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brilliants [131]
3 years ago
8

The distance from Earth of the red supergiant Betelgeuse is approximately 643 light-years. If it were to explode as a supernova,

it would be one of the brightest stars in the sky. Right now, the brightest star other than the Sun is Sirius, with a luminosity of 26LSun and a distance of 8.6 light-years. Part A How much brighter in our sky than Sirius would the Betelgeuse supernova be if it reached a maximum luminosity of 1.1×1010 LSun? Express your answer using two significant figures.
Physics
1 answer:
valkas [14]3 years ago
5 0

Answer:

1.5 Times.

Explanation:

The first step in order to solve the problem is to define our values, from this we can proceed to find the apparent brightness relationship for the two objects,

So,

B_i = \frac{L}{4 \pi R^2}

B=Apparent Brightness

L= Luminosity

R=Radius

Then,

\frac{B_{Berelgeuse}}{B_{Sirius}} = \frac{\frac{L_{Betelgeuse}}{4\pi (D_{Betelgeuse})^2}}{\frac{L_{Sirius}}{4\pi (D_{Sirius})^2}}

\frac{B_{Berelgeuse}}{B_{Sirius}} = \frac{\frac{1.1*10^{10}*L_{sun}}{4\pi (643)^2}}{\frac{26L_{Sun}}{4\pi(8.6)^2}}

\frac{B_{Berelgeuse}}{B_{Sirius}} = \frac{1.1*10^{10}}{643^2}*\frac{8.6^2}{26}

\frac{B_{Berelgeuse}}{B_{Sirius}} = 15682.3

B_{Betelgeuse} = 15682.3B_{Sirius}

Under this relationship we can conclude that in the case of a Supernova, in our sky Betelgeuse Supernova would be 1.5 * 10 ^ 4 times brighter than Sirius seen from the earth to the sky.

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Given parameters:

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One wire carries a current of I1= 2 Amps, and the other carries a parallel current (same direction, wires are side by side) of I
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A 4-kg object falls vertically a distance of 5 m. its potential energy has changed by approximately how much?
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Given:
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Neglect air resistance.

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