All categories

3 years ago

Beyond what point must an object be squeezed for it to become a black hole

Physics

1 answer:

Naddik [55]3 years ago

4 0

Answer:

Condition

$\frac{m}{R} \geq \frac{c^{2} }{2*g}$

Explanation:

Basically black hole is an object from which light rays can not escape it means to go out from gravitational field , that body should thrown with speed greater then light.

Let's do some calculation

Gravitational potential at surface = $-\frac{G*M*m}{R}$

If we give kinetic energy equal to magnitude of Potential energy as on surface it will escape.

$\frac{m*v^{2} }{2}$ = $-\frac{G*M*m}{R}$

⇒ $\frac{m}{R} =\frac{c^{2} }{2*G}$

It will be more better for black hole if above ratio (analogous to density ) is more then above calculated

You might be interested in

What is the speed of each neutron as they crash together? keep in mind that both neutrons are moving?

Sophie [7]

This involves shooting electrons (from an accelerator) at a target or protons. This technique provided evidence for the existence of quarks. proton-antiproton scattering as well.
hope this helps

7 0

3 years ago

Read 2 more answers

A pendulum is used in a large clock. The pendulum has a mass of 2 kg. If the pendulum is moving at a speed of 2.9 m/s when it re

Rufina [12.5K]

This is a classic example of conservation of energy. Assuming that there are no losses due to friction with air we'll proceed by saying that the total energy mus be conserved.
$E_m=E_k+E_p$
Now having information on the speed at the lowest point we can say that the energy of the system at this point is purely kinetic:
$E_m=Ek=\frac{1}{2}mv^2$
Where m is the mass of the pendulum. Because of conservation of energy, the total energy at maximum height won't change, but at this point the energy will be purely potential energy instead.
$E_m=E_p$
This is the part where we exploit the Energy's conservation, I'm really insisting on this fact right here but it's very very important, The totam energy Em was
$E_M=\frac{1}{2}mv^2$
It hasn't changed! So inserting this into the equation relating the total energy at the highest point we'll have:
$E_p=mgh=E_m=\frac{1}{2}mv^2$
Solving for h gives us:
$h=\frac{v^2}{2g}.$
It doesn't depend on mass!

4 0

3 years ago

Read 2 more answers

Two clouds collide and form another, more massive cloud. One cloud is stationary, while the other is traveling at 1 m/s. After t

sweet [91]

Answer: 3

Explanation:

Given

One cloud is traveling at rate of $u_2=1\ m/s$

combined velocity of the two is $v=0.25\ m/s$

Suppose the masses of the clouds be $m_1,m_2$

Conserving momentum

$\Rightarrow m_1u_1+m_2u_2=\left(m_1+m_2\right)v\\\text{Divide whole equation by }m_2\\\Rightarrow \dfrac{m_1}{m_2}u_1+u_2=\left(\dfrac{m_1}{m_2}+1\right)v\\\\\Rightarrow 0+1=0.25\dfrac{m_1}{m_2}+0.25\\\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{0.75}{0.25}\\\\\Rightarrow \dfrac{m_1}{m_2}=3$

6 0

3 years ago

Will name brainliest please pleas answer

Greeley [361]

3 is the answer teeeeeeeeeheeeeeeeee

6 0

3 years ago

The following are the Earth–Sun distance at the equinoxes and solstices: March equinox 149.0 million km June solstice 152.0 mill

Mice21 [21]

Answer:

During winter (late December/early January) the Earth is closest to the Sun and during summer (late June/early July) the Earth is farthest from the Sun.

Explanation:

In the northern hemisphere, the earth usually comes closer to the sun during the time of winter season, mostly in late December or early January.

On the other hand, the earth is farthest from the sun during the time of summer season, mostly in late June or early July.

When the earth is closer to the sun, during the winter, it is comparatively cold. It is due to the absorption of a lesser amount of incoming solar radiation. The tilt of the earth is also responsible for this low temperature.

But, when the earth is farthest from the sun, during the summer, it is comparatively hot. It is due to the absorption of a large amount of incoming solar radiation.

5 0

3 years ago