Answer:
0.049168726 light-years
Step-by-step explanation:
The apparent brightness of a star is
where
<em>L = luminosity of the star (related to the Sun)
</em>
<em>d = distance in ly (light-years)
</em>
The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.
Hence the apparent brightness of Alpha Centauri A is
According to the inverse square law for light intensity
where
light intensity at distance
light intensity at distance
Let
be the distance we would have to place the 50-watt bulb, then replacing in the formula
Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.
2p + 7p = 747
9p = 747 divded by 9 = 83
2 * 83 = 166 7 * 83 = 581
Hope it helps
First, you have to find what 80% of 15 is. In order to do that you have to change 80% percent to decimal, which would be .80. Then, multiply .80 and 15 to get 12. Subtract 12 from 15 and your answer is 3.
Answer:
x = 33
Step-by-step explanation:
sin 29° = 16/x
x = 33