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Alexxandr [17]
3 years ago
8

Approximately what core temperature is required before hydrogen fusion can begin in a star?

Physics
1 answer:
valkas [14]3 years ago
8 0

Answer:

The temperature required is near about 3 million kelvin

Explanation:

The brilliance of the star results from the nuclear reaction that take place in the core of the star and radiate a huge amount of thermal energy resulting from the fusion of hydrogen into helium.

For this reaction to take place, the temperature of the star's core must be near about 3 million kelvin.

The hydrogen atoms collide and starts and the energy from the collision results in the heating of the gas cloud. As the temperature comes to near about 1.5\times 10^{7 {\circ}C, the nuclear fusion reaction takes place in the core of the gas cloud.

The huge amount of thermal energy from the nuclear reaction gives the gas cloud a brilliance resulting in a protostar.

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A force acts on a 5kg object at rest. How fast will the object accelerate on a frictionless surface?
MakcuM [25]

Answer: The answer is C.) 25 m/s^2.

Explanation: If you input 5 as s, you would have to use the exponent 2.  This means that you have to multiply 5 by 5.  5 x 5= 25.  

Edit: Also, because the surface is frictionless, it will make the object go faster too.  Nothing can really slow it down unless something blocks it.

8 0
3 years ago
Can someone help with thsi? i will give brainliest
kogti [31]

Answer:

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Explanation:

4 0
3 years ago
A(n) _____ is a gap in the geologic record where some rock layers have been lost because of erosion.
Elenna [48]
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4 years ago
Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103
son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

PE = -\frac{GMm}{R}

As M=m, then

PE = -\frac{Gm^2}{R}

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

U = -\frac{2Gm^2}{R}

dU = -\frac{2Gm^2}{R}

dKE = -dU

Therefore replacing we have that,

\frac{1}{2}mv^2 =\frac{Gm^2}{2R}

Re-arrange to find v,

v= \sqrt{\frac{Gm}{R}}

v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}

v = 816.7m/s

Therefore the  velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

2r = 2*10^3m

Therefore

dU = Gm^2(\frac{1}{R}-\frac{1}{2r})

v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}

v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}

v = 1.83*10^7m/s

Therefore the velocity when they are about to collide is 1.83*10^7m/s

7 0
3 years ago
In your own words, explain what conservation of energy means. Also, give an example of the conservation of energy using somethin
Genrish500 [490]
Energy can not be created or destroyed but can change from one form to another.

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