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Pachacha [2.7K]
3 years ago
5

A sun-sized star will spend most of its lifetime as a:

Physics
2 answers:
nekit [7.7K]3 years ago
6 0
A sun-sized star will spend most of its lifetime as a main sequence star. It is the stage in a star's life cycle where it maintains as a stable nuclear reaction. The sun in our solar system is a clear example of this. The number of years that these stars would spend at this stage depends largely on the mass of the star. These stars would fuse atoms of hydrogen forming them into helium atoms inside their cores. Fusion reactions are continually happening as days and years pass at this stage. Approximately ninety percent of the stars found in the universe are main sequence stars.
s2008m [1.1K]3 years ago
6 0

Basically, a sun-sized star will spend most of its lifetime as main sequence star

Notably, a main sequence star is fairly average in size, and tends to spend close to 10 billion years at this stage, where its main function is continuously fusing hydrogen to helium which generates light and heat on earth, which are essential for every human and all other living organisms.

<h2>Further Explanation</h2>

At this stage, a sun-sized star maintains a steady nuclear reaction, and the amount of time it spends is determined by its mass.

All stars have starting phase, including the Sun, and their existence commences when a dense region of clouds or cold gas and dust called nebulae begins to shrink and warm up. Notably, these clouds are relatively large in size, and stars like the sun are enormous balls of plasma that unavoidably fill every space around with heat and light.

Just like humans, stars also live, and like humans, not forever. Although, a typical star could live up to several billions years are specified on the first paragraph. A star starts as a gas cloud and ends a star remnant. The life cycle of a star is in seven (7) phases. These phases are listed below:

  • The Giant gas cloud phase
  • The Protostar phase
  • The T-Tauri phase
  • The Main Sequence Star Phase
  • The expansion into red giant phase
  • The fusion into heavier elements phase
  • The supernovae and planetary nebulae phase

LEARN MORE:

  • the Sun brainly.com/question/1319415

KEYWORDS:

  • the sun
  • sun sized star
  • lifetime
  • hydrogen
  • helium
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The drawing shows a large cube (mass = 21.0 kg) being accelerated across a horizontal frictionless surface by a horizontal force
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The blocks must be pushed with a force higher than 359 Newtons horizontally in order to accomplish this friction levitation feat.

Explanation:

The first step in resolving any physics problem is to draw the given scenario (if possible), see the attached image to have an idea of the objects and forces involved.

The large cube in red is being pushed from the left by a force \vec{P} whose value is to be found. That cube has its own weight \vec{w}_1=m_1\vec{g}, and it is associated with the force of gravity which points downward. Newton's third law stipulates that the response from the floor is an upward pointing force on the cube, and it's called the normal force \vec{N}_1.

A second cube is being pushed by the first, and since the force \vec{P} is strong enough it is able to keep such block suspended as if it were glued to the first one, due to friction. As in the larger cube, the smaller one has a weight \vec{w}_2=m_2\vec{g} pointing downwards, but the normal force in this block doesn't point upwards since its 'floor' isn't below it, but in its side, therefore the normal force directs it to the right as it is shown in the picture. Normal forces are perpendicular to the surface they contact. The final force is the friction between both cubes, that sets a resistance of one moving parallel the other. In this case, the weight of the block its the force pointing parallel to the contact surface, so the friction opposes that force, and thus points upwards. Friction forces can be set as Fr=\mu~N, where \mu is the coefficient of static friction between the cubes.

Now that all forces involved are identified, the following step is to apply Newton's second law and add all the forces for each block that point in the same line, and set it as equal its mass multiplied by its acceleration. The condition over the smaller box is the relevant one so its the first one to be analyzed.

In the vertical component: \Sigma F^2_y=Fr-w_2=m_2 a_y Since the idea is that it doesn't slips downwards, the vertical acceleration should be set to zero a_y=0, and making explicit the other forces: \mu N_2-m_2g=0\quad\Rightarrow (0.710)N_2-(4.5)(10)=0\quad\Rightarrow N_2=(4.5)(10)/(0.710)\approx 63.38 [N]. In the last equation gravity's acceleration was rounded to 10 [m/s^2].

In its horizontal component: \Sigma F^2_x=N_2=m_2 a_x, this time the horizontal acceleration is not zero, because it is constantly being pushed. However, the value of the normal force and the mass of the block are known, so its horizontal acceleration can be determined: 63.38=(4.5) a_x \quad \Rightarrow a_x=(63.38)/(4.5)\approx 14.08 [m/s^2]. Notice that this acceleration is higher than the one of gravity, and it is understandable since you should be able to push it harder than gravity in order for it to not slip.

Now the attention is switched to the larger cube. The vertical forces are not relevant here, since the normal force balances its weight so that there isn't vertical acceleration. The unknown force comes up in the horizontal forces analysis: \Sigma F_x=P=m a_x, since the force \vec{P} is not only pushing the first block but both, the mass involved in this equation is the combined masses of the blocks, the acceleration is the same for both blocks since they move together; P=(21.0+4.5) 14.08\approx 359.04 [N]. The resulting force is quite high but not impossible to make by a human being, this indicates that this feat of friction suspension is difficult but feasable.

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