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icang [17]
3 years ago
9

What pushes against gravity in: a main sequence star, a white dwarf, a neutron star, and a black hole? electron degeneracy, neut

ron degeneracy, nothing, and heat pressure nothing in all cases. Gravity always wins. nothing, heat pressure, electron degeneracy, and neutron degeneracy heat pressure, neutron degeneracy, electron degeneracy, and nothing heat pressure, electron degeneracy, neutron degeneracy, and nothing
Physics
1 answer:
Inga [223]3 years ago
3 0

Answer:

heat pressure, electron degeneracy, neutron degeneracy, and nothing

Explanation:

Main Sequence Star: It is a star in which nuclear fusion is happening in the core of the star. Hydrogen molecules fuse together to generate Helium. This nuclear fusion generates outward gas pressure and radiation pressure which balances the inward gravity thus creating an equilibrium which keeps the stars in shape.

White dwarf: It is the end stage of a medium sized star like the Sun. Outer layers of the star are thrown in the form a shell/bubble leaving a small and dense core in the center called as white dwarf. This core consists of carbon and oxygen. Nuclear fusion doesn't occur in the core of white dwarfs. The inward gravity is balanced by the electron degeneracy pressure. Thus these stars will keep on radiating the remaining heat and will turn in to a black dwarf at the end.

Neutron Star: This is the end stage of a supermassive star (1-3 times the mass of the Sun). At the last stage of the life the core collapses. In these stars the inward gravity is so huge that the pressure overcomes the electron degeneracy pressure and crushes together the electron and proton to form neutron. The neutron then stops the collapse and balances the inward gravity.

Black Hole: This is the end stage of a hyper massive stars weighing more than 3 times the mass of the Sun. The inward gravitational force is so huge that even the neutrons are not able to stop the collapse the core. thus the mass of the star collapses into a very small area of immense gravity. There is nothing that can balance this inward gravity.

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In a certain region of space, the electric field is constant in direction (say horizontal, in the x direction), but its magnitud
horsena [70]

Answer:

8.3 x 10⁻⁷ C

Explanation:

Electric flux will enter the face at x=0 and exit at face x= 25 m

On the other faces , field lines are parallel so no flux will enter or exit .

Flux entering the face at x = 0

= electric field x face area

= 560 x 25 x 25 = 350000 weber

Flux exiting  the face at x = 25

= 410 x 25 x25

= 256250 weber

Net flux exiting from cube ( closed face )

350000 - 256250  = 93750 web

Apply gauss'es theorem

Q / ε = Flux coming out

Q is charge inside the closed cube

Q / ε = 93750

Q = 8.85 x 10⁻¹² x 93750

= 8.3 x 10⁻⁷ C

7 0
3 years ago
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Irina18 [472]

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6 0
3 years ago
Johan's race time was 45.03 seconds. Kyle's race time was 0.1 second less than Johan's time. What was Kyle's race time?
frutty [35]
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6 0
4 years ago
An overnight rainstorm has caused a major roadblock. Three massive rocks of mass m1=528 kg, m2=838 kg and m3=311 kg have blocked
JulsSmile [24]

Answer:

FA=419,25N

F12=287,25N

Explanation:

The bulldozer is moving the rocks as one system, the weigth for the complete system is:

M=m1+m2+m3=1677kg

From newton´s second law the acceleration can be related to the force by:

FA=m*a=1677kg*0,250m/s^{2}=419,25N

on the middle rock we have the force F12 from the first to the middle rock on the direction of movement and F32 from the las rock to the middle rock on the opposite direction of movement.

For the last rock to accelerate at the same rate it must be subjected to a force:

F23=m3*a=311kg*0,250m/s^{2} =77,75N

This equals the force F32 on the opposite direction. the resultant force on the middle rock to mantain this acceleration should be:

F2=m2*a=838kg*0,250m/s^{2} =209,5N

The sum of all the forces applied to the middle rock is:

F2=F12-F32

Solving for F12:

F12=F2+F32=209,5N+77,75N=287,25N

3 0
3 years ago
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Artyom0805 [142]

Answer:

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8 0
3 years ago
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