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3 years ago

12

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

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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Who would hire a forensic toxicologist

romanna [79]

Answer:

Scientists or doctors

Explanation:

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3 years ago

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5. A cheetah with a mass of 70 kg was clocked running at 72 mph (32 m's). How many joules of

blsea [12.9K]

Answer:

Kinetic energy = 35840 Joules

Explanation:

Given the following data;

Mass = 70kg

Velocity = 32m/s

To find the kinetic energy;

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

$K.E = \frac{1}{2}MV^{2}$

Where, K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

$K.E = \frac{1}{2}MV^{2}$

Substituting into the equation, we have;

$K.E = \frac{1}{2}*70*32^{2}$

$K.E = \frac{1}{2}*70*1024$

$K.E = 35 * 1024$

K.E = 35840 Joules.

Therefore, the kinetic energy possessed by the cheetah is 35840 Joules.

7 0

3 years ago

A girl with a mass of 27 kg is playing on a swing. There are three main forces

N76 [4]

The tension in the swing's chain at the bottom of the swing is 178.35 N.

The given parameters:

Mass of the girl, m = 27 kg
Speed of the girl, v = 3 m/s
Radius of the circle, r = 4 m

The tension in the swing's chain at the bottom of the swing is calculated as follows;

$T = mg + ma_c\\\\ T= mg + \frac{mv^2}{r} \\\\T = (12 \times 9.8) + (\frac{27 \times 3^2}{4} )\\\\T = 117.6 \ N \ + \ 60.75 \ N\\\\T = 178.35 \ N$

Thus, the tension in the swing's chain at the bottom of the swing is 178.35 N.

Learn more about tension in vertical circle here: brainly.com/question/19904705

3 0

2 years ago

Want brainlyist whats number am i thinking of from 1 to 100

tatuchka [14]

Answer:

50 out of 100 (ik its right)

4 0

3 years ago

Read 2 more answers

How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33×10−21 n

qwelly [4]

Answer:

The number of excess electrons on each sphere is 759

Explanation:

Given that,

distance , d = 20 cm

= 0.20 m

let the number of electrons is n

Electric force (F) = k × (n × e)² /d²

3.33 × $10^{-21}$ = 9 × $10^{9}$ × (n × 1.602 × $10^{-19}$ )² /0.2²

solving for n

n = 759

4 0

3 years ago