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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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Fertilization occurs in the fallopian tubes.
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What’s the question? Is it true or false?
7 0
3 years ago
Please help me guys never mind the calculations ​
vlada-n [284]

The shape is connected in parallel so;

5.1) Ans;

\frac{1}{R}  =  \frac{1}{R1} +  \frac{1}{R2}   \\  \frac{1}{R}  =  \frac{1}{2}  +  \frac{1}{3}  \\  \frac{1}{R}  =  \frac{3 + 2}{6}  =  \frac{5}{6}  \\ R =  \frac{6}{5}  = 1.2 \:  \: ohm

5.2) Ans;

\frac{1}{R}  =  \frac{1}{R1} +  \frac{1}{R2}   \\  \frac{1}{R}  =  \frac{1}{8}  +  \frac{1}{10}  \\  \frac{1}{R}  =  \frac{5 + 4}{40}  =  \frac{9}{40}  \\ R =  \frac{40}{9}  = 4.4 \:  \: ohm

I hope I helped you^_^

7 0
3 years ago
The motion of a free falling body is an example of __________ motion​
Sunny_sXe [5.5K]

Answer:

uniformly accelerated motion

Explanation:

The motion of the body where the acceleration is constant is known as uniformly accelerated motion. The value of the acceleration does not change with the function of time.

3 0
3 years ago
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AysviL [449]

Answer:

C and E...................

4 0
3 years ago
An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. Aft
Arte-miy333 [17]

Answer:

The electric potential at point B is 83.33 V

The magnitude of the electric field at point B is 166.67 N/C  

Explanation:

Given;

charge of an object, q = −6.00×10⁻⁹ C

Kinetic energy  = 5.00×10⁻⁷ J

Electric potential at point A = +30.0 V

To determine the electric potential at point B, we apply the following formula;

Fd = qEd

Also, Fd is work done in moving the charge to 0.5 m = K.E

Again, Ed is the electric potential at point B (VB)

Substitute for V and K.E in the above equation, we will have;

K.E = q(VB)

VB = K.E/q

V_B = \frac{5*10^{-7}}{6*10^{-9}} = 83.33 \ V_B

To determine the magnitude of the electric field at B, we use equation below;

V = Ed

E = \frac{V}{d}  =  \frac{83.33}{0.5} = 166.67 \ N/C

3 0
3 years ago
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