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1 year ago

5

what are the main morphological types of galaxies, and what is the difference in their rotation patterns?

1 answer:

ankoles [38]1 year ago

3 0

The main morphological types of galaxies are elliptical, spiral, and irregular.

Based on their morphology , galaxies have been classified into 3 types namely elliptical, spiral, and irregular.

These galaxies have various sizes and shapes ranging from dwarf galaxies to giant galaxies.

Elliptical Galaxy:

The shape of it is generally circular
These are the largest among all the types of galaxies because according to astronomers, it is formed by the merger of other small galaxies.
Their rotational pattern is symmetric.

Spiral Galaxy:

A spiral galaxy consists of a bright nucleus surrounded by a thin outer disk forming a spiral shape.
This type of galaxy is the most common in our universe.
It is divided into three classes: Spiral a, Spiral b, and Spiral c.
Their rotational pattern has circular symmetry.

Irregular Galaxy:

These types of galaxies have no central nucleus and irregular arms which are bluish.
They don’t have any rotational symmetry.

To know more about "galaxies", refer to the following link:

brainly.com/question/24836631?referrer=searchResults

#SPJ4

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30. The length of mercury thread when it is at 0°C, 100°C and at an unknown temperature 0 is 25mm, 225mm and 175mm respectively.

yulyashka [42]

Answer:

75°C

Explanation:

175−25/225−25=x−0/100−0

150/200=x/100

x=150×100/200

= 75°C

3 0

2 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0

3 years ago

Consider a hot-air balloon. The deflated balloon, gondola, and two passengers have a combined mass of 315 kg. When inflated, the

vladimir2022 [97]

This will help you.

3 0

2 years ago

The focal length of a lens is inversely proportional to the quantity (n-1), where n is the index of refraction of the lens of th

Ainat [17]

Answer:

46.22 cm

Explanation:

The focal refraction, fr is given by

$fr = \frac {c}{(1.572 -1)} = \frac {c}{0 .572}$

The focal red light is given by

$fv = \frac {c}{(1.605 - 1)} = \frac {c}{0.605}$

$\frac {fv}{fr} = \frac {0.572}{0 .605} = 0.945455$

$\frac {1}{fr} = \frac{1}{image} + \frac {1}{object}$ and making fr the subject we obtain

$fr = \frac {image * object}{(image + object)} = \frac {24.00 * 55} {(24.0 + 55)} = 16.70886 cm$

fv = 0.945455* 16.70886 cm = 15.79747 cm

$image = \frac {object * f} {(object - f)} = \frac {15.79747 * 24.0}{(24.0 - 15.79747)} = 46.22222 cm$

Therefore, violet image is approximately 46.22 cm

5 0

3 years ago

A neon light consists of a glass tube with metal wires at each end. When connected to a high-voltage source, the gas glows.

ivolga24 [154]

Can confirm C is correct

8 0

3 years ago

Read 2 more answers