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uranmaximum [27]
3 years ago
6

A magnetic field is applied to a freely floating uniform iron sphere with radius R = 2.00 mm. The sphere initially had no net ma

gnetic moment, but the field aligns 12% of the magnetic moments of the atoms (that is, 12% of the magnetic moments of the loosely bound electrons in the sphere, with one such electron per atom).
The magnetic moment of those aligned electrons is the sphere’s intrinsic magnetic moment μx−→


What is the sphere’s resulting angular speed ω?
Physics
1 answer:
grigory [225]3 years ago
5 0

Answer:

\omega=4.2704*10^-^5 rad/s

Explanation:

Angular Momentum Formula For atoms=L_{atom}=0.12Nm_{s}h

Where:

m_{s}h is the momentum for one atom (m_s is the spin quantum number)

N is the number of atoms=\frac{N_{A}*m}{M}

Where:

N_A is Avogadro Number

m is the mass of sphere

M is the molar mass of iron

Angular Momentum Formula For atoms will be=L_{atom}=0.12\frac{N_{A}m}{M} m_{s}h

Angular Momentum of Sphere=L_{sphere}=I\omega

where:

I=\frac{2mR^{2}}{5}

So,Angular Momentum of Sphere=L_{sphere}=\frac{2mR^{2}}{5}\omega

Angular Momentum of sphere=Angular Momentum of atoms

L_{sphere}=L_{atom}

\frac{2mR^{2}}{5}\omega=0.12\frac{N_{A}m}{M} m_{s}h

For iron, m_s =\frac{1}{2}. So above equation will become:

\omega=\frac{0.12*5*N_{A}h}{4*M*R^{2} }

Where R=2mm, M=0.0558Kg/mol (Molar Mass of iron),h=Planck's Constant/2π

\omega=\frac{0.12*5*(6.022*10^{23})(6.63*10^{-34}/2*\pi)}{4*0.0558*(2*10^{-3})^{2}}

\omega=4.2704*10^-^5 rad/s

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A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten
babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

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8 0
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Find the energy in joules required to lift a 55.0 megagram object a distance of 500cm
pantera1 [17]

1,000 grams = 1 kilogram
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100 cm = 1 meter
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Acceleration of gravity on Earth = 9.8 m/s²

Weight = (mass) x (gravity)

========================================

Work = increase in potential energy =

               (weight) x (height) =

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             (55,000 kg) x (9.8 m/s²) x (5 m) =

                     2,695,000 joules .

5 0
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