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kykrilka [37]
3 years ago
9

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]3 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

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The difference in electric potential between a thunder cloud and the ground is 1.53 108 V. Electrons move from the ground which
Kryger [21]

Answer:

=−2.451 330 152 1*10^27J

Explanation:

The electric potential=the Voltage * Charge: 

E = VQ

V = 1.53x10^8 V (positive, because the cloud has a higher potential)

Q = -1.60217657 x10^19 C (the charge of an electron)

E = (1.53x10^8 V )* (-1.60217657 x10^19 C)

E=−2.451 330 152 1*10^27J

The negative sign indicates that the potential energy is decreased by the movement of the electron.

8 0
4 years ago
A 06-C charge and a 07-C charge are apart at 3 m apart. What force attracts them?​
Vlad [161]

Answer:

Force of 37.8 × 10^(6) N attracts the two charges

Explanation:

The force between two charges is given by

F = k*q1*q2/r²

Where q1 and q2 are 0.06 C and 0.07 C.

r is the distance between q1 and q2 which is equal to 3 m

k is a constant = 9 × 10^(9) N.m²/C²

F = (9 × 10^(9) × 0.06 × 0.07)/3²

F = 37.8 × 10^(6) N

5 0
3 years ago
MathPhys Pls see this Thank you in Advance MathPhys Is the best
umka2103 [35]

Answer:

70 N

21°

1.1 m/s²

Explanation:

Draw a free body diagram of the block.  There are three forces:

Weight pulling straight down

Normal force pushing perpendicular to the incline

Friction force pushing parallel to the incline

Part 1

Sum the forces in the perpendicular direction:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

The block is at rest, so F = N μs:

F = N μs

F = mg μs cos θ

F = (20 kg) (9.8 m/s²) (0.38) (cos 19°)

F = 70 N

Part 2

Sum the forces in the parallel direction (down the incline is positive):

∑F = ma

mg sin θ − F = 0

mg sin θ = N μs

mg sin θ = mg μs cos θ

tan θ = μs

θ = atan μs

θ = atan 0.38

θ = 21°

Part 3

Sum the forces in the parallel direction (this time, acceleration is not 0).

∑F = ma

mg sin θ − F = ma

mg sin θ − N μk = ma

mg sin θ − mg μk cos θ = ma

a = g (sin θ − μk cos θ)

a = (9.8 m/s²) (sin 24° − 0.32 cos 24°)

a = 1.1 m/s²

4 0
3 years ago
Read 2 more answers
What does the atomic number tell us about an atom of a certain element?
Setler79 [48]
It tells us the number of protons that are present in the nucleus, the positively charged region of that atom.
4 0
3 years ago
An ice skater is spinning at 6.00 rev/s with his moment of inertia being 0.400 kg/m2. Calculate his new moment of inertia if he
labwork [276]

Answer:

New moment of inertia will be I=1.92kgm^2

Explanation:

It is given initially angular velocity \omega =6rev/sec=6\times 2\pi =37.68rad/sec

Moment of inertia I=0.4kgm^2

Angular momentum is equal to L=I\omega =37.68\times 0.4=15.072kgm^2/sec

Now angular velocity is decreases to \omega =1.25rev/sec=1.25\times 2\times 3.14=7.85rad/sec

As we know that angular momentum is conserved

So 15.072=I\times 7.85

I=1.92kgm^2

So new moment of inertia will be I=1.92kgm^2

4 0
3 years ago
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