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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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A dentist’s drill starts from rest. After 1.46 sof constant angular acceleration, it turns at arate of 27000 rev/min.Find the dr

Black_prince [1.1K]

Answer:

616.3 rad/s²

Explanation:

Given that

t= 1.46 s

Initial angular velocity ,ωi = 0 rad/s

Final angular velocity ωf= 27000 rev/min

Angular speed in the rad/s given as

$\omega_f=\dfrac{2\pi N}{60}\ rad/s$

Now by putting the values

$\omega_f=\dfrac{2\times 27000}{60}\ rad/s$

ωf=900 rad/s

We know that (if acceleration is constant)

ωf=ωi + α t

α=Angular acceleration

900 = 0 + α x 1.46

$\alpha=\dfrac{900}{1.46}\ rad/s^2\\\alpha=616.43\ rad/s^2$

Therefore the acceleration will be 616.3 rad/s²

4 0

3 years ago

Urippe Went to start his lawnmower and was having trouble. At the same moment, Uwi was starting her identical lawnmower in the y

Ivan

Answer:

As each mower presumably needs the same torque to start, and torque is a product of force and moment arm, the longer moment arm of 10.42 cm on Uwi's mower means lower force is required when compared to Urippe's shorter moment arm of 1.35 cm

350 rev/min = 350(2π) / 60 = 36.652 rad/s

36.652 rad/s / 0.294 s = 124.66... 125 rad/s²

a = αR = 125(0.1042) = 12.990... 13 m/s²

a = αR = 125(0.0135) = 1.68299... 1.7 m/s²

I am GUESSING that we are supposed to model these mowers as a uniform disk

τ = Iα

FR = (½mr²)α

F = mr²α/2R

Urippe's pull = (3.56)(0.2041²)(124.66) / (2(0.0135)) = 702.008... 702 N

Usi's pull = (3.56)(0.2041²)(124.66) / (2(0.1042)) = 90.9511...91.0 N

L = Iω = (½(3.56)(0.2041²))36.652 = 2.71771...2.72 kg•m²/s down

using the right hand rule

8 0

3 years ago

A student walks 3 north and 4 m west. The magnitude of the resultant displacement for the student is

evablogger [386]

Answer:

Explanation:

Using Pythagoras theorem,

a^2+ b^2=c^2

3^2+4^2=c^2

25=c^2

√(25)=c

5m=c

6 0

2 years ago

Five lamp, each labbled "6V,3W" are operated at normal brightness. What is the total energy supplied to the lamps in five second

REY [17]

Answer:

E = 75 J

Explanation:

First, we will calculate the total power consumed by the five lamps:

$Total\ Power = P = (5)(Power\ of\ one\ lamp)\\P = (5)(3\ W)\\P = 15\ W$

Now, the energy supply can be calculated as follows:

$E = Pt$

where,

E = Energy = ?

t = time = 5 s

Therefore,

E = (15 W)(5 s)

E = 75 J

8 0

2 years ago

In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the gr

NikAS [45]

A) 2.64t

B) 2.64h

C) 2.64D

Explanation:

The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The time of flight of a projectile can be found from the equations of motion, and it is found to be

$t=\frac{2u sin \theta}{g}$

where

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the time of flight is $t$ .

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:

$t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t$

The maximum height reached by a projectile can be also found using the equations of motion, and it is given by

$h=\frac{u^2 sin^2\theta}{2g}$

where

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the maximum height is $h$ .

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:

$h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h$

The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by

$D=\frac{u^2 sin(2\theta)}{g}$

where:

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the horizontal distance covered is D.

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:

$D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D$

7 0

3 years ago