For a star that has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 is mathematically given as
L=2.7*10^30w
<h3> What is its luminosity?</h3>
Generally, the equation for the luminosity is mathematically given as
L=4*\pi^2*b
Therefore
L=4*\pi^2*b
L=4* \pi *(2.83*10^{18})*2.7*10^{-8}
L=2.7*10^30
In conclusion, the luminosity
L=2.7*10^30w
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Answer:
≈ 22¢
Explanation:
240 / 1000 = 0.240 kV
0.240 kV(2.5 A)(3 hr) = 1.8 kW•hr
1.8 kW•hr($0.12/kW•hr) = $0.216
Answer:
48.6°
Explanation:
The forward force, F equals the component of the weight along the slope.
So mgsinθ = ma where a = acceleration and θ = angle between the slope and the horizontal.
So a = gsinθ
Since we are given that a = 75%g = 0.75g,
0.75g = gsinθ
sinθ = 0.75
θ = sin⁻¹(0.75)
= 48.6°
Well if it was traveling for an hour then the answer is 8 miles.
V = f(wavelength)
22.0 = 0.0680 (wavelength)
wavelength = 323.52 m