground emits 28.0W of sound power. A small microphone with a 0.900cm^2 area is 51.0m from the speaker.

A) What is I at the microphone?

B) How much sound energy impinges on the microphone in J?

Homework Equations

I = P/A
I = 4pir^2

The Attempt at a Solution

I have solved the answer to part A):

I = 28W/4pi(51)^2 = 8.57*10^-4 W/m^2The answer to part B is what's eluding me:

I know that at the microphone I = 8.57*10^-4 W/m^2

So:

8.57*10^-4 = P/.009m^2

P = 7.713*10^-6 W which is equal to J/s

Therefore my answer would be 7.713*10^-6 J of sound energy are impinged each second.

Mastering Physics is disagreeing with me. Any advice?

 

[vertabatt]

If I am doing something wrong or have not provided enough information, please let me know.

I must be missing something with this problem, because I can't seem to come up with a different answer than what I have shown.

Do I have the right approach? Am I making it too complicated by using the Intensity equation twice? Just a small nudge in the right direction would do wonders for me.

 

[vertabatt]

Okay, took me long enough, but I found my error! As usual, it's a stupid one!

They reported the area of the mic as .9cm^2, originally I converted the dimension linearly saying that that was equal to .009m^2. In actuality the area is .00009m^2.

Duh.