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2 years ago

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

Physics

1 answer:

ankoles [38]2 years 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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Two cars collide and come to a complete stop. where did all of their energy go?

MA_775_DIABLO [31]

Most of the energy will be absorbed by the materials that make up the cars, causing them to deform. The energy will also be converted into sound energy, causing a loud bang upon collision. Also, some energy will be converted to thermal energy, which will cause the cars to heat up slightly.

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3 years ago

Which best describes fusion nuclear reactors? They use rods containing uranium for fuel. They are currently used for electricity

Ludmilka [50]

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3 years ago

Read 2 more answers

You have two square metal plates with side lengths of (6.50 C) cm. You want to make a parallel-plate capacitor that will hold a

gtnhenbr [62]

Answer:

The necessary separation between the two parallel plates is 0.104 mm

Explanation:

Given;

length of each side of the square plate, L = 6.5 cm = 0.065 m

charge on each plate, Q = 12.5 nC

potential difference across the plates, V = 34.8 V

Potential difference across parallel plates is given as;

$V = \frac{Qd}{L^2 \epsilon_o} \\\\d = \frac{V L^2 \epsilon_o}{Q}$

Where;

d is the separation or distance between the two parallel plates;

$d = \frac{VL^2 \epsilon_o}{Q} \\\\d = \frac{34.8*(0.065)^2 *8.854*10^{-12}}{12.5*10^{-9}} \\\\d = 0.000104 \ m\\\\d = 0.104 \ mm$

Therefore, the necessary separation between the two parallel plates is 0.104 mm

6 0

4 years ago

A 35.4 kg girl is riding on the back of a 15.23 kg cart. the cart and the kid are both moving eastward at 4.25 m/s when she step

Grace [21]

Answer:

The final velocity of the cart is $v_c = 7.02 \ m/s$

Explanation:

From the question we are told that

The mass of the girl is $m_g = 35.4 \ kg$

The mass of the cart is $m_c = 15.23 \ kg$

The speed of the cart and kid(girl) is $v = 4.25 \ m/s$

The final velocity of the girl is $v_g = 3.06 \ m/s$

Let assume that velocity eastward is positive and velocity westward is negative (Note that if we assume vise versa it wouldn't affect the answer )

The total momentum of the system before she steps off the back of the cart

is mathematically evaluated as

$p__{T1}} = (m_g + m_c) * v$

substituting values

$p__{T1}} = (35.4 + 15.23) * 4.25$

$p__{T1}} =215.17 \ kg m /s$

The total momentum after she steps off the back of the cart is mathematically evaluated as

$p__{T2}} = (m_g * v_g ) +( m_c * v_c )$

Where $v_c$ is the final velocity of the cart

substituting values

$p__{T2}} = (35.4 * 3.06 ) +( 15.23 * v_c )$

$p__{T2}} = 108. 324 + 15.23 v_c$

Now according to the law of conservation of momentum

$p__{T1}} =p__{T2}}$

$215.17 \ kg m /s = 108. 324 + 15.23 v_c$

=> $v_c = 7.02 \ m/s$

Since the value is positive it implies that the cart moved eastward

7 0

3 years ago

The uniform 100-kg I-beam is supported initially by its end rollers on the horizontal surface at A and B. by means of the cable

Ann [662]

I attached a free body diagram for a better understanding of this problem.

We start making summation of Moments in A,

$\sum M_A = 0$

$P(6cos\theta)-981(4cos\theta)=0$

$P=654N$

Then we make a summation of Forces in Y,

$\sum F_y = 0$

$654+R-981 = 0$

$R=327N$

At the end we calculate the angle with the sin.

$sin\theta = \frac{3m}{4m+2m+2m} = \frac{3m}{8m}$

$\theta = 22.02\°$

8 0

3 years ago