A recording engineer works in a soundproofed room that is 44.0 dB quieter than the outside. If the sound intensity in the room is 1.20*10^-10 W/m^2, what is the intensity outside?
how do you work this?
2007-07-03 13:50:03 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Xdb = 10Log(X/Xo)
X/(1.20*10^-10 W/m^2) = 10^(95/10)
X = (1.20*10^-10 W/m^2)10^(95/10)
X ≈ 0.379 W/m^2
2007-07-03 14:06:24 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋
44 dB is a ratio, unitless. It is a ratio equal to 25,118 (or it's inverse, depending on which magnitude is compared to which. This comes from inv log (44/10). The divide by 10 is a convention used in all dB calculations. (dB was defined somewhat arbitrarily).
So, 1.20E-10 * 25,118 = 3.01E-6
Since dBs are always unitless, the units of the intensity outside are the same as inside.
2007-07-03 22:09:19 · answer #2 · answered by Anonymous · 2⤊ 0⤋