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NeX [460]
3 years ago
7

What is the energy difference between the states with the nuclear spin angular momentum components parallel and antiparallel to

the field?
Physics
2 answers:
mixas84 [53]3 years ago
6 0

Answer:

Explanation:

The main difference is that states with spin angular momentum in the same direction of the field are more high energy that the states with antiparallel spin agular momentum.

The energy for a particular level is

E = -(γ*h/2*π)*m*B

where γ is a constant about magnetic moment, h is the Planck constant, m is the state level and B is the strength of the magnetic field. Its clear that for antiparallel spins m=-1/2 the energy is lower in comparisson with states with m=1/2

In fact

ΔE = γ*h*B/2*π

klemol [59]3 years ago
3 0

Answer:

Refer below for the explanation.

Explanation:

The principle distinction is that states with turn precise force a similar way of the field are all the more high vitality that the states with antiparallel turn agular energy.

E = - (γ*h/2*π)*m*B,

where γ is a consistent about attractive minute, h is the Planck steady, m is the state level and B is the quality of the attractive field. Plainly for antiparallel twists m=-1/2 the vitality is lower in comparisson.

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At its peak, a tornado is 53.0 m in diameter and carries 465-km/h winds. What is its angular velocity in revolutions per second?
expeople1 [14]

Answer:

0.7757 rev/s

Explanation:

d = Diameter of the tornado = 53 m

r = Radius of the tornado = 53/2 = 26.5 m

v = Velocity of wind = 465 km/h

Converting velocity to m/s

465=465\times \frac{1000}{3600}=\frac{775}{6}

Angular velocity

\omega=\frac{v}{r}\\\Rightarrow \omega=\frac{\frac{775}{6}}{26.5}\\\Rightarrow \omega=4.87\ rad/s

\omega=4.87\frac{1}{2\pi}=0.7757\ rev/s

∴ Angular velocity is 0.7757 rev/s

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3 years ago
A basketball is shot. After the ball leaves the player's hand, in which direction does the ball accelerate?
azamat

Answer: Always accelerates in a downward direction

Explanation: Gravity is pulling the ball downward

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3 years ago
How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?
Lapatulllka [165]

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

T_{f} =  (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

T_{f}  = 466.8 K

so change in temperature ΔT

ΔT =  466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x_{i\\}² - x_{f\\}² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q =  16 J

Therefore, the required heat energy is 16 J

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