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IgorC [24]
3 years ago
9

Why do astronomers hypothesize that a massive black hole lies at the center of M87? Historical records show that a supermassive

star at the center of M87 exploded as a supernova, leaving behind a black hole. Time-lapse images from space telescopes show stars falling to the center of M87 and then disappearing from view. Images of M87 made with powerful telescopes show a well-defined black region devoid of any stars. A very small region at the center of M87 releases an enormous amount of energy.
Physics
1 answer:
antiseptic1488 [7]3 years ago
7 0
<h2>Answer: A very small region at the center of M87 releases an enormous amount of energy.</h2>

According to Einstein's theory of relativity, a<u> black hole is a "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, <u>it is possible to determine where a black hole is if it is "observed" an enormous amount of energy released.</u> 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 <u>Virgo A,</u> 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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A 60kg student traveling in a 1000kg car with a constant velocity has a kinetic energy of 1.2 x 10^4 J. What is the speedometer
777dan777 [17]

Answer:

17.64 km/h

Explanation:

mass of car, m = 1000 kg

Kinetic energy of car, K = 1.2 x 10^4 J

Let the speed of car is v.

Use the formula for kinetic energy.

K = \frac{1}{2}mv^{2}

By substituting the values

1.2\times 10^{4} = \frac{1}{2}\times 1000\times v^{2}

v = 4.9 m/s

Now convert metre per second into km / h

We know that

1 km = 1000 m

1 h = 3600 second

So, v = \left (\frac{4.9}{1000}   \right )\times \left ( \frac{3600}{1} \right )

v = 17.64 km/h

Thus, the reading of speedometer is 17.64 km/h.

4 0
3 years ago
A 50-V de voltage source was connected in series with a resistor and capacitor. Calculate the current in A to two significant fi
Naddik [55]

Answer:

Current through circuit will be 0.2706\times 10^6A

Explanation:

We have given source voltage v = 50 volt

Resistance R=25Mohm=25\times 10^6ohm

Capacitance C=0.1\mu F=0.1\times 10^{-6}F

Time t = 5 sec

Time constant of RC circuit is given by \tau =RC=25\times 10^6\times 0.1\times 10^{-6}=2.5sec

We know that voltage across capacitor is given by v_c=v_s(1-e^{\frac{-t}{\tau }})

v_c=50(1-e^{\frac{-5}{2.5 }})=43.2332v

So current will be =\frac{v_s-v_c}{R}=\frac{50-43.2332}{25\times 10^{-6}}=0.2706\times 10^6A

So current through circuit will be 0.2706\times 10^6A

4 0
4 years ago
An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force
sweet-ann [11.9K]

Answer:

1) A

2) C

3) B

4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

7) Incomplete information(picture missing)

8) A

9) C

10) C

Explanation:

1) m = 15kg, a = 10ms^{-2}, F = ma = 15*10 = 150N

2) m = 3kg, v = 4ms^{-1}, r = 4m, F = \frac{mv^{2} }{r}

\frac{3*4^{2} }{4} = 12N

3) a = 10ms^{-2}, r = 10m, v=?

F = \frac{mv^{2} }{r} and F = ma

equating the two equations and cancelling a, we have:

\frac{v^{2} }{r} = a

making v the subject of formula, we have:

v = \sqrt{ar}

= \sqrt{100}

= 10ms^{-1}

4) r = 10m, v = 5ms^{-1}, a = ?

F = \frac{mv^{2} }{r}

F = ma

equating the above equations and making a subject of formula, we get:

a = \frac{v^{2} }{r}

a = 25/10 = 2.5ms^{-2}

5) I can't find the picture associated with this question

6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

8) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is tripled

F = (3v)^{2}

F = 9v^{2}

We can see that the force will be 9X greater than it was.

9) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is doubled

F = (2v)^{2}

F = 4v^{2}

We can see that the force will be 4X greater than it was.

10) F = \frac{mv^{2} }{r}

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

7 0
3 years ago
Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a p slope at constant speed, as sho
kondor19780726 [428]

The work done by friction to move the sled is  - 1,323 J.

<h3>What is Coefficient of friction?</h3>
  • The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them.
  • Typically, it is represented by the Greek letter µ. In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
  • The coefficient of friction has no dimensions because both F and N are measured in units of force (such as newtons or pounds). For both static and kinetic friction, the coefficient of friction has a range of values.
  • When an object experiences static friction, the frictional force resists any applied force, causing the object to stay at rest until the static frictional force is removed. The frictional force opposes an object's motion in kinetic friction.

Solution:

Given that

Coefficient of friction (µ) = 0.10

Mass (m) = 90kg

distance covered (d) = 30m

We use the formula:

friction work = -µmgdcos∅

friction work = -0.100 × 90 kg × 9.8 m/s² × 30 m × cos 60°

friction work = - 1,323 J

Know more about Coefficient of friction numerical brainly.com/question/19308401

#SPJ4

5 0
2 years ago
What is electricity?​
const2013 [10]

Explanation:

electricity is the flow of electrical power or charge

4 0
3 years ago
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