B) Explain the method of preparing electromagnet. How do you test the

Physics

1 answer:

kap26 [50]3 years ago

5 0

Answer:

An electromagnet is made by forming a coil around a soft iron bar (known here as the metal) such as a nail or screw and connect with an insulated copper wire (known here as the electric current conductor) the ends of the wound copper is then connected separately to the positive and negative terminals of a battery (known here as the source of electric current)

The north seeking needle of the magnetic compass will move away when brought close to the north pole of the formed electromagnet which can then be labelled N

The magnetic compass needle will be attracted to the south pole of the electromagnet which can then be labelled S

Explanation:

An electromagnet is an electric powered magnet that is formed (temporarily) by the perpendicular movement of electric current with respect to a metal core

The magnitude and the poles of an electromagnet can be changed by changing the magnitude and the direction of flow of the electric current respectively.

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When the saw slices wood, the wood exerts a 104-N force on the blade, 0.128 m from the blade’s axis of rotation. If that force i

Soloha48 [4]

Answer:

Explanation:

When saw slices wood by exerting a force on the wood , wood also exerts a reaction force on the saw in opposite direction which is equal to the force of action that is 104 N.

So torque exerted by wood on the blade

= force x perpendicular distance from the axis of rotation

= 104 x .128

=13.312 Nm.

Since this torque opposes the movement of blade , it turns the blade slower.

5 0

2 years ago

Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$