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BabaBlast [244]
3 years ago
9

A battery of e.m.f. 10 V is connected to resistance as shown in figure. The potential

Physics
1 answer:
jonny [76]3 years ago
8 0
D correct answer for que
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Suppose that the dipole moment associated with an iron atom of an iron bar is 2.8 × 10-23 J/T. Assume that all the atoms in the
masya89 [10]

To solve this exercise it is necessary to apply the equations related to the magnetic moment, that is, the amount of force that an image can exert on the electric currents and the torque that a magnetic field exerts on them.

The diple moment associated with an iron bar is given by,

\mu = \alpha *N

Where,

\alpha = Dipole momento associated with an Atom

N = Number of atoms

\alpha y previously given in the problem and its value is 2.8*10^{-23}J/T

L = 5.8cm = 5.8*10^{-2}m

A = 1.5cm^2 = 1.5*10^{-4}m^2

The number of the atoms N, can be calculated as,

N = \frac{\rho AL}{M_{mass}}*A_n

Where

\rho = Density

M_{mass} = Molar Mass

A = Area

L = Length

A_n =Avogadro number

N = \frac{(7.9g/cm^3)(1.5cm)(5.8cm^2)}{55.9g/mol}(6.022*10^{23}atoms/mol)

N = 7.4041*10^{23}atoms

Then applying the equation about the dipole moment associated with an iron bar we have,

\mu = \alpha *N

\mu = (2.8*10^{-23})*(7.4041*10^{23})

\mu = 20.72Am^2

PART B) With the dipole moment we can now calculate the Torque in the system, which is

\tau = \mu B sin(90)

\tau = (20.72)(2.2)

\tau = 45.584N.m

<em>Note: The angle generated is perpendicular, so it takes 90 ° for the calculation made.</em>

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During which month is the bacteria population just under 1 million?
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What are the potential out comes of force
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Answer:

F in the definition of potential energy is the force exerted by the force field, e.g., gravity, spring force, etc. The potential energy U is equal to the work you must do against that force to move an object from the U=0 reference point to the position r.

Explanation:

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An Apple falls from a tree and one-half second later hits the ground
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Calculate the kinetic energy of a 4.00kg object traveling at 16.0 m/s
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KE = ½mv² = ½(4.00 kg)(16.0 m/s)² = 512 J
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