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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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If a 375 watt heater has a current of 5.0 A, what is the resistance of the heating element?

liberstina [14]

P=IV
V=IR

P=I(IR)
P=I²R
375=5²R
R=375/25
R=15

7 0

3 years ago

Based on Kepler's work, which best describes the orbit If a planet around the Sun?

krek1111 [17]

Answer:

an ellipse with the Sun at one focus or D

Explanation:

Answer for edgenuity

7 0

3 years ago

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)

ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration= $2.40m/s^{2}$ , c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = $2.40m/s^{2}$

time = 12.0 s

initial position = 0

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

$x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}$

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

$x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}$

after cancelling the pertinent units, I get that my answer will be given in meters. So I get:

$x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}$

which solves to:

$x=172.8m$

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

$a=\frac{V_{f}-V_{0} }{t}$