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QveST [7]
3 years ago
7

Describe the main distinguishing features of spiral, elliptical, and irregular galaxies.

Physics
1 answer:
hram777 [196]3 years ago
5 0

Answer:

Spiral galaxies consist of a flat, rotating disk of stars, gas and dust, and a central concentration of stars known as the bulge. These are surrounded by a much fainter halo of stars, many of which reside in globular clusters.

Elliptical galaxies have smooth, featureless light-profiles and range in shape from nearly spherical to highly flattened, and in size from hundreds of millions to over one trillion stars. In the outer regions, many stars are grouped into globular clusters. Most elliptical galaxies are composed of older, low-mass stars, with a sparse interstellar medium and minimal star formation activity They are often chaotic in appearance, with neither a nuclear bulge nor any trace of spiral arm structure. Collectively they are thought to make up about a quarter of all galaxies.

irregular galaxies were once spiral or elliptical galaxies but were deformed by gravitational action. they are shapeless.

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What was your train of thought as you navigated the picture of the candle?
defon

Answer:

Where is the picture

Explanation:

WHERE IS THE PICTURE

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3 years ago
Which of the following types of particles can be categorized into different kind of elements?
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The answer would be atoms

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3 years ago
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Which statement best describes the relationship between atoms and nuclear forces? (2 points)
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Answer:

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Explanation:

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2 years ago
A playground merry-go-round has a mass of 50 kg and a diameter of 4.0 m. There are 4 children who want to ride on it. They have
mixer [17]

Answer:

B) I1 = 1680 kg.m^2          I2 = 1120 kg.m^2

C) V = 0.84m/s      T = 29.92s

D) ω2 = 0.315 rad/s

Explanation:

The moment of inertia when they are standing on the edge:

I1 = 1/2*M*R^2 + (m1+m2+m3+m4)*R^2   where M is the mass of the merry-go-round.

I1 = 1680 kg.m^2

The moment of inertia when they are standing half way to the center:

I2 = 1/2*M*R^2 + (m1+m2+m3+m4)*(R/2)^2

I2 = 1120 kg.m^2

The tangencial velocity is given by:

V = ω1*R = 0.84m/s

Period of rotation:

T = 2π / ω1 = 29.92s

Assuming that there is no friction and their parents are not pushing anymore, we can use conservation of the angular momentum to calculate the new angular velocity:

I1*ω1 = I2*ω2    Solving for ω2:

ω2 = I1*ω1 / I2 = 0.315 rad/s

5 0
3 years ago
6300 divided by 32.8
Ilya [14]
<h2>6300÷32.8</h2><h3>=192.07317073</h3>

<h2>6300÷33</h2><h3>=190.90909091</h3>

please mark this answer as brainlist

7 0
3 years ago
Read 2 more answers
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