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34kurt
3 years ago
10

Compare and contrast the life cycle of a low mass star and a high mass star

Physics
1 answer:
True [87]3 years ago
6 0
Low mass: Live for billions (trillions?) of years. The first low mass red dwarfs in this universe still haven't died off yet, so we aren't completely sure what happens when they "die." 

<span>Very High Mass: Run through their fuel exceedingly fast. *Die* relatively quickly (in the range of tens to hundreds of millions of years instead of billions and beyond) and go out with style, Supernova that will leave behind a neutron star (the *kind of very high mass stars" end this way) or a black hole (the *very very high mass stars* end this way.)</span>
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The two main states of mechanical energy are ___________ and potential energy.
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<span>Potential energy and Kinetic energy</span>
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3 years ago
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What is meant by the statement,the linear expansivity of copper is 0.000017k
Aliun [14]

Answer:

The change in length per unit length per degree rise in temperature of copper is 0.000017k

Explanation:

Given that :

The linear expansivity of copper is 0.000017k. This simply means that ; for a given copper length, the length of such copper will increase by 0.000017k for every degree rose in temperature of the copper rod.

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What would happen if mass is continually added to a 1.4 solar mass neutron star?
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A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with
Andrej [43]

Answer:

5.4 J.

Explanation:

Given,

mass of the object, m = 2 Kg

initial speed, u = 5 m/s

mass of another object,m' = 3 kg

initial speed of another orbit,u' = 2 m/s

KE lost after collusion = ?

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Using conservation of momentum

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Initial KE = \dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2

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              = 31 J

Final KE = \dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J

Loss in KE = 31 J - 25.6 J = 5.4 J.

4 0
3 years ago
The peak intensity of radiation from a star named sigma is 2 x 10^6 mmkay. What is the average surface temperature of Sigma roun
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I had the same question and i got it right.

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