One one-thousandth is expressed by which prefix?

If two parallel conductors are both free to move and are carrying current in the same direction, what would happen if the curren

pishuonlain [190]

The wires would remain attracted to each other.

Option D.

Explanation:

It is known that magnetic flux will be generated in conductors with varying emf. So when current is flowing in two parallel conductors, the magnetic flux will be generated in those wires. If the current is flowing in same direction in both the wires, then the magnetic flux will be generated towards inside and outside the wires. Thus, the wire will get attracted to each other till the time the current is flowing in the same direction in both the wires. So if the current flow in each wire was reversed at the same time, then the wire would remain attracted to each other.

8 0

3 years ago

An office building has a 24-volt branch circuit installed for landscape lighting around the front of the building. The circuit w

Arturiano [62]

The circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.

<h3>UF cable</h3>

UF cable is used as an underground feeder cable to distribute power from an existing building to outdoor equipment. UF cable can also be used as direct burial cable.

For the 24-volt branch circuit installed, the minimum burial depth will be 6 inches.

Thus, the circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.

Learn more about UF cable here: brainly.com/question/8591560

5 0

2 years ago

A race car circles 10 times around a circular 8.0-km track in 20 min.

Otrada [13]

Find the average speed and the average velocity.

Average speed = distance / time

distance = 10 x 8000 m = 80,000 m
time = 20 min * 60 s/min = 1200 s

Average speed = 80,000 m / 1200 s = 66.67 m/s

Average velocity = displacement / time

Given that the race car made complete circles the final poin is the same initial point, then its displacement is zero and the average velocity is zero too.

6 0

3 years ago

Read 2 more answers

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

a)

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

b)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

c)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is: