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2 years ago

The force that pushes charges through a wire is called _____.

2 answers:

8 0

The force that pushes charges through a wire is called voltage. Electric current is the movement of the electrons caused by potential difference. For example, in a circuit, the electrons in the bulb will not move without the potential difference or voltage. When there is voltage, electric current occurs because there is a driving force for the electrons in the wire to move.

vodka [1.7K]2 years ago

8 0

Answer:

Electromotive

Explanation:

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13. Direct instruction takes place when

jekas [21]

A skill set is explicitly taught or an activity is completed

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2 years ago

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

6 0

2 years ago

Modify how could you charge the electric circuit shown below to allow lightbulb a to stay lit even if lightbulb b is removed fro

shepuryov [24]

When a circuit is complete, or closed, electrons can flow from one end of a battery all the way around, through the wires, to the other end of the battery. Along its way, it will carry electrons to electrical objects that are connected to it – like the light bulb – and make them work!

5 0

2 years ago

A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a

Stella [2.4K]

Answer:

y = 67.6 feet, y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

tan θ = y / L

For the small triangle. Projector up to the person

tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

y₀ / (L -d) = y / L

y = y₀ L / (L-d)

The distance from the screen (d), we look for it with kinematics

v = d / t

d = v t

we replace

y = y₀ L / (L - v t)

y = 5.2 22 / (22 - 3 t)

y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

y = 114.4 / (22-13)

y = 67.6 feet

7 0

2 years ago

10 N pushes a 10 kg crate to the right. Determine the acceleration of the crate.

oksano4ka [1.4K]

The rate of acceleration of the crate would be 1 m/s^2 because the equation for force is F=ma and when you plug in your numbers you get 10=10a so a=1

8 0

2 years ago