Answer:
Change in electric potential energy is -28.0 J
Explanation:
Electric potential energy is defined as the work is done to move a charge particle from one position to another in space in the presence of other charge particle or electric potential.
OR
Electric potential energy is also equal to the change in the configuration of the charge particles.
Thus,
Change in electric potential energy = - Work Done
According to the problem, Work Done is equal to 28 J. Thus,
Change in electric potential energy = -28 J
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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Answer:
20 N/m
Explanation:
From the question,
The ball-point pen obays hook's law.
From hook's law,
F = ke............................ Equation 1
Where F = Force, k = spring constant, e = compression.
Make k the subject of the equation
k = F/e........................ Equation 2
Given: F = 0.1 N, e = 0.005 m.
Substitute these values into equation 2
k = 0.1/0.005
k = 20 N/m.
Hence the spring constant of the tiny spring is 20 N/m
