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Masja [62]
4 years ago
5

Why do star clusters provide excellent tests for theories of stellar evolution?

Physics
1 answer:
cestrela7 [59]4 years ago
6 0

Explanation:

The star clusters are a great place to prove the theories related to stellar evolution. Over the period of Astronomical evolution we have studied a lot of stars and developed various theories related to how they are born, how they grow up to be main sequence stars, how they go supernova and then turn into a white dwarf or pulsars.

Stars are thought to be born out of a nebula. One nebula can result in thousands of stars which will spread all over with time. Open star clusters are proof of this theory. These contains young stars which were born out of same nebula and now slowly moving apart.

Globular star clusters tell us a lot about the older and massive stars of the galaxy.

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Drawing a ray diagram for an object far from convex lens requires how many rays?
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Answer

4

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four rays can also explain the ray diagram if the object is at infinity

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3 years ago
A lab cart is loaded with different masses and moved at various velocities
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A lab cart is loaded with different masses and moved at various constant velocities? the anser should be

1.0m/s → 4kg
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A snail and an inchworm are in a race. Their race track heads north for a distance of 2 m. If the inchworm comes to the end of t
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Early psychologists were predominately all of the following except __________. A. white B. male C. European D. wealthy
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Answer: D. wealthy

Explanation: on the e2020 test its right

7 0
4 years ago
Read 2 more answers
Consider a well-insulated rigid container with two chambers separated by a membrane. The total volume is 5.0 cubic meters. The f
mamaluj [8]

Answer:

The Entropy generated by the steam = 2.821 kJ/K

Explanation:

Total volume of container = 5m³

Heat transfer does not exist between system and surrounding, dQ = 0

At the first chamber, temperature of water at saturated liquid is 300°C

From the steam table:

Specific enthalpy of saturated liquid at 300°C , h_{f} = 1344.8 kJ/kg

Specific internal energy of saturated liquid at 300°C, U_{f1} =  1332.7 kJ/kg

For closed system, the first law of thermodynamics state that:

dQ = dw + dU..................(1)

work done for free expansion, dw =0

0 = 0 + dU

dU = 0 , i.e. U₁ = U₂

At the second chamber,

The final pressure, P₂ = 50 kPa

From the steam table, at P₂ = 50 kPa,  U_{f2} = 340.49 kJ/kg

(U_{fg} )_{2} =  2142.7 kJ/kg

Let the dryness fraction at the second chamber = x

U_{2} = U_{f2} + U_{fg2}

U_{2} = 340.49 + x2140.7Since U₁ = U₂

1332.7 = 340.49 + x2140.7

Dryness fraction, x = 0.463

From steam table, the specific volume is, u_{f2} = 0.00103 m^{3} /kg\\

u_{2} = u_{f2} + xu_{fg2}

u_{2} = 0.00103 + 0.463(3.2393)\\u_{2} = 1.5 m^{3} /kg\\

u_{2} = \frac{v_{2} }{m_{2} }

V₂ = 5 m³

1.5 = 5/m₂

m₂ = 3.33 kg

At 300°C S_{1} = S_{f} = 3.2548 kJ/kg-k\\

S_{2} = S_{f2} + xS_{fg2}

From the steam table,

S_{f2} = 1.0912 kJ/kg-k\\S_{fg2} = 6.5019 kJ/kg-k\\S_{2} = 1.0912 + 0.463(6.5019)\\S_{2} = 4.102 kJ/kg-k

Therefore the entropy generated will be :

Entropy = mass* (S₂ - S₁)

Entropy = 3.33* (4.102 - 3.2548)

Entropy = 2.821 kJ/K

5 0
4 years ago
Read 2 more answers
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