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lora16 [44]
3 years ago
13

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

Physics
1 answer:
mixas84 [53]3 years ago
3 0

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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The pressure exerted by a gas is 2.0 atm while it has a volume of 350 mL. What would be the volume of this sample of gas at stan
PSYCHO15rus [73]

Answer:

volume is 700 mL

Explanation:

pressure = 2 atm

volume = 350 mL = 0.350 L

to find out

volume

solution

we will apply here equation that is

P1×V1 = P2×V2   ..............1

here P1 = 2 and V1 = 0.350 and P2 = 1 for standard atmospheric pressure

so put all value here  in equation 1 and get V2 volume

2 × 0.350 = 1 × V2

V2 = 0.700 L

V2 = 700 mL

so volume is 700 mL

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An object is dropped from a height of 75.0 m above ground level. (a) determine the distance traveled during the first second.
barxatty [35]

The relevant formula we can use in this case would be:

h = v0 t + 0.5 g t^2

where,

h = height or distance travelled

v0 = initial velocity = 0 since it was dropped

t = time = 1 seconds

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So calculating for height h:

h = 0 + 0.5 * 9.8 m/s^2 * (1 s)^2

<span>h = 4.9 meters</span>

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valentina_108 [34]

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