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1 year ago

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What evidence supports the existence of a very massive black hole at the center of our galaxy?

1 answer:

faltersainse [42]1 year ago

5 0

<u>The motions of the gas and stars at the center indicate that it contains 4 million solar masses within a region no larger than our solar system</u> is the evidence supports the existence of a very massive black hole at the center of our galaxy.

<h3>What is black hole?</h3>

Black holes are points in space that are so dense they create deep gravity sinks. Beyond a certain region, not even light can escape the powerful tug of a black hole's gravity. And anything that ventures too close—be it star, planet, or spacecraft—will be stretched and compressed like putty in a theoretical process aptly known as spaghettification.

There are four types of black holes: stellar, intermediate, supermassive, and miniature. The most commonly known way a black hole forms is by stellar death. As stars reach the ends of their lives, most will inflate, lose mass, and then cool to form white dwarfs. But the largest of these fiery bodies, those at least 10 to 20 times as massive as our own sun, are destined to become either super-dense neutron stars or so-called stellar-mass black holes.

Learn more about black holes

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a toy car travels from A to B at constant speed 30km/hr and without stopping at B return A at constant speed v. if the average s

adelina 88 [10]

Answer:

Speed BA = 18 km/hr.

Explanation:

Given the following data;

Speed AB = 30km/hr

Speed BA = x km/hr

Average speed = 24 km/hr

To find the value of x;

Average speed = (Speed AB + Speed BA)/2

Substituting into the equation, we have

24 = (30 + x)/2

48 = 30 + x

x = 48 - 30

x = 18 km/hr

4 0

3 years ago

Type the correct answer in the box. Round your answer to the nearest whole number. Calculate the man’s mass. (Use PE = m × g × h

Blababa [14]

Answer:

56 kg

Explanation:

The change in potential energy of the man is given by:

$\Delta U = mg \Delta h$

where

m is the man's mass

g is the gravitational acceleration

$\Delta h$ is the change in height of the man

In this problem, we have:

$\Delta U=4620 J$ is the gain in potential energy

g = 9.8 m/s^2 is the gravitational acceleration

$\Delta h=8.4 m$ is the change in height

Re-arranging the equation and substituting the numbers, we find the mass:

$m=\frac{\Delta U}{g\Delta h}=\frac{4620 J}{(9.8 m/s^2)(8.4 m)}=56 kg$

6 0

3 years ago

Read 2 more answers

A 0.350kg bead slides on a curved fritionless wire,

LuckyWell [14K]

Answer:

h2 = 0.092m

Explanation:

From a balance of energy from point A to point B, we get speed before the collision:

$m1*g*h-\frac{m1*V_B^2}{2}=0$ Solving for Vb:

$V_B=\sqrt{2gh}=6.56658m/s$

Since the collision is elastic, we now that velocity of bead 1 after the collision is given by:

$V_{B'}=V_B*\frac{m1-m2}{m1+m2} = \sqrt{2gh}* \frac{m1-m2}{m1+m2}=-1.34316m/s$

Now, by doing another balance of energy from the instant after the collision, to the point where bead 1 stops, we get the distance it rises:

$m1*g*h2-\frac{m1*V_{B'}^2}{2}=0$ Solving for h2:

h2 = 0.092m

6 0

3 years ago

hichkok12 [17]

Gravitational potential energy i think

8 0

3 years ago

The mass of a truck with its cargo is 1800 kg. The truck is coasting to the right with a speed of 10 m/s. As it coasts its cargo

OverLord2011 [107]

Here we apply conservation of linear momentum. The momentum of the truck with cargo and without cargo remains constant. That is,

$mv=m'v'$ .

Here $m,v$ are initial mass and velocity. $m',v'$ are final mass and velocity. Here $m=1800,v=10$ and $m'=1800-300=1500$ .

The velocity of the truck be after its cargo is taken off is

$v'=\frac{mv}{m'} \\ v'=\frac{1800*10}{1500} \\ v'=12 m/s$

6 0

3 years ago

Read 2 more answers