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sergij07 [2.7K]
3 years ago
15

Suppose there are single-particle energy eigenvalues of 0, e, 2.::, and 3.:: which are non-degenerate. A total of 6.0 is to be s

hared between four particles. List the configuration of the particles and their degeneracies for: distinguishable particles; indistinguish3:ble Bose particles; indistinguishable Fermi particles

Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
8 0

Answer:

so 40 ways for distinguishable

4 ways for bosons

1 way for fermions

Explanation:

The explanation is attached below

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A bar magnet is dropped toward a conducting ring lying on the floor. As the magnet falls toward the ring, does it move as a free
Maslowich

Consider that the bar magnet has a magnetic field that is acting around it, which will imply that there is a change in the magnetic flux through the loop whenever it moves towards the conducting loop. This could be described as an induction of the electromotive Force in the circuit from Faraday's law.

In turn by Lenz's law, said electromotive force opposes the change in the magnetic flux of the circuit. Therefore, there is a force that opposes the movement of the bar magnet through the conductor loop. Therefore, the bar magnet does not suffer free fall motion.

The bar magnet does not move as a freely falling object.

3 0
3 years ago
A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a
sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0
3 years ago
The particles of a substance stay close together but slide past one another as they move when thermal energy is added to the sub
Katen [24]

Answer: A liquid to gas

Explanation: I just got it wrong :(

6 0
3 years ago
Determine the binding energy per nucleon of an mg-24 nucleus. the mg-24 nucleus has a mass of 24.30506. a proton has a mass of 1
My name is Ann [436]

The mass of Mg-24 is 24.30506 amu, it contains 12 protons and 12 neutrons.

Theoretical mass of Mg-24:

The theoretical mass of Mg-24 is:

Hydrogen atom mass = 12 × 1.00728 amu = 12.0874 amu

Neutron mass = 12 x 1.008665 amu = 12.104 amu

Theoretical mass = Hydrogen atom mass + Neutron mass = 24.1913 amu

Note that the mass defect is:

Mass defect = Actual mass - Theoretical mass : 24.30506 amu- 24.1913 amu= 0.11376 amu

Calculating the binding energy per nucleon:

\frac{B.E.}{nucleon}=\frac{(0.11376amu)(931Mev/amu}{24nucleons}  = 4.41294 Mev/nucleon

So approximately 4.41294 Mev/necleon


3 0
3 years ago
How do I do this question?
Mkey [24]

a. 27

after that all you need to do is use the work formula as you will be able to acheive results

5 0
3 years ago
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