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anygoal [31]
3 years ago
8

If the solar system shrank so that the sun were located just one centimeter from Earth, about how far away could you find the ne

xt nearest star?
Physics
1 answer:
DedPeter [7]3 years ago
8 0
E                  S                                                               *

The "E" represents Earth, "S" represent Sun, and the "*" represents the nearest star(which is Proxima Centauri).

The main thing to worry about here is units, so ill label everything out.
D'e,s'(Distance between earth and sun) = .<span>00001581 light years
D'e,*'(Distance between earth and Proxima) = </span><span>4.243 light years

Now this is where it gets fun, we need to put all the light years into centimeters.(theres alot)
In one light year, there are </span>9.461 * 10^17 centimeters.(the * in this case means multiplication) or 946,100,000,000,000,000 centimeters.

To convert we multiply the light years we found by the big number.
D'e,s'(Distance between earth and sun) = 1.496 * 10^13 centimeters<span>
D'e,*'(Distance between earth and Proxima) = </span><span>4.014 * 10^18 centimeters
</span>
Now we scale things down, we treat 1.496 * 10^13 centimeters as a SINGLE centimeter, because that's the distance between the earth and the sun. So all we have to do is divide (4.014 * 10^18 ) by (<span>1.496 * 10^13 ).
Why? because that how proportions work.

As a result, you get a mere 268335.7 centimeters.

To put that into perspective, that's only about 1.7 miles

A lot of my numbers came from google, so they are estimations and are not perfect, but its hard to be on really large scales.</span>
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Answer:

Sound Intensity at microphone's position is 9.417\times 10^{- 4} W/m^{2}

The amount of energy impinging on the microphone is 9.417\times 10^{- 8} W/m^{2}

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Emitted Sound Power, P_{E} = 32.0 W

Area of the microphone, A_{m} = 1.00 cm^{2} = 1.00\times 10^{- 4} m^{2}

Distance of microphone from the speaker, d = 52.0 m

Now, the intensity of sound, I_{s} at a distance away from the souce of sound follows law of inverse square and is given as:

I_{s} = \frac{P_{E}}{Area} = \frac{P_{E}}{4\pi d^{2}}

I_{s} = \frac{32.0}{4\pi (52.0)^{2}} = 9.417\times 10^{- 4} W/m^{2}

Now, the amount of sound energy impinging on the microphone is calculated as:

If I_{s} be the Incident Energy/m^{2}/s

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