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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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If frequency increases, then wavelength _________.
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Answer

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Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:
Natasha2012 [34]

The work done by \vec F along the given path <em>C</em> from <em>A</em> to <em>B</em> is given by the line integral,

\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r

I assume the path itself is a line segment, which can be parameterized by

\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath

with 0 ≤ <em>t</em> ≤ 1. Then the work performed by <em>F</em> along <em>C</em> is

\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}

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When Maggie applies the brakes of her car, the car slows uniformly from 15.3 m/s to 0 m/s in 2.23 s. How far ahead of a stop sig
nirvana33 [79]

Answer:

The answer is 34.119m

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where:

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