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sergiy2304 [10]
3 years ago
12

Why do astronomers hypothesize that a massive black hole lies at the center of m87?

Physics
1 answer:
lana66690 [7]3 years ago
4 0
<h2>Answer:</h2>

According to <u>Einstein's theory of relativity</u>, a black hole is a <u>"singularity"</u> that consists of a region of the space in which the density of matter tends to infinity. In consequence, this huge massive body has a gravitational pull so strong that not even light can escape from it.

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

In other words, 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.  

In this sense, and according to the relativity, it is possible to determine where a black hole is if it is "observed" an <em><u>enormous amount of energy released</u></em>. So, in accordance to this, galaxies like ours must have a black hole in its center.

On the other hand, the elliptical galaxy Mesier 87 (also called Virgo A, but from now on M87) was showing the above described behaviour, with enormous jets of high-energy particles shooting away from its vicinity . This was imaged by the Hubble Space Telescope years ago; that is why astronemers were hypothesizing about the existence of a massive black hole there.

Well now, on April, 10th 2019 this was demonstrated with the publication of the image, for the first time, of the event horizon of the black hole in M87. This is the first time in human history a picture of a black hole is taken.

This was done by the huge effort of diverse scientist and by the syncronization of eight radio telescopes scattered across the Earth (located at: Hawaii, Spain, Chile, Mexico, Arizona and the South Pole), which took the same point of the sky at the same time.

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Water being turned into ice cubes in a freezer is an example of _____.
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2 years ago
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If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a
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This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

b) the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s w_o = 12 rad/s

Angular Velocity at time 0.15s w_f = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( w_f - w_o ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

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