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2 years ago

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If a star collapses to a tenth its size, gravitation at its surface increases by?

1 answer:

tino4ka555 [31]2 years ago

8 0

If a star collapses to a tenth its size, gravitation at its surface increases by 100 times as much.

The contraction of an astronomical object caused by its own gravity, which tends to pull stuff inward toward the center of gravity, is known as gravitational collapse. A cloud of interstellar matter gradually collapses under the influence of gravity to form a star. The temperature rises as a result of the compression brought on by the collapse until thermonuclear fusion takes place in the star's core. At this point, the collapse gradually comes to an end as the outward heat pressure equalizes the gravitational forces. Following that, the star is in a condition of dynamic equilibrium. A star will repeatedly collapse once all of its energy sources have been used up until it reaches a new equilibrium condition.

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Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

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$- N - m g = \dfrac{-mv^2}{R}$

$N + m g = \dfrac{mv^2}{R}$

$N = \dfrac{mv^2}{R}- m g$

$N = m(\dfrac{v^2}{R}- g)$

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Solution :

(1) The equation used is,

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$U_{initial}$ = initial internal energy

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$U_{final}-(2000J)=(1000J)+(500J)$

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