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3 years ago

What's a diode..??? How does it act as a rectifier??

1 answer:

6 0

A diode is a semiconductor device that essentially acts as a one-way switch for current. It allows current to flow easily in one direction, but severely restricts current from flowing in the opposite direction.

A rectifier is a device that converts an Alternating Current (AC) into a Direct Current (DC) by using one or more contact diodes. ... In simple words, a diode allows current in just one direction. This unique property of the diode allows it to act sort of a rectifier by converting an alternating current to a DC source

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If the moon rises around 3 a.m., its phase must be

Luden [163]

New moon or cresant moon i believe

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3 years ago

How is displacement different from distance?

Anni [7]

Answer:

i do belive its C

Explanation:

i remeber this question from somewhere also it makes the most sense

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3 years ago

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BIGGGGGGG POINTS 22 POINTS ANSWER NOW. Tungsten wire is used as the filament in light bulbs. It glows white-hot as current passe

Mashutka [201]

Answer:

Tungsten wire is used as the filament in light bulbs. It glows white-hot as current passes through it.

Tungsten is used in light bulbs because its high RESISTANCE converts electric energy into light and heat.

Explanation:

7 0

3 years ago

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Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

3 years ago

When a magnetic field is first turned on, the flux through a 20-turn loop varies with time according to Φm=5.0t2−2.0t, where Φm

goldfiish [28.3K]

Answer:

A) ( - 200t + 40 ) volts

B) b) anticlockwise , c) anticlockwise , d) clockwise , e) clockwise

Explanation:

Given data:

magnetic flux (Φm) = 5.0t^2 − 2.0t

number of turns = 20

a) determine induced emf

E = - N $\frac{d\beta }{dt}$

= - N ( 10t - 2 ) = - 20 ( 10t - 2 )

= - 200t + 40 volts

b) Determine direction of induced current

i) at t = 0

E = - 0 + 40 ( anticlockwise direction )

ii) at t = 0.10

E = -20 + 40 = 20 ( anticlockwise direction )

iii) at t = 1

E = - 200 + 40 = - 160 ( clockwise direction)

iv) at t = 2

E = -400 + 40 = - 360 ( clockwise direction )

8 0

2 years ago