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Anastasy [175]
3 years ago
8

Thank you so much i might have failed my test if it wasn't for you thanks

Mathematics
2 answers:
user100 [1]3 years ago
5 0

Answer:

I didn't help but no problem

antoniya [11.8K]3 years ago
3 0

Answer:

okay

Step-by-step explanation:

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Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul
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0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

\bf B=\displaystyle\frac{L}{4\pi d^2}

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\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}

where

\bf I_1= light intensity at distance \bf d_1  

\bf I_2= light intensity at distance \bf d_2  

Let \bf d_2  be the distance we would have to place the 50-watt bulb, then replacing in the formula

\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

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