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Tema [17]
3 years ago
15

Suppose that a white dwarf is gaining mass through accretion in a binary system. what happens if the mass someday reaches the 1.

4 solar mass limit? suppose that a white dwarf is gaining mass through accretion in a binary system. what happens if the mass someday reaches the 1.4 solar mass limit? the white dwarf will collapse to become a black hole. the white dwarf will undergo a nova explosion. the white dwarf will collapse in size, becoming a neutron star. the white dwarf will explode completely as a white dwarf supernova.
Physics
2 answers:
lukranit [14]3 years ago
8 0

Answer:

The white dwarf will explode completely as a white dwarf supernova.

Explanation:

When a white dwarf gains mass to the Chandrasekhar limit of 1.44 solar masses, it would begin to collapse and, in a few seconds, the matter in the white dwarf will undergo nuclear fusion, and will become in a supernova.  

Only when the core of the white dwarf is composed of neon, magnesium, and oxygen, the gaining mass to 1.44 solar mass, will result in a neutron star.      

Therefore, the answer is the white dwarf will explode completely as a white dwarf supernova.              

I hope it helps you!                

Soloha48 [4]3 years ago
7 0
It would blow up turning into a supernova.
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The gravitational field strength is approximately equal to 10 N.

<u>Explanation:</u>

Gravitational field strength is the measure of gravitational force acting on any object placed on the surface of the planet. Generally, the mass of the object is considered as 1 kg.

So the gravitational field strength will be equal to the gravitational force acting on the object.

The formula for gravitational field strength is

g = \frac{F}{m}

Here g is the gravitational field strength, m is the mass of the object placed on the surface and F is the gravitational force acting on the object.

Since, the mass of any object placed on the surface of earth will be negligible compared to the mass of Earth, so the mass of the object is considered as 1 kg.

Then the g = F

And F =\frac{GMm}{r^{2} }

Here G is the gravitational constant, M is the mass of Earth and m is the mass of the object placed on the surface, while r is the radius of the Earth.

g = F = \frac{6 \times 10^{24} \times 6.67 \times 10^{-11}  \times 1}{(6.6 \times 10^{6}) ^{2} }

g = 0.977 \times 10^1= 9.77\ N

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5 0
3 years ago
A solid cylinder of mass 7 kg and radius 0.9 m starts from rest at the top of a 20º incline. It is released and rolls without sl
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Answer:

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Using conservation of energy

Total Energy at the bottom = Gravitational potential energy at the top  

TE = PE

TE = m g h

TE = (7) (9.8) (2.3)

TE = 157.8 J

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