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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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Every winter I fly home to Chicago. It takes 3 hours. What is my average speed?
Tanya [424]

It depends on where you live when you're not visiting Chicago. We need to know the distance of the trip.

Your average speed on the trip is . . .

(total distance in miles) / (3 hours)

miles per hour

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1 year ago
Two fun-loving otters are sliding toward each other on a muddy (and hence frictionless) horizontal surface. One of them, of mass
zvonat [6]

Answer:

(a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

Explanation:

Given that,

Mass of one otter = 8.50 kg

Speed = 6.00 m/s

Mass of other = 5.75 kg

Speed = 5.50 m/s

(a). We need to calculate the magnitude and direction of the velocity of these free-spirited otters right after they collide

Using conservation of momentum

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v

Put the value into the formula

8.50\times(-6.00)+5.75\times5.50=(8.50+5.75)\times v

v=\dfrac{-19.375}{14.25}

v=-1.35\ m/s

Negative sign shows the direction of motion of the object after collision is toward left.

(b). We need to calculate the initial kinetic energy

Using formula of kinetic energy

K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2

Put the value into the formula

K.E_{i}=\dfrac{1}{2}\times8.50\times(6.00)^2+\dfrac{1}{2}\times5.75\times(5.50)^2

K.E_{i}=239.96\ J

We need to calculate the final kinetic energy

Using formula of kinetic energy

K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2

Put the value into the formula

K.E_{f}=\dfrac{1}{2}\times(8.50+5.75)\times(-1.35)^2

K.E_{f}=12.98\ J

We need to calculate the mechanical energy dissipates during this play

Using formula of loss of mechanical energy

\Delta K.E=K.E_{f}-K.E_{i}

Put the value into the formula

\Delta K.E=12.98-239.96

\Delta K.E=-226.98\ J

Negative sign shows the loss of mechanical energy

Hence, (a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

8 0
1 year ago
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A particle of charge 2.3 ✕ 10−8 C experiences an upward force of magnitude 4.6 ✕ 10−6 N when it is placed in a particular point
Marysya12 [62]
<h2>Answer:</h2>

(a) +2 x 10² N/C (upwards)

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<h2>Explanation:</h2>

The force (F) acting on a particle of charge (Q) at a particular point is related to its electric field (E) by the following;

F = Q x E   ----------------------(i)

This means that the force acting on the charged particle is the product of its charge and the electric field at that point.

<em>(a) Given</em>;

Q = charge of the particle = 2.3 x 10⁻⁸ C

F = force acting on the particle = 4.6 x 10⁻⁶N

<em>Substitute these values into equation (i) as follows;</em>

=> 4.6 x 10⁻⁶  = 2.3 x 10⁻⁸ x E

=> E = 4.6 x 10⁻⁶ ÷ 2.3 x 10⁻⁸

=> E =  2 x 10² N/C

Since the value is positive, the electric field is directed upwards.

Therefore, the electric field at that point is +2 x 10² N/C

<em>(b) If a charge of q = -1.1 x 10⁻⁸ is placed there, the force (F) acting is calculated as follows;</em>

<em>Substitute Q = q into equation (i) as follows;</em>

F = q x E

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F = -1.1 x 10⁻⁸ x 2 x 10²

F = -2.2 x 10⁻⁶ N

F = -2.2μN

Since the value of the force, F, is negative, it means it is directed downwards.

Therefore the force on the charge is -2.2μN

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2 years ago
Three objects of the same mass begin their motion at the same height. One object falls straight down, one slides down a low-fric
erik [133]

Answer:

D. Same

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1 year ago
An uncharged spherical conducting shell surrounds a charge –q at the center of the shell. Then charge +3q is placed on the outsi
Veronika [31]

Answer:

a) The the charges on the inner and outer surfaces of the shell are respectively +q, -q

Explanation:

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This must be zero so that no charge will be moving since the conductor is in static equilibrium.

Also, since Electric field, E is zero, then flux through the surface will zero.

From Gauss' law, the total charge enclosed is zero.

Given –q as the  charge at the center of the shell, then the opposite charge on inner surfaces  will be +q, so that the total charge enclosed will be zero.

Since the charge is in static equilibrium, then opposite charge will be on the surface, that is –q.

Therefore, the the charges on the inner and outer surfaces of the shell are respectively +q, -q

8 0
1 year ago
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