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

Answers

Answer 1

Answer:

Answer:

Schwarzschild radius

Explanation:

A black hole is defined as that object from which light cannot escape form the surface of the object because object is very small an very dense. There are numerous black hole sin the universe.

Black hole is well understood by the concept of "escape velocity". Escape velocity is the speed of any object to break the gravitational pull from another object.

Escape velocity depends on mass of the object and the distance from the center of the object ( how big or small the object is ). If the object is smaller or more denser, than the escape velocity is more. If we squeeze our earth to a radius of about 9 mm sphere, then the escape velocity of any object becomes equal to the velocity of light.

And the radius of any object which have an escape velocity which is same as the velocity of light is known as the 'Schwarzschild radius'.

Any object which is smaller than the "Schwarzschild radius' has an escape velocity more than the speed of light and is termed as a black hole.

Thus, any object which is squeezed beyond the Schwarzschild radius be becomes a black hole.

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Why can't we breath under water?

Answers

We can certainly draw water into our lungs, even though
our brain screams "Don't do that !".

But our lungs can only separate oxygen out of air, not out of
water. So if there's only water in our lungs, we pass out after
a short time because we're not getting any oxygen.

Our lungs are not designed breathe oxygen from water to be able to breathe under water.

A stealth bomber is flying straight and level at a constant speed of 150 m/s at an altitude of 3500 m above the ground. It drops a bomb intended to hit a specific target on the ground. a) How long will it take for the bomb to hit the ground?

b) At what horizontal distance away from its target must the bomber release the bomb in order to strike its target?

**disregard air resistance**

Answers

$Height = (1)/(2) gt^2 \n \n 3500 = (1)/(2) (9.8)(t^2)\n \n3500 = 4.9t^2\n\nt^2 =714.285714286\n\n\boxed {t=26.73~seconds}$

$\sf Horizontal ~distance = velocity * time\n\n Horizontal distance =150 * 26.73 \n\nHorizontal ~distance =4009.5$

red giant white dwarf black hole supernova a shock wave produced by extremely high temperatures a low-mass star that has shed its outer layers a star whose outer shell has a high-mass star that has collapsed from gravity expanded and cooled

Answers

Answer:

Asymptotic-giant-branch stars have helium-burning shells inside the hydrogen-burning shells, whereas red-giant-branch stars have hydrogen-burning shells only. In either case, the accelerated fusion in the hydrogen-containing layer immediately over the core causes the star to expand.

A dock worker loading crates on a ship finds that a 24 kg crate, initially at rest on a horizontal surface, requires a 65 N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 50 N is required to keep it moving with a constant speed. The acceleration of gravity is 9.8 m/s 2 . Find the coefficient of static friction between crate and floor.

Answers

Answer:

0.276

Explanation: Here we have to consider the kinetic friction force acting on the body. If it is going at a constant speed according to newton's 1 st law. There was'not net force induced at the system. There for kinetic friction foce must be equal to the 50N .

But here we have been asked to get cofficient of static friction . There for you have to get static friction force. It should be 65N.

Mass of the object is 24. Then the reaction between surface and the object would be 24*9.8 = 235.2 N

There for using this,

Static friction force = Cofficient of static friction * Reaction

Cofficient of static friction = 65/235.2

= 0.276

Answer:

$\mu_s=0.276$

$\mu_k=0.213$

Explanation:

It is given that,

Mass of the crate, m = 24 kg

Force acting on the crate when it is at rest, $F_r=65\ N$

When the crate is in motion, force acting on the crate, $F_m=55\ N$

To find,

The coefficient of static friction between crate and floor

Solution,

When the crate is at rest, the force acting on the crate is given by :

$F_r=\mu_sN$

$65=\mu_s* 24* 9.8$

$\mu_s=0.276$

When the crate is in motion, the force acting on the crate is given by :

$F_s=\mu_kN$

$50=\mu_k* 24* 9.8$

$\mu_k=0.213$

Hence, this is the required solution.

A star's apparent magnitude is most closely related to which of the following? Select all that apply.a. radius.
b. luminosity.
c. distance from Earth.
d. absolute magnitude. E. surface temperature.

Answers

Answer:

Luminosity

Explanation:

The apparent magnitude is a measure of the star's apparent luminosity, the apparent luminosity is directly related to the star's apparent magnitude.

The star's luminosity is the measure of the star's brightness and this diminishes with distance, the luminosity of a star also depends on the size of the star, this means that a star's magnitude and luminosity are directly related.

If you want the "most", then you can't have more than one.

A star's apparent magnitude is determined by both its intrinsic luminosity
and its distance from us.

A charge of +1.4 × 10Ò coulombs moves from point A to B. What is the potential difference between the two points if the work done on the charge is +3.4 × 10¥ joules?

Answers

The exponents in both data quantities needed to calculate the answer have been corrupted and misprinted, so no answer is possible. Proofreading was REALLY needed for this one, but alas . . .

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