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2 years ago

Why were quasars and active galaxies not initially recognized as being “special” in some way?

Physics

1 answer:

djyliett [7]2 years ago

8 0

Answer:

In the past, astronomers look into the sky and the universe as a whole with an idea that it is a place where stars are born, transition through their life stages and ultimately die and this is because they couldnt differentiate between stars, quasars and active galaxies because with advanced equipment, they all look similar but as they technological ages arrived, they were able see that they are not the same.

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What are two different units that represent work?

Zina [86]

Work can be represented by Wd (Work done) or Ew (Energy, work). Since it is a form of energy, it is measured in joules

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2 years ago

Which component of the galaxy is shown in this image? Dust

Westkost [7]

Answer:

stars

those are stars in the galaxy

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2 years ago

10 kg of R-134a fill a 1.115-m^3 rigid container at an initial temperature of -30∘C. The container is then heated until the pres

Sidana [21]

Answer : The final temperature an the initial pressure of gas is, 273.6 K and 177.6 kPa

Explanation :

First we have to calculate the initial pressure of gas.

As we know that, R-134a is 1,1,1,2-Tetrafluoroethane.

Using ideal gas equation:

$PV=nRT\\\\PV=\frac{w}{M}RT$

where,

P = pressure of gas = ?

V = volume of gas = $1.115m^3$

T = temperature of gas = $-30^oC=273+(-30)=243K$

R = gas constant = $8.314m^3Pa/mole.K$

w = mass of gas = 10 kg = 10000 g

M = molar mass of R-134a gas = 102.03 g/mole

Now put all the given values in the ideal gas equation, we get:

$P\times 1.115m^3=\frac{10000g}{102.03g/mole}\times (8.314m^3Pa/mole.K)\times (243K)$

$P=177587.9687Pa=177.6kPa$

Thus, the initial pressure of gas is, 177.6 kPa

Now we have to calculate the final temperature of gas.

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$

or,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 177.6 kPa

$P_2$ = final pressure of gas = 200 kPa

$T_1$ = initial temperature of gas = $-30^oC=273+(-30)=243K$

$T_2$ = final temperature of gas = ?

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

$\frac{177.6kPa}{243K}=\frac{200kPa}{T_2}$

$T_2=273.6K$

Thus, the final temperature an the initial pressure of gas is, 273.6 K and 177.6 kPa

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2 years ago

What is a greater force 10 Newtons or 5lbs?

Alexxandr [17]

5lbs is greater.
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According to newton's third law of motion, when a hammer strikes and exerts force to push it into a piece of wood, the nail

Darya [45]

According to newton's third law of motion, when a hammer strikes and exerts force to push it into a piece of wood, the nail <span>C. exerts an equal or opposite force on the hammer. The third law of motion states that every action has an equal BUT opposite reaction. This means that the nail exerts the same force the hammer exerts on it.</span>

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3 years ago

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