What is the Sound Intensity at 7.1 Meters?

[yankeesfan08]

A loudspeaker supported well above the ground on a post is supplied with
8.5 W of electrical power. The speaker is able to convert
5.6% of that into sound power. (That number is known as
the speaker's efficiency.) If we assume that sound
radiates equally in all directions, what is the sound intensity at a
distance of 7.1 m?

Please help? :shy:

 

[Tide]

What, precisely, is "intensity?" Look it up in your textbook. Understanding the definition will lead you to the answer!

 

[wais]

Based on the given information, we can calculate the sound power of the loudspeaker by multiplying the electrical power (8.5 W) by the efficiency (5.6% or 0.056). This gives us a sound power of 0.476 W.

We can then use the inverse square law to calculate the sound intensity at a distance of 7.1 m. The inverse square law states that sound intensity decreases in proportion to the square of the distance from the source.

So, at a distance of 7.1 m, the sound intensity would be (0.476 W) / (4 * π * (7.1 m)^2) = 0.0002 W/m^2. This is a very low sound intensity, as most humans can only hear sounds with intensities above 0.0001 W/m^2.

I hope this helps with your question! Remember to always check your units and use the appropriate formulas when solving physics problems. Good luck!