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fiasKO [112]
1 year ago
13

Stellar nurseries, such as the orion nebula, contain hundreds or more fragmenting and contracting regions, as well as many proto

stars and stars. What condition would allow a protostar to become a stable star?.
Physics
1 answer:
Natasha_Volkova [10]1 year ago
3 0

The hydrogen fusion process will begin after the protostar reaches a temperature of 10 million degrees kelvin, and it will then turn into a stable star.

<h3>How does a protostar become a stable star?</h3>

The interstellar medium can sometimes be gathered into a large nebula, which is a cloud of gas and dust. A nebula can span a number of light years. These nebulae are where gas and dust can combine to produce stars. Until a star can combine hydrogen into helium, it cannot be considered a star. They are referred to as protostars before then. As gravity starts to gather the gases into a ball, a protostar is created. Accrution is the term for this procedure.

Gravitational energy starts to heat the gasses as gravity draws them into the ball's core, which causes the gasses to radiate radiation. Radiation initially just dissipates into space. However, much of the radiation is retained inside the protostar as it draws in stuff and becomes denser, which causes the protostar to heat up even more quickly.

The hydrogen fusion process will begin after the protostar reaches a temperature of 10 million degrees kelvin, and it will then turn into a star.

Learn more about a protostar here:

brainly.com/question/12534975

#SPJ4

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An 8.00- W resistor is dissipating 100 watts. What are the current through it and the difference of potential across it?
hjlf

Answer:

I= 3.5 amps

Explanation:

Step one:

given data

rating of resistor R= 8 ohms

power P= 100W

Required

The current I

Step two

Yet this power is also given by

P = I^2R

make I subject of the formula we have

I= \sqrt{\frac{P}{R} }

substitute

I= \sqrt{\frac{100}{8} }\\\\I=\sqrt{12.5}\\\\I= 3.5 amps

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First, we assume this as an ideal gas so we use the equation PV=nRT. Then, we use the conditions at STP that would be 1 atm and 273.15 K. We calculate as follows:

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1 atm (.245 L) =1.30(0.08206)(273.15) / MM
MM = 118.94 g/mol <--- ANSWER
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