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Arturiano [62]
3 years ago
9

Electric potential is associated with both electric fields due to static charges and induced electric fields. Electric potential

is associated with magnetic fields but not with electric fields due to static charges. Electric potential is associated with both electric fields due to static charges and magnetic fields. Electric potential is associated with electric fields due to static charges but not with induced electric fields. Electric potential is associated with induced electric fields but not with electric fields due to static charges.
Physics
1 answer:
maksim [4K]3 years ago
4 0

Answer:

Electric potential is associated with electric fields due to static charges but not with induced electric fields.

Explanation:

An electric potential is the amount of work needed to move a unit charge from a reference point to a specific point inside the field without producing an acceleration. Typically, the reference point is the Earth or a point at infinity. Induced electricity is as a result of changing magnetic flux linkage (no charge is involved)

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A metal ring 4.60 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend
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Answer:

A. Ein = 8.05*10^-4 V/m

B. Clockwise sense

Explanation:

A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:

\int E_{in} \cdot ds=-\frac{d\Phi_B}{dt}            (1)

Ein: induced electric field

ds: differential of a path of the ring

ФB: magnetic flux in the ring

The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):

\int E_{in}ds=E_{in}(2\pi r)=-A\frac{dB}{dt}\\\\E_{in}=-\frac{A}{2\pi r}\frac{dB}{dt}   (2)

dB/dt = -0.280T/s     (it is decreasing)

A: area of the ring = π(r/2)^2= (π/4) r^2

r: radius of the ring = 4.60/2 = 2.30 cm

Then, you replace the values of all variables in the equation (2):

E_{in}=-\frac{(\pi/4)r^2}{2\pi r}\frac{dB}{dt}=\frac{r}{8}\frac{dB}{dt}\\\\E_{in}=-\frac{0.0230m}{8}(-0.280T)=8.05*10^{-4}\frac{V}{m}

hence, the induced electric field is 8.05*10^-4 V/m

B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.

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A student must design an experiment to determine the gravitational mass of an object. Which of the following experiments could t
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Answer: a. Place the object on one side of a lever at a known distance away from the fulcrum. Place known masses on the other side of the fulcrum so that they are also paced on the lever at known distance from the fulcrum. Move the known masses to a known distance such that the lever is in static equilibrium.

d. Place the object on the end of a vertically hanging spring with a known spring constant. Allow the spring to stretch to a new equilibrium position and measure the distance the spring is stretched from its original equilibrium position.

Explanation:

The options are:

a. Place the object on one side of a lever at a known distance away from the fulcrum. Place known masses on the other side of the fulcrum so that they are also paced on the lever at known distance from the fulcrum. Move the known masses to a known distance such that the lever is in static equilibrium.

b. Place the object on a surface of negligible friction and pull the object horizontally across the surface with a spring scale at a non constant speed such that a motion detector can measure how the objects speed as a function of time changes.

c. Place the object on a surface that provides friction between the object and the surface. Use a surface such that the coefficient of friction between the object and the surface is known. Pull the object horizontally across the surface with a spring scale at a nonconstant speed such that a motion detector can measure how the objects speed as a function of time changes.

d. Place the object on the end of a vertically hanging spring with a known spring constant. Allow the spring to stretch to a new equilibrium position and measure the distance the spring is stretched from its original equilibrium position.

Gravitational mass simply has to do with how the body responds to the force of gravity. From the options given, the correct options are A and D.

For option A, by balancing the torque, the mass can be calculated. Since the known mass and the distance has been given here, the unknown mass can be calculated.

For option D, here the gravitational force can be balanced by the spring force and hence the mass can be calculated.

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