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N76 [4]
3 years ago
10

Evolutionary sequence of our sun

Physics
1 answer:
n200080 [17]3 years ago
8 0
<span>A cloud of gas and dust begins to contract under the force of gravity. In regions of star birth, we find gaseous nebulae and molecular clouds. These sites of pre-birth are dark patches called globules.The protosun collapsed. As it did, its temperature rose to about 150,000 degrees and the sun appeared very red. Its radius was about 50 present solar radii.When the central temperature reaches 10 million degrees, nuclear burning of hydrogen into helium commences.The star settles into a stable existence on the Main Sequence, generating energy via hydrogen burning. This is the longest single stage in the evolutionary history of a star, typically lasting 90% of its lifetime. Thermonuclear fusion within the Sun is a stable process, controlled by its internal structure.</span><span>The hydrogen in the core is completed burned into helium nuclei. Initially, the temperature in the core is not hot enough to ignite helium burning. With no additional fuel in the core, fusion dies out. The core cannot support itself and contracts; as it shrinks, it heats up. The rising temperature in the core heats up a thin shell around the core until the temperature reaches the point where hydrogen burning ignites in this shell around the core. With the additional energy generation in the H-burning shell, the outer layers of the star expand but their temperature decreases as they get further away from the center of energy generation. This large but cool star is now a red giant, with a surface temperature of 3500 degrees and a radius of about 100 solar radii.<span>The helium core contracts until its temperature reaches about 100 million degrees. At this point, helium burning ignites, as helium is converted into carbon (C) and oxygen (O). However, the core cannot expand as much as required to compensate for the increased energy generation caused by the helium burning. Because the expanion does not compensate, the temperature stays very high, and the helium burning proceeds furiously. With no safety valve, the helium fusion is uncontrolled and a large amount of energy is suddenly produced. This<span>helium flash </span>occurs within a few hours after helium fusion begins.The core explodes, the core temperature falls and the core contracts again, thereby heating up. When the helium burns now, however, the reactions are more controlled because the explosion has lowered the density enough. Helium nuclei fuse to form carbon, oxygen, etc..</span>The star wanders around the red giant region, developing its distinct layers, eventually forming a carbon-oxygen core.When the helium in the core is entirely converted into C, O, etc., the core again contracts, and thus heats up again. In a star like the Sun, its temperature never reaches the 600 million degrees required for carbon burning. Instead, the outer layers of the star eventually become so cool that nuclei capture electrons to form neutral atoms (rather than nuclei and free electrons). When atoms are forming by capturing photons in this way, they cause photons to be emitted; these photons then are readily available for absorption by neighboring atoms and eventually this causes the outer layers of the star to heat up. When they heat up, the outer layers expand further and cool, forming more atoms, and releasing more photons, leading to more expansion. In other words, this process feeds itself.The outer envelope of the star blows off into space, exposing the hot, compressed remnant core. This is a <span>planetary nebula </span>.</span><span>The core contacts but carbon burning never ignites in a one solar mass star. Contraction is halted when the electrons become degenerate, that is when they can no longer be compressed further. The core remnant as a surface temperature of a hot 10,000 degrees and is now a <span>white dwarf </span>.With neither nuclear fusion nor further gravitational collapse possible, energy generation ceases. The star steadily radiates is energy, cools and eventually fades from view, becoming a black dwarf.</span>
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A light ray travels in the +x direction and strikes a slanted surface with an angle of 62° between its normal and the ty axis. T
Elena-2011 [213]

Answer:

- 0.6

Explanation:

Given that angle between normal y axis is 62° so angle between  normal

and x axis will be 90- 62 = 28 °. Since incident ray is along x axis , 28 ° will be the angle between incident ray and normal ie it will be angle of incidence

Angle of incidence = 28 °

angle of reflection = 28°

Angle between incident ray and reflected ray = 28 + 28 = 56 °

Angle between x axis and reflected ray = 56 °

x component of reflected ray

= - cos 56 ( it will be towards - ve x axis. )

- 0.6

6 0
3 years ago
Um elétron é lançado entre duas placas eletrizadas como mostra a figura. Sejam v= 6x10^6 m/s, ângulo 45°, E= 2x10^3 N/C, d= 3 cm
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3 0
3 years ago
(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5
Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is 3.02\times10^{-11}.

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

a=\dfrac{v^2}{r}

Put the value into the formula

a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}

Put the value into the formula

a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}

a=2.32\times10^{15}\ m/s^2

We need to calculate the rate at which it emits energy because of its acceleration is

\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}

Put the value into the formula

\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}

\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s

The energy in ev/s

\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s

\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s

We need to calculate the fraction of its energy that it radiates every second

\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}

\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}

Hence, The fraction of its energy that it radiates every second is 3.02\times10^{-11}.

5 0
3 years ago
How big is the universe?<br> sorry this is so random
prisoha [69]

Answer:

Explanation: 46 billion light years I think

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6 0
3 years ago
A truck speed from rest to 120km/h in 13s. Find its acceleration. (convert to m/s)
ioda

answer:

2.5 m/s²

Explanation:

120 km/h = 120 ÷ 3.6 = 100/3 ≈ 33 m/s

a = (v2 - v1)/∆t = (33m/s - 0)/ 13s = 33/13 m/s²≈ 2.5 m/s²

4 0
3 years ago
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