Question

1. What is the intensity of a sound when an observer is 10.0 m from the speaker

that has power of 200. W?

Answer 1

About 133 db.

Step-by-step explanation:

Sound Intensity Level in db (SIL db) is equal to 10log (base 10) times the ratio of the sound intensity at 200 watts (I) relative to the sound intensity of the reference sound intensity (I sub 0), which by default is equal to 10⁻¹² W/m² or 0 dB.

I = 200 w / 10 m^2 = 20 w per square meter

I sub 0 = 10^-12 w per square meter

SIL = 10log ( I / I sub o) = 20 / 10^-12 = 10log ( 20^12) = 10 ( 13.3 ) = 133 db

Hope I typed this part correctly. Hard to get it in without being able to do exponents, etc. :D