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iris [78.8K]
3 years ago
6

Does time stop in a black hole

Physics
2 answers:
lbvjy [14]3 years ago
4 0

Answer:

In standard GR, nothing exists at the center of a black hole. The center of a black hole is a singularity, and because GR fails at that point it is simply removed from the manifold. That means that the singularity is not part of spacetime.

To answer your question more realistically, we believe that GR is an approximate theory that fails well before you reach the center. Unfortunately, we have no good alternative theory with which to answer the question in the region where GR fails. We simply don’t have any data from that regime and it is very hard to formulate a good theory without data. So there very well could be time at the center, but we simply don’t have a good way to even guess.

kifflom [539]3 years ago
4 0

Answer:

<u>Black</u><u> </u><u>hole</u><u> </u><u>is</u><u> </u><u> </u><u>a region of space having a gravitational field so intense that no matter or radiation can </u><u>escape</u><u> </u>

<u>Actually</u><u> </u><u>time</u><u> </u><u>can't</u><u> </u><u>be</u><u> </u><u>stop</u><u>.</u>

<u>although</u><u> </u><u>black</u><u> </u><u>hole</u><u> </u><u>slower</u><u> </u><u>a</u><u> </u><u>time</u><u> </u>

Explanation:

Time never cannot be stop totally.

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0.0239364 N

0.0057879 N

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V = Volume

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Buoyant force is given by

F_b=\rho gV\\\Rightarrow F_b=1.22\times 9.81\times 2\times 10^{-3}\\\Rightarrow F_b=0.0239364\ N

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Net vertical force is given by

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The value of the moment of inertia when the person is on the edge of the merry-go-round is I2=1/3 (m*L^2)

Moment of Inertia refers to:

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The moment of inertia of single particle rotating in a circle I1 = 1/2 (m*r^2)

here We note that the,

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The value of the moment of inertia when the person is on the edge of the merry-go-round is determined with parallel-axis theorem:

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I2(edge) = 1/3 (m*L^2)

learn more about moment of Inertia here:

<u>brainly.com/question/14226368</u>

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