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alukav5142 [94]
2 years ago
7

What are two ways you can see that there is a smaller load going to the base of the transistor?

Engineering
1 answer:
Lyrx [107]2 years ago
7 0

Answer:

<h3><em>Transistor switches can be used to switch a low voltage DC device (e.g. LED’s) ON or OFF by using a transistor in its saturated or cut-off </em><em>state</em></h3>

  1. <em>. Cut-off </em><em>Region</em>

<em>Here the operating conditions of the transistor are zero input base current ( IB ), zero output collector current ( IC ) and maximum collector voltage ( VCE ) which results in a large depletion layer and no current flowing through the device. Therefore the transistor is switched “Fully-OFF”.</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>2</em><em>.</em><em>Saturation </em><em>Region</em>

<em>Here the transistor will be biased so that the maximum amount of base current is applied, resulting in maximum collector current resulting in the minimum collector emitter voltage drop which results in the depletion layer being as small as possible and maximum current flowing through the transistor. Therefore the transistor is switched “Fully-ON”.</em>

Explanation:

hope it helps

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Answer:

Life critical systems or computer systems that cause harm to people

4 0
3 years ago
Disk A has a mass of 8 kg and an initial angular velocity of 360 rpm clockwise; disk B has a mass of 3.5 kg and is initially at
garri49 [273]

Answer:

A. αa= 9.375 rad/s^2, αb= 28.57 rad/s^2

B. ωa=251rpm, ωb=333rpm

Explanation:

Mass of disk A, m1= 8kg

Mass of disk B, m2= 3.5kg

The initial angular velocity of disk A= 360rpm

The horizontal force applied = 20N

The coefficient of friction μk = 0.15

While slipping occurs, a frictional force is applied to disk A and disk B

T= 1/2Mar^2a

T=1/2 (8kg) (0.08)^2

T= 0.0256kg-m^2

N=P=20N

F= μN

F= 0.15 × 20

F=3N

We have

Summation Ma= Summation(Ma) eff

Fra= Iaαa

(3N)(0.08m)= (0.0256kg-m^2)α

αa= 9.375 rad/s^2

The angular acceleration at disk A is αa= 9.375 rad/s^2 is acting in the anti-clockwise direction.

For Disk B,

T= 1/2Mar^2a

T= 1/2(3.5) (0.06)^2

=0.0063kg-m^2

We have,

Summation Mb= Summation(Mb) eff

Frb= Ibαb

(3N)(0.06m)= (0.0063kg-m^2) α

αb= 28.57 rad/s^2

B) ( ωa)o= 360rpm(2 pi/60)

= 1 pi rad/s

The disk will stop sliding where

ωara=ωbrb

(ωao-at)ra=αtr

(12pi-9.375t) (0.08)=28.57t(0.06)

(37.704-9.375t)(0.08)= 1.7142t

3.01632-0.75t= 1.7142t

t=1.22s

Now,

ωa=(ωa)o- t

12pi - 9.375(1.22)

37.704-11.4375

=26.267 rad/s

26.267× (60/2pi)

= 250.80

251rpm

The angular velocity at a, ω= 251rpm

Now,

ωb= αbt

=28.57(1.22)

=34.856rad/s

34.856rad/s × (60/2pi)

=332.807

= 333rpm

Therefore the angular velocity at b ω=333rpm

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3 years ago
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Can the United States defeat Iranian forces
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Answer:

yes for sure

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Problem 1 (10 points) In the first homework you were instructed to design the mechanical components of an oscillating compact di
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Answer:

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