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serg [7]
3 years ago
5

How are galaxies organized and distributed within the universe

Physics
2 answers:
den301095 [7]3 years ago
3 0
The galaxies aren't distributed randomly throughout the universe but they are grouped in gravitationally bound clusters.
Mars2501 [29]3 years ago
3 0

Answer:

<u><em>The answer is</em></u>:<em><u> </u></em><u>Many are associated in pairs, trios, groups or clusters, and their distribution in filamentary structures.</u>

<u></u>

Explanation:

<em>Galaxies rarely appear isolated and their distribution in the Universe is not uniform</em>. Many are associated in pairs, trios, groups of some tens or clusters of up to a few thousand. <em>These groups are held together by gravitation. </em>

The large-scale distribution of galaxies seems to indicate that they are concentrated along large filamentary structures.

<u><em>The answer is</em></u>:<em><u> </u></em><u>Many are associated in pairs, trios, groups or clusters, and their distribution in filamentary structures.</u>

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Use your data (attachment) from Part 3 and Newton’s laws to explain why the force meter measures a force if the cart is moving a
vazorg [7]
The cart experiences a frictional force which is directly proportional to its weight. This means that there must be a force applied on the car to balance the forces on the car to produce a net force of 0.
This is in accordance to Newton's first law which states that an object at rest will remain at rest and an object in motion will remain in motion unless an external force acts on it. The force must be a resultant force.
Therefore, the force needed increases with the total weight of the cart as well as with the added mass in a linear manner.
4 0
3 years ago
Jack and Jill have made up since the previous HW assignment, and are now playing on a 10 meter seesaw. Jill is sitting on one en
Airida [17]

Answer: 3 m.

Explanation:

Neglecting the mass of the seesaw, in order the seesaw to be balanced, the sum of the torques created by  gravity acting on both children  must be 0.

As we are asked to locate Jack at some distance from the fulcrum, we can take torques regarding the fulcrum, which is located at just in the middle of the length of the seesaw.

If we choose the counterclockwise direction as positive, we can write the torque equation as follows (assuming that Jill sits at the left end of the seesaw):

mJill* 5m -mJack* d = 0

60 kg*5 m -100 kg* d =0

Solving for d:

d = 3 m.

6 0
3 years ago
Being able to see a square in the middle of the image despite not having lines to form a square represents the Gestalt principle
vredina [299]

Answer: The correct answer for the blank is -

C. closure.

Gestalt principle of closure describes how we perceive complete figures even when the information that form the figure is missing.

This is due the fact that our brain responds to the familiar patterns inspite of getting incomplete information.

For instance, in the given question, we are able to perceive an image of square in the center despite not having actual lines that form a square.

Thus, it represents principle of closure.

7 0
3 years ago
Read 2 more answers
A block with mass m 2.00 kg is placed against a spring on a frictionless incline with angle 30 degrees (Fig. B-43). (The block i
guajiro [1.7K]

Answer:

Explanation:

a )

The stored elastic energy of compressed spring

= 1 / 2 k X²

= .5 x 19.6 x (.20)²

= .392 J

b ) The stored potential energy will be converted into gravitational potential energy of the block earth system when the block will ascend along the incline . So change in the gravitational potential energy will be same as stored elastic potential energy of the spring that is .392 J .

c ) Let h be the distance along the incline which the block ascends.

vertical height attained ( H ) =h sin30

= .5 h

elastic potential energy = gravitational energy

.392 = mg H

.392 = 2 x 9.8 x .5 h

h = .04 m

4 cm .

=

7 0
3 years ago
Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c
Travka [436]

Answer:

22.1 V

Explanation:

We are given that

A_1=0.7 cm^2=0.7\times 10^{-4} m^2

A_2=2.5 cm^2=2.5\times 10^{-4} m^2

A_3=2.2 cm^2=2.2\times 10^{-4} m^2

A_4=3 cm^2=3\times 10^{-4} m^2

Using 1cm^2=10^{-4} m^2

We know that

R=\frac{\rho l}{A}

In series

R=R_1+R_2+R_3+R_4

R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}

R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}

Substitute the values

R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})

R=\rho l(2.62\times 10^4)

V=145 V

I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}

Voltage across the 2.5 square cm wire=IR=I\times \frac{\rho l}{A_2}

Voltage across the 2.5 square cm wire=\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V

Voltage across the 2.5 square cm wire=22.1 V

6 0
3 years ago
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