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:
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Answer:
Explanation:
(a) It is given that Joseph jogs on a straight road of 300m in a time interval of 2 minutes and 30 seconds, which is equal to 150seconds. Therefore, when Joseph jogs from point A to point B, he covers a distance of 300m in time of 150seconds. Hence, his average speed is 300m/150s=2ms^−1. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m.
Hence, his average velocity is 300m/150s=2ms^−1
(b) Then it is given that he turns back and points B and jogs on the same road but in the opposite direction for a time interval for 1 minute and covers a distance of 100m.If we consider the whole motion of Joseph, i.e. from point A to point C, then he covers a total distance of 300m+100m=400m. And he covers this total distance in a time interval of 2.5min+1min=3.5min=210s.
Therefore, his average speed for this journey is 400m210s=1.9ms−1.
For the same journey is displacement is equal to the distance between the points A and C,i.e. 300m−100m=200m.
Hence, his average velocity for this case is 200m/210s=0.95ms^−1
That would be answer B
hope this helped you
Answer:
Computer A is 1.41 times faster than the Computer B
Explanation:
Assume that number of instruction in the program is 1
Clock time of computer A is 
Clock time of computer B is 
Effective CPI of computer A is 
Effective CPI of computer B is
CPU time of A is
CPU time of B is
Hence Computer A is Faster by 
Computer A is 1.41 times faster than the Computer B
