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Nesterboy [21]
4 years ago
7

The following figures show four stages that occur during the formation of a one-solar-mass star. Rank these stages based on thei

r rotation rate, from fastest to slowest. (Assume that the angular momentum of the forming star is conserved throughout the formation process, though in fact it may shed some angular momentum by ejecting material into interstellar space.)
Physics
1 answer:
Stolb23 [73]4 years ago
6 0

Answer:

  • The four stages that occur during the formation of a one-solar-mass star are as follows:                                                                                

Explanation:

  1. The formation of the Red dwarf which is also one of the celestial entities inside the galaxies as, the formation of the more gradual body of the mixture of different gasses having a imaginable low temperature. As this whole process leads to the formation of the main sequence star.
  2. When the different masses of the mixtures of the gases combines to form a very celestial body then it leads to the composition of each component inside the celestial body to acquire a very flattened form of structure leading to the flattened disk with radiation properties. As the Protostar with jets is formed which means that still the star is in the process of collecting the right amount of materials for its bodily development.
  3. The star formation does take a very swift form of development when all the required number of factors are there to satisfy the formation of the star, as these star nurseries are able to collect the right level or optimum amount of energy for its formation.As, these contracting cloud trapping the infrared light or radiation inside it.
  4. Now, after the different conditions and terms for the formation of the star are satisfied then the mass or shape which the different components of the star body takes place is the molecular cloud fragment.
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A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,
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The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
S= \frac{1}{2} a t^2
where a is the acceleration. Using the data of the problem, we can find a:
a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J

and so the power output is
P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W

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and so the power output is
P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W

c) Power output at t=6.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J

and so the power output is
P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W
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