# Physics

This time I have two questions:
#1.A recording engineer works in a soundproofed room that is 49.7 dB quieter than the outside. If the sound intensity in the room is 2.66 x 10-10 W/m2, what is the intensity outside?

#2.A listener increases his distance from a sound source by a factor of 3.86. Assuming that the source emits sound uniformly in all directions, what is the change in the sound intensity level?

can i plz get some help?

2: Change in intensity level= 1/3.86^2
Now if you want it in db, it is -10log 1/3.86 db or 5.86 db less intense.

1: 49.7 = 10log (I/Io)

I/Io= 10^4.97
I= 2.66 x E-10 * 9.33E4

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1. 1. db = 10*Log (I/Io) =
10*Log (2.66*10^-10/1*10^-12) =
10*Log (266) = 24.2 db Inside.

Outside sound intensity level =
24.2 + 49.7 = 73.9 db.

db=10*Log I/Io=10*Log(I/10^-12)=73.9
10*Log (I/10^-12) = 73.9.
Log (I/10^-12) = 7.39.
I/10^-12 = 10^7.39. = 24.55*10^6.
I = 2.46*10^-5 W/m^2.

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posted by Henry

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