sound problem

[echoyjeff222]

Same question as here:

http://forums.studentdoctor.net/threads/absolute-sound-intensity-question.1074725/

If the absolute sound intensity from firing the bullet is 2.0 W/m^2 at a distance of 2 m from the rifle, what will be the absolute sound intensity at a distance of 100 m from the rifle?
A. 2.0 x ^-4 W/m^2
B. 8.0 x 10^-4 W/m^2
c. 2.0 x 10^-2 W/m^2
D. 2.8 x 10^-1 W/m^2

I did this using the "trick" from TBR. I remember that for every 3x greater distance, the sound intensity goes up by a factor of ten. So, I did 2m * (3^x) and found that it was between 3 and 4 to get to 100 meters. So, the intensity has to decrease by factor of 10^3 to 10^4, so that got me down to B. Is this OK reasoning?

---

Members don't see this ad.

 

[justadream]

@echoyjeff222

"I remember that for every 3x greater distance, the sound intensity goes up by a factor of ten."

I think you mean goes DOWN?

Anyways, the TBR "trick" is overcomplicating things.

Just remember the intensity is related to 1 / r^2.

So do: 2 / (50^2) and get the right answer.

 

[echoyjeff222]

@echoyjeff222

"I remember that for every 3x greater distance, the sound intensity goes up by a factor of ten."

I think you mean goes DOWN?

Anyways, the TBR "trick" is overcomplicating things.

Just remember the intensity is related to 1 / r^2.

So do: 2 / (50^2) and get the right answer.

Yeah sorry I meant down ... haha. And do you mean 2^2 / 50^2?

But I just wanted to make sure my method works.

 

[justadream]

Yeah sorry I meant down ... haha. And do you mean 2^2 / 50^2?

But I just wanted to make sure my method works.

Why 2^2?

Doesn't 2 / (50^2) get you to B?

 

[echoyjeff222]

Why 2^2?

Doesn't 2 / (50^2) get you to B?

one sec... meant 2 * (2^2/100^2). similar deal, nevermind ignore me