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taurus [48]
3 years ago
7

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

Physics
1 answer:
Fynjy0 [20]3 years ago
6 0

The correct answer is: C. It is the point beyond which neither light nor anything else can escape.

You might be interested in
Two objects having masses m1 and m2 are connected to each other as shown in the figure and are released from rest. There is no f
tankabanditka [31]

Answer:m_1g

Explanation:

Given

Two masses m_1  and m_2  is released and there is tension T in the string

Suppose a is the acceleration of the system

Therefore from Diagram

For m_1

T-m_1g=m_1a

T=m_1(g+a)------1

for m_2 body

m_2g-T=m_2a

T=m_2(g-a)-------2

From above two Equation it is said that Tension is greater than m_1g and less than m_2g

m_1g

4 0
2 years ago
Read 2 more answers
Prompt
Mamont248 [21]

Answer:

1.bandages

2.cottons

3.scissor

4.medicines

6 0
2 years ago
Can Hotspots can be found at the edges of tectonic plates?
PolarNik [594]

Answer:

No.

Explanation:

Hot spots don't appear at the barriers of tectonic plates, instead, they originate at hot centres known as mantle plumes. Mantle plumes exist below the tectonic plates and may develop a string of volcanoes on the Earths surface.

6 0
2 years ago
3. A 10-centimeter diameter solid sphere made of a conducting material has 10 micro-Coulombs of charge placed upon it. What is t
bazaltina [42]

Answer:

zero

Explanation:

For a solid conducting sphere, charges are present on the surface of the sphere due to a phenomenon known as electrostatic sheilding. This affects the charge present in the body and makes it zero. However, the electrostatic potential appears to be equal to the whole present point that shows on the surface. The surface of a spherical conducting solid sphere is known as an equipotential surface. Thus, the potential difference between the two opposite points on the surface of the sphere will also be zero.

4 0
2 years ago
A 2.4 kg block is dropped onto a spring and platform of negligible mass. The block is released
statuscvo [17]

The speed of the block when the compression is 15 cm is 9.85 m/s.

The given parameters;

  • <em>mass of the block, m = 2.4 kg</em>
  • <em>height of the block, h =  5 m</em>
  • <em>compression of the spring, x = 25 cm = 0.25 m</em>

The spring constant is calculated as follows;

F = kx\\\\mg = kx\\\\k = \frac{mg}{x} \\\\k = \frac{2.4 \times 9.8}{0.25} \\\\k = 94.08 \ N/m

The speed of the block when the compression is 15 cm can be determined by applying the principle of conservation of energy;

\Delta K.E = \Delta P.E\\\\\frac{1}{2} m(v^2  - v_{0 }^2 ) = mgh - \frac{1}{2} kx^2\\\\\frac{1}{2} mv^2  = mgh -   \frac{1}{2} kx^2\\\\mv^2   = 2mgh - kx^2\\\\v^2 = \frac{2mgh - kx^2}{m} \\\\v = \sqrt{\frac{2mgh - kx^2}{m}} \\\\v = \sqrt{\frac{(2 \times 2.4 \times 9.8 \times 5) - (94.08 \times 0.15^2)}{2.4}} \\\\v = 9.85 \ m/s

Thus, the speed of the block when the compression is 15 cm is 9.85 m/s.

Learn more here:brainly.com/question/14289286

8 0
2 years ago
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