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wel
3 years ago
7

A localized electron has been polarized so that its spin is oriented in the positive z- direction. It is now subject to the appl

ication of a constant uniform magnetic field B = Bî along x over a period of time of duration T. After that, it is subject to the application of another magnetic field of the same magnitude B but along y: B2 = Bý, also with duration T. (a) What is the probability P that the spin-flip would occur as a result? That is, what is the probability that the spin of the electron would be found oriented in the negative z-direction after the application of the magnetic fields is over? (b) Is it possible to find such duration that the spin-flip would occur with certainty: P = 1? If yes, what would be that time r?

Physics
1 answer:
ioda3 years ago
4 0

Answer: The answer is attached

Explanation:

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A batter hits two baseballs with the same force. One hits the ground near third base. The other is a home run out of the park. W
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3 years ago
horizontal circular platform rotates counterclockwise about its axis at the rate of 0.919 rad/s. You, with a mass of 73.5 kg, wa
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Answer:

The total angular momentum is 292.59 kg.m/s

Explanation:

Given that :

Rotation of the horizontal circular platform \omega = 0.919 rad/s

mass of the platform (m) = 90.7 kg

radius (R) = 1.91 m

mass of the poodle m_p = 20.5 kg

Your mass  m' = 73.5 kg

speed v = 1.05 m/s with respect to the platform

V_p = \frac{1.05}{2} \ m /s \\ \\ = 0.525  \ m /s  \\  \\r = \frac{R}{2}

r = 0.955

Mass of the mutt m_m = 18.5 kg

r' = \frac{3}{4} \ R

Your angular momentum is calculated as:

Your angular velocity relative to the platform is \omega' = \frac{v}{R}  = \frac{1.05}{1.91 } = 0.5497 \ rad/s

Actual \ \omega_y = \omega - \omega ' = (0.919 - 0.5497) \ rad/s = 0.3693 \ rad/s

I_y = m'R^2 = 73.5 *1.91^2= 268.14 \ kgm^2

L_Y = I_y*\omega_y= 268.14*0.3683= 98.76 \ kg.m/s

For poodle :

Relative \ \ \omega' = \frac{V_p}{R/2} = 0.550 \  rad/s

Actual \omega_p = \omega - \omega' =  0.919 -0.550 = 0.369 \ rad/s

I_p = m_p(\frac{R}{2} )^2  = 20.5(\frac{1.91}{2} )^2 = 18.70 \ kgm^2

L_p = I_p *\omega_p = 18.70*0.369 = 6.9003 \  kgm/s

I_M = m_m(\frac{3}{y4} R)^2 = 18.5 (\frac{3}{4} 1.91)^2 = 37.96 \  kgm^2

L_M = I_M \omega = 37.96 * 0.919= 34.89 \ kg.m/s

Disk I = \frac{mr^2}{2} =  \frac{90.7*1.91^2}{2}= 165.44 \ kgm^2

L_D = I \omega = 165.44*0.919 =152.04 \ kg.m/s

Total angular momentum of system is:

L = L_D +L_Y+L_P+L_M

= (152.04 + 98.76 + 6.9003 + 34.89) kg.m/s

= 292.59 kg.m/s

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Answer:

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Answer:

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