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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Answer:
Centre of mass is the point at which the distribution of mass is equal in all directions, and does not depend on gravitational field. Centre of gravity is the point at which the distribution of weight is equal in all directions, and does depend on gravitational field.
momm=massxvelocity
momm=1200x2.5=120x25=600x5=3000kgm/s
<span>Assume: neglect of the collar dimensions.
Ď_h=(P*r)/t=(5*125)/8=78.125 MPa ,Ď_a=Ď_h/2=39 MPa
τ=(S*Q)/(I*b)=(40*〖10〗^3*π(〖0.125〗^2-〖0.117〗^2 )*121*〖10〗^(-3))/(π/2 (〖0.125〗^4-〖0.117〗^4 )*8*〖10〗^(-3) )=41.277 MPa
@ Point K:
Ď_z=(+M*c)/I=(40*0.6*121*〖10〗^(-3))/(8.914*〖10〗^(-5) )=32.6 MPa
Using Mohr Circle:
Ď_max=(Ď_h+Ď_a)/2+âš(Ď„^2+((Ď_h-Ď_a)/2)^2 )
Ď_max=104.2 MPa, Ď„_max=45.62 MPa</span>
sorry its quite messy haha