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>

3 0

3 years ago

A student pushes on a 8-kg box with a force of 35 N forward. The force of sliding friction is 10 N backward. What is the acceler

Ne4ueva [31]

Answer:

(35 N - 10 N)/8kg = 3.125 m/s^2

Explanation:

The formula for Force is:

Force = Mass*Acceleration

(Force is equal to Mass times Acceleration)

Since we're told to find the acceleration of the box. We make acceleration the subject of the equation:

Acceleration = Force/Mass

(Acceleration equal to Force divided by Mass)

We know that the force are 35 N forward and 10 N backward, and the weight of the box is 8kg.

= (35 N - 10 N)/8kg

The reason that 35 N minus 10 N is because the 10 N is pushing the box backward.

= 25 N/8kg

= 3.125 m/s^2

Hope it helps :DD

3 0

3 years ago

A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. The mass is pull

sdas [7]

In spring mass system we know that angular frequency is given as

$\omega = 2\pi f$

f = 8.38 Hz

$\omega = 2\pi(8.38)$

$\omega = 52.65 rad/s$

now we know that speed of SHM at its extreme position is given by

$v = A\omega$

here we know that

A = 17.5 cm

$v = 0.175 (52.65)$

$v = 9.21 m/s$

so maximum speed is 9.21 m/s

7 0

3 years ago