B. their distances from the sun.
Explanation:
Absolute Magnitude:
Astronomers defines the absolute magnitude of a stars brightness in terms of how bright a star appears from a standard distance of 10 parsecs. Parsec is a unit of distance in astronomy. 10 parsecs is equal to 32.6 light years.
Apparent Magnitude:
Apparent magnitude of a star refers to how bright the star appears at its distance from the Earth.
If two stars have the same absolute magnitude but their apparent magnitude differs, the reason is that the distance of both the stars from the Earth varies. Hence their brightness differs when measured from Earth. The farther a star is from the Earth, the fainter its brightness.
Keywords: star, brightness, parsec, light years, apparent magnitude, absolute magnitude
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The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light , is calculated as follows: 3×
/ 1.72×
or approximately 1.74×
m.
The distance between the two positive, two negative, or two minimal points on the waveform is known as the wavelength of the wave. The following formula expresses the relationship between the frequency and wavelength of light:
f = c / λ
where, f = frequency of light
c = speed of light
λ = wavelength of light
Given data = f = 1.72×Hz
Therefore, λ = 3× / 1.72×
λ = 1.74×m
The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light , is calculated as follows: 3× / 1.72× or approximately 1.74×m.
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Answer:
Explanation:
Given that,
Initial angular velocity is 0
ωo=0rad/s
It has angular velocity of 11rev/sec
ωi=11rev/sec
1rev=2πrad
Then, wi=11rev/sec ×2πrad
wi=22πrad/sec
And after 30 revolution
θ=30revolution
θ=30×2πrad
θ=60πrad
Final angular velocity is
ωf=18rev/sec
ωf=18×2πrad/sec
ωf=36πrad/sec
a. Angular acceleration(α)
Then, angular acceleration is given as
wf²=wi²+2αθ
(36π)²=(22π)²+2α×60π
(36π)²-(22π)²=120πα
Then, 120πα = 8014.119
α=8014.119/120π
α=21.26 rad/s²
Let. convert to revolution /sec²
α=21.26/2π
α=3.38rev/sec
b. Time Taken to complete 30revolution
θ=60πrad
∆θ= ½(wf+wi)•t
60π=½(36π+22π)t
60π×2=58πt
Then, t=120π/58π
t=2.07seconds
c. Time to reach 11rev/sec
wf=wo+αt
22π=0+21.26t
22π=21.26t
Then, t=22π/21.26
t=3.251seconds
d. Number of revolution to get to 11rev/s
∆θ= ½(wf+wo)•t
∆θ= ½(0+11)•3.251
∆θ= ½(11)•3.251
∆θ= 17.88rev.
The proper difference between hor Rse and horse is both shows the same thing that is the horse
If you were given distance & period of time, you would be able to calculate the speed.
Hope this helps!
