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konstantin123 [22]
2 years ago
8

Define a machine in the physics subject ​

Physics
1 answer:
mezya [45]2 years ago
3 0

Answer:

Explanation:

Different Parts of an Electric Motor and Their Function

A simple motor has the following parts:

A power supply – mostly DC for a simple motor

Field Magnet – could be a permanent magnet or an electromagnet

An Armature or rotor

Commutator

Brushes

Axle

Power Source: A simple motor usually has a DC power source.  It supplies power to the motor armature or field coils.

Commutator: It is the rotating interface of the armature coil with a stationary circuit.

Field Magnet: The magnetic field helps to produce a torque on the rotating armature coil by virtue of Fleming’s left-hand rule.

Armature Core: Holds the armature coil in place and provides mechanical support.

Armature Coil: It helps the motor to run.

Brushes: It is a device that conducts current between stationary wires and moving parts, most commonly the rotating shaft.

What Is The Working Principle of An Electric Motor?

The working of an electric motor is based on the fact that a current carrying conductor produces a magnetic field around it. To better understand, imagine the following situation.

Take two bar magnets and keep the poles facing each other with a small space in between. Now, take a small length of a conducting wire and make a loop. Keep this loop in between the space between the magnets such that it is still within the sphere of influence of the magnets. Now for the last bit. Connect the ends of the loop to battery terminals.

Once electricity flows through your simple circuit, you will notice that your loop “moves”. So why does this happen? The magnetic field of the magnets interferes with that produced due to electric current flowing in the conductor. Since the loop has become a magnet, one side of it will be attracted to the north pole of the magnet and the other to the south pole. This causes the loop to continuously rotate. This is the principle of working of electric motor.

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A disk is uniformly accelerated from rest with angular acceleration α. The magnitude of the linear acceleration of a point on th
solniwko [45]

Answer:

a = R\alpha\sqrt{1 + \alpha^2t^4}

Explanation:

As we know that the acceleration of a point on the rim of the disc is in two directions

1) tangential acceleration which is given as

a_t = R\alpha

2) Centripetal acceleration

a_c = \omega^2 R

here we know that

\omega = \alpha t

a_c = (\alpha t)^2 R

now we know that net linear acceleration is given as

a = \sqrt{a_c^2 + a_t^2}

so we have

a = \sqrt{R^2\alpha^2 + R^2(\alpha t)^4}

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4 0
3 years ago
What is the potential difference across a parallel-plate capacitor whose plates are separated by a distance of 4.0 mm where each
suter [353]

The potential difference across the parallel plate capacitor is 2.26 millivolts

<h3>Capacitance of a parallel plate capacitor</h3>

The capacitance of the parallel plate capacitor is given by C = ε₀A/d where

  • ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
  • A = area of plates and
  • d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.

<h3>Charge on plates</h3>

Also, the surface charge on the capacitor Q = σA where

  • σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
  • a = area of plates.

<h3>The potential difference across the parallel plate capacitor</h3>

The potential difference across the parallel plate capacitor is V = Q/C

= σA ÷ ε₀A/d

= σd/ε₀

Substituting the values of the variables into the equation, we have

V = σd/ε₀

V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m

V = 20.0 C/m × 10⁻³/8.854 F/m

V = 2.26 × 10⁻³ Volts

V = 2.26 millivolts

So, the potential difference across the parallel plate capacitor is 2.26 millivolts

Learn more about potential difference across parallel plate capacitor here:

brainly.com/question/12993474

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2 years ago
An athlete completes 1 laps around a track with a radius of 25 meters in 180 seconds. What is the magnitude of the athlete's tan
DanielleElmas [232]

Answer:

0.872<em>m/s</em>

Explanation:

Tangential velocity is given by the formula,

v= 2\pi r/ t

In the question given,

radius= 25meters

time= 180secs

pie= 3.14

number of laps= 1

The magnitude of tangential velocity equals;

\frac{1lap* 2 *3.14 *25m}{180secs}

<em>v </em>= 157<em>m</em>/180<em>secs</em>

Therefore, the magnitude of the tangential velocity

=0.872<em>m/secs</em>

5 0
2 years ago
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