All categories

2 years ago

If you observe an object in the universe that is roughly 8 megaparsecs in size, what is the object most likely to be?

Physics

1 answer:

Dmitrij [34]2 years ago

4 0

E. Galaxy Cluster

Explanation:

A galaxy cluster, or cluster of galaxies, is a structure that consists of anywhere from hundreds to thousands of galaxies that are bound together by mutual gravity.

A megaparsec is a million parsecs and there are about 3.3 light years in a mega-parsec. Parsec is rather a natural distance unit for astronomers. The standard abbreviation of a mega-parsec is Mpc.

A parsec is approximately 3.09 x 1016 meters, a megaparsec is about 3.09 x 1022 meters.

Hence, 8 megaparsecs is gigantic size and that can be only of a galaxy cluster consisting of hundreds and thousands of galaxies bounded together.

Keywords: galaxies, parsec, megaparsec, galaxy cluster

Learn more about galaxy clusters and astronomical units from:

https://brainly.in/question/4624292

brainly.com/question/14214806

brainly.com/question/13315988

#learnwithBrainly

You might be interested in

10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of

Rina8888 [55]

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

$W=Fd cos \theta$

where :

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

$\theta=22^{\circ}$ is the angle between the direction of the force and the displacement

Substituting, we find the work done

$W=(87)(44)(cos 22^{\circ})=3549 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0

2 years ago

The position of a particle moving along the x axis may be determined from the expression x(t) = btu + ctv, where x will be in me

KIM [24]

As per given equation we have

$x = bt^u + ct^v$

now as per the dimensional analysis we can say that dimension of right side of equation must be equal to left side of the equation

now as per left side of equation its dimension is same as length or meter

now we can say it should be meter on right side also

$bt^u = M^0L^1T^0$

$b*T^8 = M^0L^1T^0$

$b = M^0L^1T^{-8}$

similarly for other term we have

$ct^v = M^0L^1T^0$

$c*T^7 = M^0L^1T^0$

$c = M^0L^1T^{-7}$

<em>so above are the dimensions of b and c</em>

8 0

1 year ago

A baseball of mass m = 0.49 kg is dropped from a height h1 = 2.25 m. It bounces from the concrete below and returns to a final h

Brilliant_brown [7]

Answer:

Explanation:

Impulse = change in momentum

mv - mu , v and u are final and initial velocity during impact at surface

For downward motion of baseball

v² = u² + 2gh₁

= 2 x 9.8 x 2.25

v = 6.64 m / s

It becomes initial velocity during impact .

For body going upwards

v² = u² - 2gh₂

u² = 2 x 9.8 x 1.38

u = 5.2 m / s

This becomes final velocity after impact

change in momentum

m ( final velocity - initial velocity )

.49 ( 5.2 - 6.64 )

= .7056 N.s.

Impulse by floor in upward direction

= .7056 N.s

6 0

2 years ago

Sort the following forces as relevant or not relevant to this situation. The symbols are defined as follows: normal force = n⃗ ,

Irina18 [472]

Answer:

it would be least to graetest

Explanation:

10-84

7 0

2 years ago

A rigid tank is divided into two equal parts by a partition. Initially one side of the tank contains 5 kg of water at 200 kPa an

DochEvi [55]

Answer:

There is no heat transfer in the system as the temperature before and after remains same i.e, 25 C.

Explanation:

Heat energy is mathematically expressed as,

Q=mCΔT

By looking at the above equation we can say that heat gain or loss by any system is equal to the mass times specific heat multiplied by the change in temperature.

Now if we consider the water system in view of this equation then ΔT would be 0 therefore it is evident that there is no transfer of heat.

If we consider the empty half of the tank so it is mentioned that there is nothing no air and giving us m=0, hence Q=0.

Mathematically we can express this as

Q=mCΔT

Q=(5)(4200)(0)

Q= 0 KJ <em>(which indicates that there was no heat gain or loss by the system)</em>

8 0

2 years ago