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JulijaS [17]
3 years ago
12

The diagram shows waves of sound travelling through the air. By which factor is the sound intensity decreased at 3 meters?

Physics
2 answers:
SIZIF [17.4K]3 years ago
8 0
The diagram is missing; however, we know that the intensity of a sound wave is inversely proportional to the square of the distance from the source:
I(r)= \frac{1}{r^2}
where I is the intensity and r is the distance from the source.

We can assume for instance that the initial distance from the source is r=1 m, so that we put 
I= \frac{1}{r^2}= \frac{1}{(1)^2}=1
The intensity at r=3 m will be
I= \frac{1}{r^2}= \frac{1}{(3)^2}= \frac{1}{9}
Therefore, the sound intensity has decreased by a factor 1/9.
RSB [31]3 years ago
3 0

its a  on edge just finished my exam

heres the diagram

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The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

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