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3 years ago

What is the outcome of a star that runs out of hydrogen

Physics

1 answer:

givi [52]3 years ago

3 0

<h2>Answer: The nucleus becomes entirely helium.</h2>

To shine, the stars transform their hydrogen into helium by means of <u>nuclear fusion</u>. When at half of its life a star is without hydrogen, the nucleus becomes entirely helium and <u>the star declines, becoming colder and brighter due to the energy generated by the nuclear reactions, then the star begins to contract. </u>

Nevertheless, if the star is bigger, this helium will be also consumed and the nucleus transformed in Carbon, then in Oxigen, and so on. Being the last transformation Iron until the star delivers all its energy.

Note this process will depend on how massive the star is.

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2) A constant net force acts on an object. Describe the motion of the object.?

77julia77 [94]

Constant acceleration

7 0

3 years ago

During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0

oksian1 [2.3K]

Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

$d=\frac{1,280,000,000km}{37,000}=34,594.59km$

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

$v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}$

hence, the speed of the Hubble is approximately 384km/min

5 0

3 years ago

Pre-questioning identifies a purpose for reading.

nadya68 [22]

Answer:

True

Explanation:

Pre-questioning may help a reader focus on information s/he hopes to find in the reading selection.

8 0

3 years ago

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Imagine that you pushed a box, applying a force of 30 Newton’s , over a distance of 5 meters. How much would you do have done ?

kati45 [8]

30 ),which is the force,divided by 5
30÷5=6

7 0

3 years ago

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Two masses, each weighing 1.0 × 103 kilograms and moving with the same speed of 12.5 meters/second, are approaching each other.

juin [17]

A perfectly elastic<span> collision is defined as one in which there is no loss of </span>kinetic energy<span> in the collision. Therefore, we just add the kinetic energies of each system. We calculate as follows:

KE = 0.5(</span>1.0 × 10^3)(12.5 )^2 + 0.5(1.0 × 10^3)(12.5 )^2
KE = 156250 J = 1.6 x 10^5 J -------> OPTION A

5 0

3 years ago