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yuradex [85]
3 years ago
13

b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul

d be one of the brightest stars in the sky. Right now, the brightest star in the sky other than the Sun is Sirius (which has a luminosity of 26LSun and is 26 light years away). How much brighter than Sirius would the Betelgeuse supernova be (from our point of view) if it reached a maximum luminosity of 10^10LSun? c) There have been some claims that when Betelgeuse explodes it will be like having a second Sun in the sky. Compare Betelgeuse’s brightness to the Sun’s brightness at Earth. Is this likely to be correct?
Physics
1 answer:
Nat2105 [25]3 years ago
4 0

Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is \approx 1.37 \cdot 10^{-5} } the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

B = \displaystyle{\frac{L}{4\pi d^2}} (1)

where B is the brightness, L is the star luminosity and d, the distance from the star to the point where the brightness is calculated (measured). Thus:

b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

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Question 23
bonufazy [111]

If the gymnast mass were doubled, her height (h) from the top of the board would be as follows,

с  Stay the same

Explanation:

  • The Mass of an object or body does not affect the acceleration due to gravity in any kind of way.
  • Light weight objects accelerate more slowly than the heavy objects because when the forces other than the gravity also plays a major role.
  • Mass increases of a body when an object has higher velocity or the speed.
  • The greater the force of gravity, it would give a direct impact on the object's acceleration; thus considering only a force, the heavier the object is, it would accelerate faster. But an acceleration depends upon the two factors which are  force and mass.
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3 0
3 years ago
If pressure is increased from 200 kPa to 300 kPa, and the original volume of gas was 1.5 L, what is the new volume? Assume the t
Oxana [17]

Answer:

The answer to your question is:      V2 = 1 l

Explanation:

Data

P1 = 200 kPa

P2 = 300 kPa

V1 = 1.5 l

V2 = ?

Formula

                          P1V1 = P2V2

                          V2 = (P1V1) / P2

                          V2 = (200 x 1.5) / 300

                          V2 = 1 l

6 0
3 years ago
1.Which of the following is evidence that the Earth is round?
PtichkaEL [24]

Answer:

1\D. Both A and C

2\A true

5 0
3 years ago
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Please help!!
HACTEHA [7]

Answer:

An element is a pure substance that contains only one type of atom. TRUE. 2. ... FALSE, they are all carbon atoms. 7. The smallest unit of matter is an atom. TRUE. 8. When an ... TRUE. 10. Most of the mass of an atom is found in the electron layers. FALSE. 11. The nucleus of an atom contains protons and neutrons. TRUE ...

Explanation:

7 0
3 years ago
Please help!!!!!!!!!​
Elena-2011 [213]

Answer:

v = 12.86 km/h

v = 3.6 m/s

Explanation:

Given,

The distance, d = 13.5 km

The time, t = 21/20 h

                  =  1.05 h

The velocity of a body is defined as the distance traveled by the time taken.

                                     v = d / t

                                        =  13.5 km / 1.05 h

                                        = 12.86 km/h

The conversion of km/h to m/s

                                   1 km/h = 0.28 m/s

                                     12.86 km/h = 12.86 x 0.28 m/s

                                                         = 3.6 m/s

Hence, the velocity in m/s is, v = 3.6 m/s

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3 years ago
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