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kykrilka [37]
2 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]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

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when you drop a pebble from height h, it reaches the ground with kinetic energy k if there is no air resistance. from what heigh
marysya [2.9K]

Answer:

From the initial height h

Explanation:

When a material or substance is drop from a height h, it possesses potential energy, immediately it is dropped from that height, the potential energy is gradually converted to kinetic energy, it gets to a point where the potential energy equals the kinetic energy, as the material touches the ground, all potential energy has been converted to kinetic energy already

6 0
1 year ago
If a sound with frequency fs is produced by a source traveling along a line with speed vs. If an observer is traveling with spee
Alexus [3.1K]

Answer:

457.81 Hz

Explanation:

From the question, it is stated that it is a question under Doppler effect.

As a result, we use this form

fo = (c + vo) / (c - vs) × fs

fo = observed frequency by observer =?

c = speed of sound = 332 m/s

vo = velocity of observer relative to source = 45 m/s

vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).

fs = frequency of sound wave by source = 459 Hz

By substituting the the values to the equation, we have

fo = (332 + 45) / (332 - (-46)) × 459

fo = (377/ 332 + 46) × 459

fo = (377/ 378) × 459

fo = 0.9974 × 459

fo = 457.81 Hz

7 0
2 years ago
Imagine you derive the following expression by analyzing the physics of a particular system: v2=v20+2ax. The problem requires so
lisabon 2012 [21]

As per kinematics equation we are given that

v^2 = v_o^2 + 2ax

now we are given that

a = 2.55 m/s^2

v_0 = 21.8 m/s

v = 0

now we need to find x

from above equation we have

0^2 = 21.8^2 + 2(2.55)x

0 = 475.24 + 5.1 x

x = 93.2 m

so it will cover a distance of 93.2 m

7 0
2 years ago
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The distance of the earth from the sun is 93 000 000 miles. if there are 3.15 × 107 s in one year, find the speed of the earth i
faltersainse [42]

The angular velocity of the orbit about the sun is:

w = 1 rev / year = 1 rev / 3.15 × 10^7 s

 

Now in 1 rev there is 360° or 2π rad, therefore:

w = 2π rad / 3.15 × 10^7 s

 

To convert in linear velocity, multiply the rad /s by the radius:

v = (2π rad / 3.15 × 10^7 s) * 93,000,000 miles

<span>v = 18.55 miles / s = 29.85 km / s</span>

5 0
2 years ago
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Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d
Step2247 [10]
Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

9 0
2 years ago
Read 2 more answers
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