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3 years ago

Describe what causes the flow of electric current.

2 answers:

5 0

An electrical current is caused by flow of free electrons from one atom to another.

VMariaS [17]3 years ago

3 0

My answer -

Electrical pressure, called "electro-motive force" (EMF), is what causes current to flow in an electrical circuit. EMF is measured in volts. Pressure and the quantity of electricity transmitted can be considered analogous to pressure and quantity of water flowing in pipes: the longer the distance, the higher is the flow of water (called the "flow of electrical charge, which is also known as the "electric current"). That is why, for long distance transmission, high pressure (voltage for electricity) is required, failing which, the current - and therefore the power - will not reach the intended destination. Instead, it will be lost (also called "dissipate") along the way. We can think of electrical
current as the quantity of electricity which will be drawn from the pipeline
 cables for electricity) at the pressure (= voltage) required. 

P.S

Happy to help you have an AWESOME!!! day ^-^

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If the shoe has less mass, it will experience _______________ (more, less, the same) friction as it would with more mass.

Valentin [98]

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3 years ago

In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue

Tomtit [17]

a) p₀ = 1.2 kg m / s

b) p_f = 1.2 kg m / s

c) $v_{2f}$ = 1.278 m/s

d)1.2482 = $v_{2f}$ cos θ

e) 0.2736 = $v_{2f}$ sin θ

g) θ = 12.36

<h3>What is momentum?</h3>

Momentum is the product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction.

According to the question,

We define a system formed by the two balls, which are isolated and the forces during the collision are internal.

Hence , the momentum is conserved

a) The momentum of the system before the collision:

$p_o = m v_1_0 + 0 p_o= 0.6 2 p_o = 1.2 kg m / s$

b) The momentum of the system after the collision, as the system is isolated, the momentum is conserved so

$p_f = 1.2 kg m / s$

d. On X axis

$p_ox$ = 1.2 kg m / s

$p_{fx}$ = m v1f cos 20 + m v2f cos θ

$p_o = p_f$

1.2 = 0.6 (-0.8) cos 20+ 0.6 $v_{2f}$ cos θ

1.2482 = $v_{2f}$ cos θ

e. On Y axis

$p_{oy} = 0$

$p_{fy}$ = m $v_{1f}$ sin 20 + m $v_{2f}$ cos θ

0 = 0.6 (-0.8) sin 20 + 0.6 $v_{2f}$ sin θ

0.2736 = $v_{2f}$ sin θ

Here,

g. we write our system of equations

0.2736 = $v_{2f}$ sin θ

1.2482 = $v_{2f}$ cos θ

0.219 = tan θ (divide)

θ = tan⁻¹ 0.21919

θ = 12.36

c. Now To find the velocity

0.2736 = $v_{2f}$ sin θ

$v_{2f}$ = 0.2736 / sin 12.36

$v_{2f}$ = 1.278 m / s

Learn more about momentum here:

brainly.com/question/4956182

#SPJ1

6 0

2 years ago

A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b

Natasha_Volkova [10]

Answer:

Part a)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

Outside the outer cylinder we will again use Guass law

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.