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k0ka [10]
3 years ago
7

The intensity of the sound from a certain source is measured at two points along a line from the source. The points are separate

d by 11.0 m, the sound level is 66.40 dB at the first point and 55.74 dB at the second point. How far is the source from the first point?
Physics
1 answer:
Artemon [7]3 years ago
8 0

Answer:

The source is at a distance of 4.56 m from the first point.

Solution:

As per the question:

Separation distance between the points, d = 11.0 m

Sound level at the first point, L = 66.40 dB

Sound level at the second point, L'= 55.74 dB

Now,

L = 10log_{10}\frac{I}{I_{o}}          

I = I_{o}10^{\frac{L}{10}} = I_{o}10^{0.1L} = 10^{- 12}\times 10^{0.1\times 66.40} = 10^{- 5.36}      

L' = 10log_{10}\frac{I'}{I_{o}}

I' = I_{o}10^{\frac{L'}{10}} = 10^{- 12}\times 10^{0.1\times 55.74} = 10^{- 6.426}        

where

I_{o} = 10^{- 12} W/m^{2}

I = Intensity of sound

Now,

I = \frac{P}{4\pi R^{2}}

Similarly,

I' = \frac{P}{4\pi (R + 11.0)^{2}}

Now,

\frac{I}{I'} = \frac{(R + 11.0)^{2}}{R^{2}}

\frac{10^{- 5.36}}{10^{- 6.426}} = \frac{(R + 11.0)^{2}}{R^{2}}

R ^{2} + 22R + 121 = 11.64R^{2}}

10.64R ^{2} - 22R - 121 = 0

Solving the above quadratic eqn, we get:

R = 4.56 m

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Answer:

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*PHYSICS HELP*
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My calculator is about 1cm thick, 7cm wide, and 13cm long.

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The question wants me to assume that the density of my calculator
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<h3>What is sa Galaxy?</h3>

Sa galaxies are normal spirals that have narrow, tightly wound arms, which usually are visible because of the presence of interstellar dust and, in many cases, bright stars.

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