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3 years ago

Which of the following is a likely life cycle of a star?

Physics

1 answer:

melomori [17]3 years ago

7 0

Answer:

Explanation:

But this just happen for big stars, like more than 20x the Sun mass.

Shortly: A nebula is a cloud of gas and dust, the material starts to be acummuleted and became a protostar (is like a big planet, almost a star). With enought mass this is a star, burn hydrogen and transform it in Helium.

This occurs in Main Sequence, is about almost all the life time of a star. Then starts the lack of hydrogen. Gravity compress everything, pressure goes up and heat all. Too much energy, Helium get burned and the star grews fast, became a Red Giant. Time pass and the fuel is over, no more making fusion, gravity compress the star, too much strenght, colapses, neutron star.

If it have pretty mass, ok. If have more than like 2x Sun mass, became a blackhole.

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3 years ago

A baseball diamond is a square (don’t be fooled that it is usually shown rotated by 45°). Each side of the square is 90 feet lon

gayaneshka [121]

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2 years ago

What does the principal quantum number determine?

morpeh [17]

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Hope this helps!

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2 years ago

If the volume of a container of gas is reduced, what will happen to the pressure inside the container?.

AfilCa [17]

Answer:

The pressure of the gas will increase

Explanation:

When gas is put into a container, for example, a balloon, the gas expands to fill the space it can occupy. Since gas is not a solid or a liquid, its particles are all over the place - they are constantly moving and vibrating. As such, when too much gas is blown into a balloon, it will pop. So, when the volume of the container decreases, the pressure of the gas will increase the smaller it gets. Vice versa, the greater the space, the less pressure that will be present in the container.

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2 years ago

In an oscillating LC circuit, the total stored energy is U and the maximum current in the inductor is I. When the current in the

Deffense [45]

Answer:

The definition of that same given problem is outlined in the following section on the clarification.

Explanation:

The Q seems to be endless (hardly any R on the circuit). So energy equations to describe and forth through the inducer as well as the condenser.

Presently take a gander at the energy stored in your condensers while charging is Q.

⇒ $U =\frac{Qmax^2}{C}$

So conclude C doesn't change substantially as well as,

When,

⇒ $Q=\frac{Qmax}{2}$

⇒ $Q^2=\frac{Qmax^2}{4}$

And therefore only half of the population power generation remains in the condenser that tends to leave this same inductor energy at 3/4 U.

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3 years ago