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olga55 [171]
3 years ago
12

What do we mean by the event horizon of a black hole?

Physics
2 answers:
TEA [102]3 years ago
5 0
<h2>Answer:</h2>

The event horizon is the surface of a black hole, it is the border of space-time in which the events on one side of it can not affect an observer on the other side.  

That is, at this border also called "point of no return", nothing can escape (not even light) and no event that occurs within it can be seen from outside.

 

Flura [38]3 years ago
3 0

Answer:

The event horizon is an imaginary surface surrounding a black hole, where the escape velocity is equal to the speed of light. Therefore, nothing within that horizon can escape due to the attraction of a large gravitational field

Explanation:

No particle that falls within the event horizon comes out again, since a speed greater than light would be needed, and so far no particle can reach it. In this way, there is no way to look inside the event horizon. For this reason, black holes have no visible characteristics.

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The spinning of earth on its axis causes the apparent rising and setting of the
rodikova [14]
Sun, moon, and some stars
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3 years ago
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A hollow metal sphere has 7 cm and 11 cm inner and outer radii, respectively. The surface charge density on the inside surface i
Alexus [3.1K]

Answer:

1)  E = 0 , 2) zero, 3)    E = - 2,162 10⁴ N/C , 4) directed towards the center of the sphere , 5) E = 1.412 10⁴ N / C , 6)direction coming out of the sphere

Explanation:

The electric field is a vector quantity, therefore we can calculate the field due to each charge distribution and add vector  

          E = E₁ + E₂

To calculate each field we can use Gauss's law, which states that the flow is equal to the charge  by the Gaussian surface divided by ε₀

For this case, let's take a sphere as a Gaussian surface

          Ф = ∫E dA = q_{int} /ε₀

 The area of ​​a sphere is

        A = 4π r²

         E = 1 / 4πε₀   q_{int} / r²

1) r = 4 cm

This radius is smaller than the radius of the sphere, therefore the charge inside is zero and therefore the field is zero

            E = 0

2) there is no field

3) r = 8 cm

Let's calculate each field, for the inner surface

This radius is larger than the internal radius, so the field is

           σ = q_{int} / A

The area of ​​the sphere is

          V = 4 π R_in²

         Rho = q_{int} / 4π R_in²

          q_{int} = ρ 4π R_in²

         E₁ = 1 /ε₀ ρ r_in² / r²

For the outer surface

This radius is smaller so there is no load inside the Gaussian surface and therefore the field is zero

          E₂ = 0

Total E

         E = E₁ + 0

          E = 1 /ε₀  ρ₁ R_in² / r²

Let's calculate

           E = 1 /8,854 10⁻¹²   250 10⁻⁹ (7/8)²

           E = - 2,162 10⁴ N / C

4) as the electric field is negative, it is directed towards the center of the sphere

5) r = 12 cm

In this case the two surfaces contribute to the electric field,

Inner surface

        Q₁ = ρ₁ 4π R_in²

        E₁ = 1 /ε₀  ₁rho1 R_in² / r²

Outer surface

         Q₂ = ρ₂ 4π R_out²

          E₂ = 1 /ε₀  ρ₂ R_out² / r²

The total field is

          E = E₁ + E₂

          E = 1 /ε₀  | ρ | [- R_in² + R_out²] / r²

Let's calculate

          E = 1 /8,854 10⁻¹² 250 10⁻⁹ [- 7² + 11²] / 12²

          E = 1.412 10⁴ N / C

6) As the field is positive, it is directed radially with direction coming out of the sphere

6 0
3 years ago
What is the characteristic of a Matrix Transpose
yulyashka [42]

Answer:

columns are converted into rows , and rows are converted into columns

5 0
3 years ago
The smallness of the critical angle θc for diamond means that light is easily "trapped" within a diamond and eventually emerges
hoa [83]

Answer:

26.6°

Explanation:

refractive index of diamond, n = 2.23

When a ray of light passes from denser medium to the rarer medium and refracts at an angle of 90 degree from the normal of the surface, such angle of incidence in the denser medium is called the critical angle.

By the Snell's law

\frac{Sini}{Sin r }= n

For critical angle, angle of incidence is critical angle, i = θc and angle of refraction, r = 90

So,

Sin θc / Sin 90 = 1 / 2.23

Sin θc = 0.448

θc = 26.6°

Thus, the critical angle is 26.6°.

3 0
3 years ago
Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E= 3.
svetlana [45]

Explanation:

The given data is as follows.

     Electric field between plates without dielectric, E_{1} = 3.50 \times 10^{5} V/m

   Electric field between the plates with dielectric, E_{2} = 2.40 \times 10^{5} V/m.

  Permittivity of free space, \epsilon_{o} = 8.85 \times 10^{-12} C^{2}/Nm^{2}

Now, we will determine the charge density as follows.

            \sigma_{i} = \epsilon_{o}(E_{1} - E_{2})

                 = 8.85 \times 10^{-12} \times (3.50 \times 10^{5} - 2.40 \times 10^{5})

                 = 9.735 \times 10^{-7} C/m^{2}

Thus, we can conclude that the charge density on each surface of the dielectric is 9.735 \times 10^{-7} C/m^{2}.

6 0
3 years ago
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