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3 years ago

How are galaxies organized and distributed within the universe

2 answers:

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:

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

Explanation:

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

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

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

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A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,

shusha [124]

Answer:

$v=d\sqrt{\frac{k}{m}}$

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

$U_{0}+K_{0}=U_{f}+K_{f}$

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

$U_{0}=K_{f}$

The initial potential energy of the spring is given by the equation:

$U_{0}=\frac{1}{2}kd^{2}$

the Kinetic energy of the block is then given by the equation:

$K_{f}=\frac{1}{2}mv_{f}^{2}$

so we can now set them both equal to each other, so we get:

$=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}$

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

$kd^{2}=mv_{f}^{2}$

so now we can solve this for the final velocity, so we get:

$v=d\sqrt{\frac{k}{m}}$

6 0

3 years ago

The driver of a 1000 kg car traveling at a speed of 16.7 m/s applies the brakes. If the brakes provide a force of - 8000 N to st

RSB [31]

Answer:

Given:

m=1000kg

u= 16.7m/s

v=0m/s

F=8000N

Required:

s=?

Solution:

F=m × a

8000N=1000kg × a

a=8m/s^2

Since it decelerate a= -8m/s^2

v^2 = u^2 + 2as

s=v^2 - u^2 / 2a

s= 0 - (16.7m/s)^2 / 2 × -8m/s^2

s= -278.89/-16

s= 17.43m

The car travels approximately 17.43m before it stops

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6 0

2 years ago

Several types of radiation may be emitted during radioactive decay. The equation represents

Fittoniya [83]

I think the answer is B

5 0

4 years ago

Read 2 more answers

A force of 20. Newtons to the left exerted on a cart for 10. Seconds. For what period of time must a 50.-newton force to the rig

FinnZ [79.3K]

Impulse = (force) x (time)

The first impulse was (20 N) x (10 sec) = 200 meters/sec

The second one is (50 N) x (time) and we want it equal to the first one, so

(50 N) x (time) = 200 meters/sec

Divide each side by 50N : Time = 200/50 = 4 seconds

By the way, the quantity we're playing with here is the cart's momentum.

6 0

4 years ago

The standard measure used to compare sound intensities is the ______ .

uranmaximum [27]

The answer is Decibels.

4 0

3 years ago

Read 2 more answers