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# Betelgeuse is a star in the constellation Orion. It radiates heat at the rate of 3.00 × 1030 W and ...

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4 years ago
Betelgeuse is a star in the constellation Orion. It radiates heat at the rate of 3.00 × 1030 W and has a surface temperature of 3000 K. Assuming that it is a perfect emitter, what is the radius of Betelgeuse? The Stefan-Boltzmann constant is 5.67 x 10-8 W/(m2•K4).
A) 7.80 × 1010 m
B) 8.70 × 1010 m
C) 1.40 × 1011 m
D) 1.90 × 1011 m
E) 2.30 × 1011 m
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## Physics

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Raj.SRaj.S
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4 years ago

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3 weeks ago
 QuoteA giant star has a surface temperature of 3000 K and the radiant power emitted by the star is 6 x 1030 W. Assume that the star is a perfect emitter and that it is spherical. What is the radius of the star?Stefan Boltzmann constant $$\sigma$$= 5.67x10^-8 W/m^2.KT = 3000 K, P = 6x1030 WFrom Stefan Boltzman law for perfect emitterPower emitted by giant star is P = $$\sigma$$AsT46x1030 = (5.67*10-8)(4x3.14xR2)(3000)4Radius of stat R = 3.225x1011 m