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Delicious77 [7]
2 years ago
9

1) A capacitor consists of two metal surfaces separated by an electrical insulator with no electrically conductive path through

it. Why does a current flow in a resistor-capacitor circuit when the switch is closed?
Physics
1 answer:
ki77a [65]2 years ago
7 0

Answer:

Current flows in a resistor-capacitor circuit because of the varying electric field across the plates of a capacitor induced by an AC voltage source <em>(displacement current)</em>

Explanation:

In a capacitor, current does not flow the same way it does in a circuit, that is through conduction. This is because there is a highly resistive material in between the plates of the capacitor. Rather current flows through a phenomenon called displacement current.

Because of change in charge accumulation with time above the plates, the electric field changes causing the displacement current.

Displacement current arises due to the flow of electrons as a result of the varying magnetic fields set up on the plates of the capacitor when supplied with an AC voltage. It is important to note that a DC voltage does not induce any displacement current.

<em>Through this, phenomenon discovered by Maxwell,  current is able to flow in a resistor-capacitor circuit despite the absence of an electrically conductive path through the plates.</em>

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As the driver steps on the gas pedal, a car of mass 1 140 kg accelerates from rest. During the first few seconds of motion, the
krok68 [10]

Answer:

(a) KE=16405.215 J

(b) P = 6309.6981 W

(c) Value in above part is described as minimum because there would have been power loss in the actual system to achieve this acceleration from the state of rest.

Explanation:

Given:

mass of car, m = 1140 kg

expression of acceleration, a=1.14t-0.210t^2+0.240t^3

where "t" is time in seconds

initial time, t_i=0 s

final time, t_f=2.6 s

(a)

We know,

\frac{dv}{dt} =a

dv=a.dt

v=\int\limits^{2.6}_0 {1.14t-0.210t^2+0.240t^3} \, dt

v=5.3648 m.s^{-1}

Kinetic Energy

∴KE= \frac{1}{2} m.v^2

KE=\frac{1}{2}\times 1140\times 5.3648^2

KE=16405.215 J

(b)

We know,

Power

P= \frac{\Delta KE}{\Delta t}

P=\frac{16405.215}{2.6}

P = 6309.6981 W

(c)

Value in above part is described as minimum because there would have been power loss in the actual system to achieve this acceleration from the state of rest.

3 0
2 years ago
A proposed space elevator would consist of a cable stretching from the earth's surface to a satellite, orbiting far in space, th
NISA [10]

To solve this problem we will apply the concepts related to energy conservation. Here we will use the conservation between the potential gravitational energy and the kinetic energy to determine the velocity of this escape. The gravitational potential energy can be expressed as,

PE= \frac{GMm}{d}

The kinetic energy can be written as,

KE= \frac{1}{2} mv^2

Where,

G = 6.67*10^{-11}m^3/kg\cdot s^2Gravitational Universal Constant

m = 5.972*10^{24}kg Mass of Earth

h = 56*10^6m  Height

r = 6.378*10^6m Radius of Earth

From the conservation of energy:

\frac{1}{2} mv^2 = \frac{GMm}{d}

Rearranging to find the velocity,

v = \sqrt{\frac{2Gm}{d}} \rightarrow  Escape velocity at a certain height from the earth

If the height of the satellite from the earth is h, then the total distance would be the radius of the earth and the eight,

d = r+h

v = \sqrt{\frac{2Gm}{r+h}}

Replacing the values we have that

v = \frac{2(6.67*10^{-11})(5.972*10^{24})}{6.378*10^6+56*10^6}

v = 3.6km/s

Therefore the escape velocity is 3.6km/s

3 0
2 years ago
Inductive charging is used to wirelessly charge electronic devices ranging from toothbrushes to cell phones. Suppose the base un
denpristay [2]

Answer:

1.2*10^{-6}s

Explanation:

The problem must be addressed through the concepts of electromotive force. By Faraday's law it is defined that

\epsilon = NA \frac{dB}{dt}

Where

\epsilon = Electromotive Force

N = Number of Loops

A = Area

B = Magnetic Field (chaging through the time)

From this equation and our values, we need to find the time, then we re-arrange the equation

dt = NA \frac{dB}{\epsilon}

t = (16)(2.75*10^{-4})\frac{1.50*10^{-3}}{5.50}

t = 1.2*10^{-6}s

Therefore the time required for the magnetic field to decrease to zero from its maximum value is 1.2*10^{-6}s

8 0
2 years ago
A 150 g particle at x = 0 is moving at 8.00 m/s in the +x-direction. As it moves, it experiences a force given by Fx=(0.850N)sin
krok68 [10]

Answer:

9.98 m/s

Explanation:

The force acting on the particle is defined by the equation:

F=(0.850) sin (\frac{x}{2.00}) [N]

where x is the position in metres.

The acceleration can be found by using Newton's second law:

a=\frac{F}{m}

where

m = 150 g = 0.150 kg is the mass of the particle. Substituting into the equation,

a=\frac{0.850}{0.150}sin (\frac{x}{2.00})=5.67 sin(\frac{x}{2.00}) [m/s^2]

When x = 3.14 m, the acceleration is:

a=5.67 sin(\frac{3.14}{2.00})=5.67 m/s^2

Now we can find the final speed of the particle by using the suvat equation:

v^2-u^2=2ax

where

u = 8.00 m/s is the initial velocity

v is the final velocity

a=5.67 m/s^2

x = 3.14 m is the displacement

Solving for v,

v=\sqrt{u^2+2ax}=\sqrt{8.00^2+2(5.67)(3.14)}=9.98 m/s

And the speed is just the magnitude of the velocity, so 9.98 m/s.

4 0
2 years ago
Lightning results from ________.
just olya [345]
An imbalance between electrical charges
8 0
2 years ago
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