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zubka84 [21]
3 years ago
8

Which phrases describe an irregular galaxy?

Engineering
1 answer:
pickupchik [31]3 years ago
4 0

An irregular galaxy is the one which does not have any shape. It means it does not fit into the categories described by the Hubble. Such types of galaxies contain large amount of dust and gases. These are a site of stellar formation and hence contain young stars and are very bright. these galaxies are usually smaller than other types of galaxies. for example: Large Magellanic Clouds and Small Magellanic Cloud. Therefore, the phrases that describe an irregular galaxy are:

1. contains many young stars

2. contains a lot of gas and dust

3. is smaller than other types of galaxies

You might be interested in
Explain the two advantages and the two disadvantages of fission as an energy source.
yawa3891 [41]

Answer with Explanation:

1) The advantages of fission energy are:

a) Higher concentration of energy : Concentration of energy or the energy density is defined as the amount of energy that is produced by burning a unit mass of the fuel. The nuclear energy obtained by fission has the highest energy density among all the other natural sources of energy such as coal,gas,e.t.c.

b) Cheap source of energy : The cost at which the energy is produced by a nuclear reactor after it is operational is the lowest among all the other sources of energy such as coal, solar,e.t.c

2) The disadvantages of fission energy are:

a) Highly dangerous residue: The fuel that is left unspent is highly radioactive and thus is very dangerous. Usually the residual material is taken deep into the earth for it's disposal.

b) It has high initial costs of design and development: The cost to design a nuclear reactor and to built one after it is designed is the most among all other types of energy sources and requires highly skilled personnel for operation.

6 0
3 years ago
Javier’s class visited a power plant near his city, and they learned how it produced electricity. What does this form of power d
adelina 88 [10]

The electricity is generated by the burning of fuel in the power plant that increases the kinetic energy of water vapor which rotates the turbine and shaft of the generator.

<h3>What is a power plant?</h3>

A power plant is a place where heavy machinery objects are installed to produce electricity.

The power is generated in the power plant with the help of the burning of fuel that increases the temperature of the boiler. There are so many tubes in which water is present. Water absorbs heat and increases its kinetic energy. And this kinetic energy help to rotate the turbine and the shaft of the generator is attached to the turbine as result electricity is generated.

More about the power plant link is given below.

brainly.com/question/7670779

#SPJ4

5 0
2 years ago
A transformer has 300,000 windings in its primary coil and uses 12,000V AC input. (4 points) How many windings would be needed t
viva [34]

Answer:

  2750

Explanation:

The number of windings and the voltage are proportional.

__

Let n represent the number of windings to produce 110 Vac. Then the proportion is ...

  n/110 = 300,000/12,000

  n = 110(300/12) = 2750 . . . . multiply by 110

2750 windings would be needed to produce 110 Vac at the output.

7 0
2 years ago
The heat required to raise the temperature of m (kg) of a liquid from T1 to T2 at constant pressure is Z T2CpT dT (1) In high sc
a_sh-v [17]

Answer:

(a)

<em>d</em>Q = m<em>d</em>q

<em>d</em>q = C_p<em>d</em>T

q = \int\limits^{T_2}_{T_1} {C_p} \, dT   = C_p (T₂ - T₁)

From the above equations, the underlying assumption is that  C_p remains constant with change in temperature.

(b)

Given;

V = 2L

T₁ = 300 K

Q₁ = 16.73 KJ    ,   Q₂ = 6.14 KJ

ΔT = 3.10 K       ,   ΔT₂ = 3.10 K  for calorimeter

Let C_{cal} be heat constant of calorimeter

Q₂ = C_{cal} ΔT

Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂

Q₁ - Q₂ = m C_p ΔT

number of moles of n-C₆H₁₄, n = m/M

ρ = 650 kg/m³  at 300 K

M = 86.178 g/mol

m = ρv = 650 (2x10⁻³) = 1.3 kg

n = m/M => 1.3 / 0.086178 = 15.085 moles

Q₁ - Q₂ = m C_p' ΔT

C_p = (16.73 - 6.14) / (15.085 x 3.10)

C_p = 0.22646 KJ mol⁻¹ k⁻¹

6 0
3 years ago
A circular bar will be subjected to an axial force (P) of 2000 lbf. The bar will be made of material that has a strength (S) of
schepotkina [342]

Answer:

n = 2.36

Explanation:

The stress experimented by the circular bar is:

\sigma = \left[\frac{2000\, lbf}{\frac{\pi}{4}\cdot (0.5\,in)^{2}}\right]\cdot \left(\frac{1\,kpsi}{1000\,psi} \right)

\sigma = 10.186\,kpsi

The safety factor is:

n = \frac{24\,kpsi}{10.186\,kpsi}

n = 2.36

5 0
3 years ago
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