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g100num [7]
3 years ago
11

In a binary-star system that produces a nova, the white dwarf pulls matter from the companion star. The matter forms an accretio

n disk that orbits the white dwarf. Then a specific sequence of events must take place for a nova event to occur. Rank the steps leading up to the observed nova event in chronological order from first to last.
Physics
1 answer:
Studentka2010 [4]3 years ago
4 0

As a head-up, it is important to notice that a white dwarf only shines thanks to the stored energy and light, because a white dwarf doesn't have any hydrogen left to perform nuclear fusion.

Now the process:

First, the white dwarf accumulates all the extracted matter from its companion, onto its own surface. This extra matter increases the white dwarf's temperature and density.

After a while, the star reaches about 10 million K, so nuclear fusion can begin. The hydrogen that has been "stolen" from the other star and accumulated in the white dwarf's surface it's used for the fusion, dramatically increasing the star's brightness for a short time, causing what we know as a Nova.

As this fuel its quickly burnt out or blown into space, the star goes back to its natural white dwarf state. Since the white dwarf nor the companion star are destroyed in this process, it can happen countless of times during their lifespan.

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3. (a) Sate any three properties of an ideal gas as assumed by the kinetic theory of gas.​
Aliun [14]

Explanation:

The simplest kinetic model is based on the assumptions that: (1) the gas is composed of a large number of identical molecules moving in random directions, separated by distances that are large compared with their size; (2) the molecules undergo perfectly elastic collisions (no energy loss) with each other and with the ...

4 0
2 years ago
If Light is travels from air into pure water with an incident angle of 30°. What is the angle of refraction?
Orlov [11]

Explanation:

For air, n1 = 1.00003; for water, n2 = 1.3330

Given: θ2 = 30 degrees, then

θ1 = arcsin [(n2/n1) sin θ2]

= arcsin [(1.3330/1.0003) sin (40)]

= 58.93 degrees

Note that since, in this example, light is traveling from a medium of higher density (water; n2 = 1.3330) to a medium of lower density (air; n1 = 1.0003), then n2 > n1, and the angle of refraction (θ1) is larger than the angle of incidence (θ2), thus the light bends away from the normal (in this example, the vertical) as it leaves the water and enters the air.

7 0
3 years ago
Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into th
satela [25.4K]

The tank has a volume of \dfrac\pi3R^2H, where H=6\,\rm m is its height and R=\dfrac d2=2\,\rm m is its radius.

At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:

\dfrac26=\dfrac rh\implies r=\dfrac h3

The volume of water in the tank at any given time is

V=\dfrac\pi3r^2h

and can be expressed as a function of the water level alone:

V=\dfrac\pi3\left(\frac h3\right)^2h=\dfrac\pi{27}h^3

Implicity differentiating both sides with respect to time t gives

\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9h^2\,\dfrac{\mathrm dh}{\mathrm dt}

We're told the water level rises at a rate of \dfrac{\mathrm dh}{\mathrm dt}=20\,\frac{\rm cm}{\rm min} at the time when the water level is h=2\,\mathrm m=200\,\mathrm{cm}, so the net change in the volume of water \dfrac{\mathrm dV}{\mathrm dt} can be computed:

\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9(200\,\mathrm{cm})^2\left(20\,\dfrac{\rm cm}{\rm min}\right)=\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}

The net rate of change in volume is the difference between the rate at which water is pumped into the tank and the rate at which it is leaking out:

\dfrac{\mathrm dV}{\mathrm dt}=(\text{rate in})-(\text{rate out})

We're told the water is leaking out at a rate of 10,500\,\frac{\mathrm{cm}^3}{\rm min}, so we find the rate at which it's being pumped in to be

\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}=(\text{rate in})-10,500\,\dfrac{\mathrm{cm}^3}{\rm min}

\implies\text{rate in}\approx289,753\,\dfrac{\mathrm{cm}^3}{\rm min}

4 0
3 years ago
Which scientist was the first to conclude through experimentation that atoms have positive charges in their nuclei?
Pavel [41]
<h2>Answer: Ernest Rutherford </h2>

Ernest Rutherford was a British physicist and chemist of New Zealand origin, who conducted a series of experiments together with Hans Geiger and Ernest Marsden; where the result led him to propose a new atomic model.

It should be noted that at that time, the "accepted" atomic model was Thomson's raisin pudding atomic model<u> </u><em><u>(electrons with negative charge immersed an the atom of positive charge that counteracted the negative charge of the electrons, like raisings embedded in a pudding)</u></em>, who discovered the electron and formerly was a professor of Rutherford.  

Now, the experiment conducted under the direction of Ruherford at the laboratories of the University of Manchester during the year 1911; was for the purpose of <u>corroborating Thomson's atomic model</u>. To achieve this, a thin metal sheet was bombarded with alpha particles (nuclei of helium gas).

The idea was that these alpha particles, having positive electric charge, were attracted by the atom's negative charges and repelled by the positive charges, and it was expected that they would pass through the thin sheet without hardly deviating. Then, to observe the crash site of the particle, a phosphorescent screen was placed behind and on the sides of the metal sheet.

For according to Thomson's atomic model the positive and negative charges were evenly distributed, the sphere (the atom) had to be electrically neutral, and <u>the alpha particles would pass through the sheet without deviating. </u>

However, the results were surprising:

As expected, most of the particles went through the sheet without deviating.

<h2>But some suffered large deviations and, most importantly, <u>a small number of particles bounced backwards</u>. </h2>

That is:

<h2>The alpha particle beam was scattered (repelled) when it hit the thin metal sheet. </h2>

These facts could not be explained by Thomson's atomic model, so Rutherford developed another, suggesting that:

<h2><em>There is a concentration of charge in the center of the atom (which was later called nucleus) surrounded by electrons. </em></h2>

This new model could explain the proven fact in his experiments that some particles bounced in the direction opposite to the incident particles, because the electrical charge of this nucleus was positive, equal to the electrical charge of the alpha particles.

This is how Rutherford proposed a new atomic model and discovered the existence of the nucleus. However, this was not the definitive model, because on 1913 it was replaced by Bohr's.

8 0
3 years ago
Does which object have the compacity that is the best measured in pints?
Wewaii [24]

Answer:

1 or 2

Explanation:

8 0
3 years ago
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