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gregori [183]
2 years ago
9

What is meant by black hole ?​

Physics
2 answers:
Xelga [282]2 years ago
8 0

Answer:

<h2>A black hole is a place in space where gravity pulls so much that even light can not get out. The gravity is so strong because matter has been squeezed into a tiny space. This can happen when a star is dying. Because no light can get out, people can't see black holes</h2>

Explanation:

<h2><u>PLEASE</u><u> </u><u>MARK</u><u> ME</u><u> BRAINLIEST</u><u> AND</u><u> FOLLOW</u><u> M</u><u> E</u><u> LOTS</u><u> OF</u><u> LOVE</u><u> FROM</u><u> MY</u><u> HEART</u><u> AND</u><u> SOUL</u><u> DARLING</u><u> TEJASWI</u><u> HERE</u><u> ❤️</u></h2>
guapka [62]2 years ago
3 0

Answer:

A black hole is a region of space having a gravitational field so intense that no matter or radiation can escape.

Explanation:

Sorry if this isn't what you meant or were looking for.

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Suppose you wish to whirl a pail full of water in a vertical circle at a constant speed without spilling any of its contents (ev
Yanka [14]

Answer:

V = 2.87 m/s

Explanation:

The minimum speed required would be that at which the acceleration due to gravity is negated by the centrifugal force on the water.

Thus, we simply need to set the centripetal acceleration equal to gravity and solve for the speed V using the following equation:

Centripetal acceleration = V^2 / r

where r is the distance of water from the pivot or shoulder.

For our case, r will be 0.65 + 0.19 = 0.84 m

and solving the above equation we get:

9.81 = V^2 / 0.84

V^2 = 8.2404

V = 2.87 m/s

6 0
3 years ago
A professor, with dumbbells in his hands and holding his arms out, is spinning on a turntable with an angular velocity. What hap
Olenka [21]

Answer:

<em>His angular velocity will increase.</em>

Explanation:

According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.

The angular momentum of a system = I'ω'

where

I' is the initial rotational inertia

ω' is the initial angular velocity

the rotational inertia = mr'^{2}

where m is the mass of the system

and r' is the initial radius of rotation

Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.

we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to I.

From

I'ω' = Iω

since I is now reduced, ω will be greater than ω'

therefore, the angular velocity increases.

5 0
3 years ago
The SI unit that is used to measure time is the
Sunny_sXe [5.5K]

Answer:second

Explanation:

4 0
3 years ago
Read 2 more answers
X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to
lys-0071 [83]

Answer:

a) \Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

b) \lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

c) E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

Explanation

Part a

For this case we can use the Compton shift equation given by:

\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

3 0
3 years ago
Read 2 more answers
Helpppppppppppppppppppppppppppopppppppppp
harkovskaia [24]

Answer:

I think it's 2 the photo is hard to tell what they are exactly talking about.

7 0
3 years ago
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