The first law of thermodynamics invention that spurred its development is

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

antiseptic1488 [7]

<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.

7 0

3 years ago

a student wants to push a box of books with the mass of 50 kg in 3 m horizontally towards the location of the shelves where the

irina1246 [14]

Answer:

The work done is 360 J.

Explanation:

Given that,

Mass = 50 kg

Distance =3 m

We need to calculate the work done

The work done is equal to the product of force and displacement.

Using formula of work done

$W = F\cdot d$

$W = Fd\cos\theta$

Where, F = force

D = distance

θ = Angle between force and displacement

Put the value into the formula

$W=120\times3\cos0^{\circ}$

$W=360\ J$

Hence, The work done is 360 J.

8 0

3 years ago

Mark all of the antimatter particlesa) Proton b) Electron c)Anti-top d) Gluon e) Tau Neutrino

Bond [772]

Answer:

c) Anti-top

Explanation:

Protons are particles which belong to the hadron group of particles. The antimatter equivalent particle is antiproton

Electrons are particles that belong to the fermion group of particles. The antimatter equivalent is antielectron

Antitop quarks are particles that belong to the group family of particles. They are the antimatter equivalent of top quarks.

Gluons are particles that belong to the group of particles called bosons.

Tau neutrinos are particles that belong to the fermion group of particles. The antimatter equivalent is Tau antineutrino.

3 0

3 years ago

Answer<br> It's too short. Write at least 20 characters to get a great answer.

Roman55 [17]

Answer: I think the first option is the best option because water isn't a heat conductor and all the other options are not very accurate.

Explanation:

8 0

3 years ago

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altidude

Serjik [45]

Answer:

a) 252 ft/s

b) 1076 ft

Explanation:

The equation for motion for uniform acceleration (we can use it because the rocket is affected only by gravity) is as follows:

Y(t) = Y0 + V0 * t + 1/2 * a * t^2

Where

Y(t): altitude at a given time

Y0: initial altitude

V0: initial speed

a: acceleration, in this case -32.2 ft/s^2 (negative because gravity points down)

We set a 1 dimensional coordinate system with Y pointing up and the origin of coordinates at ground level.

We consider t=0 as the moment where powered flight ended (motor ran oou of fuel), at this moment the altitude was

Y(0) = 89.6 ft

Therefore:

Y0 = 89.6 ft

We also know that the rocket fell to ground 16 seconds later, therefore

Y(16) = 0 ft

So we can write

Y(t=16) = Y0 + V0 * t + 1/2 * a * t^2

V0 * t = Y(t=16) - Y0 - 1/2 * a * t^2

V0 =( Y(t=16) - Y0 - 1/2 * a * t^2 )/t

V0 =( 0 - 89.6 - 1/2 * (-32.2) * 16^2 )/16 = 252 ft/s

In the highest point of flight the rocket will have a speed = 0

The first derivative of the equation of motion is the equation of speed:

V(t) = V0 + a * t

If we equate this to zero we eill find the time at which the rocket achieved it's highest altitude.

0 = V0 + a * t

a * t = 0 - V0

t = -V0/a

t = -252/(-32.2) = 7.83 s

Now, we can take this time value andd plug it back into the position equation

Y(7.83) = 89.6 + 252 * 7.83 + 1/2 * (-32.2) * 7.83^2 = 1076 ft

6 0

3 years ago