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.
593-540=53
9 R8
I hope that helped u:))
Answer:
Step-by-step explanation:
Both are squares so
(3x - 4)2
Answer:
7
Step-by-step explanation:
[(1/3 – 1/9)² + (1/3)³] : (1/3)⁴
Next, we shall simplify (1/3 – 1/9)². This is illustrated below:
(1/3 – 1/9)² = ((3 – 1)/9)² = (2/9)²
[(1/3 – 1/9)² + (1/3)³] : (1/3)⁴
= [(2/9)² + (1/3)³] : (1/3)⁴
= [4/81 + 1/27] : 1/81
= [(4 + 3)/81 ] : 1/81
= 7/81 : 1/81
= 7/81 ÷ 1/81
Invert
= 7/81 × 81/1
= 7
C = 2π×r
C =2π×7
C = 14π cm
7) C = 2π r
18π = 2π r
r = 9
diameter = 2r
diameter = 18 inches
