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Korolek [52]
3 years ago
6

Under what circumstances will a binary star produce a nova?

Physics
1 answer:
Oksana_A [137]3 years ago
5 0
When a star uses up all of it's energy and begins to die, it swells up to become a red giant star. This causes its surface gravity to decrease, thereby allowing some of its mass to escape into space. A binary star is a pair of stars that orbit each other because of their gravitational attraction to each other. When one member of the binary pair uses up all of its energy and begins to die, it loses mass due to the reduction in surface gravity. But instead of escaping into space, this mass is attracted to the companion star because of its gravitational pull. That increases the mass of the companion star. In a process that takes thousands of years, enough matter is transfered that causes the temperature and pressure to increase sufficiently to result in nuclear fusion reactions on the companion star. When these nuclear reactions become extremely violent, the released nuclear energy increases the brightness of this companion star dramatically, thereby creating a nova. Therefore, it is the dying of one of the stars in a binary system along with a sufficient transfer of star mass to sustain nuclear reactions that results in a nova.
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<h2>Answer:</h2>

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<h2>Explanation:</h2>

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Remember that the magnetic flux through a given area is the total magnetic field passing through that area. Since there is not magnetic field through the cylinder, the total magnetic flux is therefore zero (0).

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3 years ago
Air with a mass flow rate of 2.3 kg/s enters a horizontal nozzle operating at steady state at 450K, 350kPa, and a velocity of 3
MissTica

Answer:

(a) A_1=0.283m^2

(b) Q=-105.5kW: From the air to the surroundings.

Explanation:

Hello,

(a) In this case, we can compute the area at the entrance by firstly computing the inlet volumetric flow:

V_1=\frac{mRT_1}{P_1M}= \frac{2.3\frac{kg}{s} *8.314\frac{kPa*m^3}{kmol* K}*450K}{350kPa*28.97\frac{kg}{kmol} } =0.849\frac{m^3}{s}

Then, with the velocity, we compute the area:

A_1=\frac{V_1}{v_1}=\frac{0.849\frac{m^3}{s} }{3\frac{m}{s} }  =0.283m^2

(b) In this case, via the following energy balance for the nozzle:

Q-W=H_2-H_1+\frac{1}{2} mV_2^2-\frac{1}{2} mV_1^2

We can easily compute the change in the enthalpy by using the given Cp and neglecting the work (no done work):

\Delta H=H_2-H_1=mCp\Delta T=2.3kg/s*1.011\frac{kJ}{kg*K}*(300K-450K)\\ \\\Delta H=-348.795kW

Finally, the heat turns out:

Q=-348.795kW+\frac{1}{2}*2.3\frac{kg}{s}*[(460\frac{m}{s})^2   -(3\frac{m}{s})^2  ]\\\\Q=-348.795kW+243329.65W*\frac{1kW}{1000W}\\ \\Q=-105.5kW

Such sign, means the heat is being transferred from the air to the surroundings.

Regards.

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