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Veseljchak [2.6K]
3 years ago
10

How do black holes grow?

Physics
1 answer:
MariettaO [177]3 years ago
3 0

Answer:

Black holes of stellar mass are expected to form when very massive stars collapse at the end of their life cycle. After a black hole has formed, it can continue to grow by absorbing mass from its surroundings.

Explanation:

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A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal
dsp73

Answer:

v\approx 8.570\,\frac{m}{s}

Explanation:

The equation of equlibrium for the box is:

\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a

The formula for the acceleration, given in \frac{m}{s^{2}}, is:

a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}

Velocity can be derived from the following definition of acceleration:

a = v\cdot \frac{dv}{dx}

v\, dv = a\, dx

\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx

\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg}  \int\limits^{17\,m}_{0\,m}\, dx  - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx

\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}

v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}]  }

The speed after the box has travelled 17 meters is:

v\approx 8.570\,\frac{m}{s}

3 0
3 years ago
5. What is the velocity of a Nolan Catholic football player if he runs 200.0 m in 24.0 s?
Degger [83]

Answer:

\boxed{ \bold{ \huge{ \boxed {\sf{8.33 \: m \:/s \: }}}}}

Explanation:

Distance travelled = 200 metre

Time taken = 24 second

Velocity = ?

<u>Finding </u><u>the</u><u> </u><u>velocity</u><u> </u>

\boxed{ \sf{velocity =  \frac{distance \: travelled \: }{time \: taken}}}

\dashrightarrow{ \sf{velocity =  \frac{200 \: m}{24 \: s} }}

\dashrightarrow{ \sf{velocity = 8.33 \: m/s}}

Hope I helped!

Best regards!

6 0
3 years ago
Suppose the magnitude of the proton charge differs from the magnitude of the electron charge by a mere 1 part in 109
Oksana_A [137]

Answer:

 A) F = 1.09 10 5 N, b) Yes  

Explanation:

Part A

For this exercise we need the number of free electrons in copper, as the valence of copper +1 there is a free electron for each atom. Let's use the concept of density to find the mass of copper in the sphere

               ρ = m / V

               .m = ρ V = ρ 4/3 π r³

The radius is half the diameter

               r = 1.9 10⁻² / 2 = 0.95 10⁻² m

               ρ = 8960 kg / m3

               m = 8960 4/3 π (0.95 10⁻²)³

               m = 3.2179 10⁻² kg

The molecular weight of copper is 63,546 g / mol which has 6,022 10²³ atoms

With this we can use a rule of proportions to enter the number of atom is this mass

             #_atom = 6.022 10²³ 3.2179 10⁻² / 63.546 10⁻³

             #_atom = 3,049 10²³

Therefore there is the same number of electrons, as they indicate that the charge of the protone and the electon differs by 1/10⁹ the total charge for each spherical is

               q = e / 10⁹    #_atom

               q = e / 10⁹    3,049 1023

               q = 3,049 10⁴  (-1.6 10⁻¹⁹)

               q = -4,878 10-5 C

Electric force is

             F = k q₁q₂ / r²

             F = k q² / r²

             

Let's calculate

             F = 8.99 10⁹ (4.878 10⁻⁵)²2 / (1.4 10⁻²)²

              F = 1.09 10 5 N

This is a force of repulsion.

Part B

 The magnitude of this force is  in very easy to detect

4 0
3 years ago
An energy resource that can be replaced in a reasonable short period of time is call an
Semenov [28]
A Renewable resource
6 0
3 years ago
Read 2 more answers
Which element contains a full 2p orbital in its valence shell
alex41 [277]
So Neon ( Ne) is the correct answer.

3 0
3 years ago
Read 2 more answers
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