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4 years ago

What are Galaxies classified by?

Physics

1 answer:

enyata [817]4 years ago

4 0

<h2>Answer: elliptical, spiral and irregular </h2>

On 1930 the astronomer Edwin Hubble classified the galaxies into elliptical, spiral and irregular, being the first two classes the most frequent.

However, it should be noted that this classification is based only on the visual appearance of the galaxy, and does not take into account other aspects, such as the rate of star formation or the activity of the galactic nucleus.

The classification is as follows:

1. Elliptical galaxies: Their main characteristic is that the concentration of stars decreases from the nucleus, which is small and very bright, towards its edges. In addition, they contain a large population of old stars, usually little gas and dust, and some newly formed stars.

They are symbolized by the letter E and subdivided into eight classes, from E0 with zero eccentricity (spherical) to E7 (called husiform).

2. Spiral galaxies: They have the shape of flattened disks containing some old stars and also a large population of young stars, enough gas and dust, and molecular clouds that are the birthplace of the stars.

They are symbolized with the letter S and depending on the minor or major development that each arm possesses, it is assigned a letter: a, b or c (for example: Sa, Sb, Sc, SBa, SBb, SBc).

These galaxies, are also divided into two types:

-Lenticular galaxies

-Barred spiral galaxies

3. Irregular Galaxies: They are symbolized by the letter I (or IR), although they are usually dwarf or rare and do not have well-defined structure and symmetry.

They are classified in:

-Irregular type 1 (Magellanic), which contain large numbers of young stars and interstellar matter.

-Regular type 2, less frequent and whose content is difficult to identify.

This type of irregular galaxies are generally located close to larger galaxies, and usually contain large amounts of young stars, gas and cosmic dust.

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Answer:

$v_{1}=1.886 \frac{m}{s}$

β= 57.99 south of east

Explanation:

$v_{1}=2.2 \frac{m}{s} \\v_{2}=1.2 \frac{m}{s} \\m_{1}=m_{2}=m\\v_{fx}=1.6 \frac{m}{s} \\v_{fy}=$ ?

Velocity in axis x the two balls come one from east and west

$m_{1}*v_{1x}+m_{2}*v_{2x}=m_{1}*v_{fx1}+m_{2}*v_{fx2}\\m*(v_{1x}+v_{2x})=m*(v_{fx1}+v_{fx2})\\v_{fx2}=0\\v_{1x}+v_{2}=v_{f1}+0\\v_{fx1}=2.2 \frac{m}{s}+(1.2\frac{m}{s})\\ v_{fx1}=1 \frac{m}{s} \\$

Velocity in axis y initial is zero so:

$v_{y1}+v_{y2}=v_{y1f}+v_{y2f}\\v_{y1}=0\\v_{y2}=0\\v_{y1f}+v_{y2f}=0\\v_{y1f}=-v_{y2f}\\v_{y2f}=1.6\frac{m}{s}$

$v=\sqrt{v_{1fx}^{2}+v_{1fy}^{2}}\\ v=\sqrt{1^{2}+1.6^{2}}\\v=1.886 \frac{m}{s}$

Angle is find using:

tan(β)= $\frac{v_{fy}}{v_{fx}}$

$\beta =tan^{-1}*\frac{1.6}{1}=57.99$

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4 years ago

A train travels due south at 20 m/s. It reverses its direction and travels due north at 20 m/s. What is the change in velocity o

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Answer:

40 m/s due north

Explanation:

Consider that the south direction a negative Y axis and north direction as + Y axis

v1 = 20 m/s South = 20 (-j) m/s

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Change in velocity = v2 - v1 = 20 j - 20 (-j) = 40 j m/s

So, change in velocity is 40 m/s due north.

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3 years ago

4. Some early mornings , dew drops can be found on grass or a car parked outside,

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In some early mornings , dew drops can be found on grass or a car parked outside, but not on other materials such as the sidewalk because the night -time temperature on grass and the car went below the dew point, but the temperature of the concrete did not drop enough to reach the dew point level

Dew can be formed on any object when the temperature of the object drop. When this happen, the object will be cool which will eventually cool the surrounding air around the object.

Dew drops is as a result of condensation in the air. When the cool air causes the air vapor to convert to liquid. The dew will form when the temperature of the object balances with the dew point in the surrounding environment.

In some early mornings , dew drops can be found on grass or a car parked outside, but not on other materials such as the sidewalk because the night -time temperature on grass and the car went below the dew point, but the temperature of the concrete did not drop enough to reach the dew point level

Therefore the correct option is therefore A

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3 years ago

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