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3 years ago

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

1 answer:

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.

shutvik [7]

<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.

Therefore, the change in length per unit length per degree rise in temperature (k) is 0.000017

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3 years ago

What would happen if mass is continually added to a 1.4 solar mass neutron star?

andrey2020 [161]

Eventually wouldn't it collapse in on itself and create black hole.

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3 years ago

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 = ?

Final velocity of the system

Using conservation of momentum

m u + m'u' = (m + m') V

2 x 5 + 3 x 2 = ( 2 + 3 )V

16 = 5 V

V = 3.2 m/s

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

= $\dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2$

= 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.

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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

Igoryamba

Answer:

1.45 K

I had the same question and i got it right.

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3 years ago

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