Question

what abt black hole?
is that specify its density is so high?

Answer 1

\[\int\limits_{0}^{\pi}sinx\]

Answer 2

gravity

Answer 3

There is a lot of explanation abt black hole.......A place of extremely high gravity and density, that means Extreme mass in very small place.......That is Black hole.......The gravity is so massive that light waves can not go out of that surface and that also attracts the light comming from outside........Though light cannot go out, it seems to be black(Really black hole don't have black color)................This is black hole.

Answer 4

Mathematically, a black hole is any region that the density has gone to infinity, meaning that the radius is R=0. In Astrophysics though that is not the case. A black hole is actually the remains of a collapsed Massive Star. The idea is that when a star loses all of its nuclear energy, ie once is had reached the nuclear Iron burning, the burning starts to require an input of energy that is greater than the output of energy so the star is unable to hold itself against gravitational collapse. The star will undergo a supernova explosion, ie expelling all of their upper layers outward and depending on the size of the remaing core may be able to establish a hydrostatic equilibrium, meaning that that is remains as a star, called a neutron star, because now it is made of neutrons rather protons, nuetrons and electrons. But if the core is too big it will fail to reach an equilibrium and result in what we call a Black Hole, ie a region of space that has a finite radius described by the total mass of the core. The reason why they are referred to as Balck Holes, is not because they are actually black, but because their mass is so large in such a small space, ie their density is so LARGE that general relativistic effects must be taken into consideration. This means that a certain distance light is basically stopped and bent, making the region black, and inobservable. We can observe black holes indirectly throught their gravitational interactions and EM interactions, but directly it is impossible to observe them.