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Brilliant_brown [7]
3 years ago
14

7. Which of the following accurately describes junction diodes? A. In a forward-biased setup, large numbers of charge carriers w

ill be pulled across the junction and result in a large current. B. Free electrons are concentrated in the P-type material and holes form in the N-type material. C. Only a small current will flow through the diode in the forward-biased setup. D. The negative terminal of the source of voltage is connected to the N-type material within a reverse-biased setup.
Physics
1 answer:
forsale [732]3 years ago
8 0

Answer: A. In a forward-biased setup, large numbers of charge carriers will be pulled across the junction and result in a large current.

Explanation:

A two terminal device which conducts current only in one-direction is known as diode. In forward bias, the p type is connected to the positive terminal and n-type is connected to the negative terminal of the source of the voltage. Thus, in forward bias, large numbers of charge carriers will be pulled across the junction resulting in a large current.

Therefore, correct option is A.

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If the period of a simple pendulum is T and you increase its length so that it is 4 times longer, what will the new period be?
goldfiish [28.3K]

Answer:

T' = 2T

Explanation:

The time period of a simple pendulum is given by the relation as follows :

T=2\pi \sqrt{\dfrac{l}{g}}

l is length of the pendulum

g is acceleration due to gravity

If the length is increased four time, new length is l' = 4l

So,

New time period is :

T'=2\pi \sqrt{\dfrac{l'}{g}}\\\\T'=2\pi \sqrt{\dfrac{4l}{g}}\\\\T'=2\times 2\pi \sqrt{\dfrac{l}{g}}\\\\T'=2\times T

So, the new time period is 2 times of the initial time period.

5 0
3 years ago
What does democratized knowledge mean?
lutik1710 [3]
I think it’s A I got that off the information I got in the internet
8 0
3 years ago
Read 2 more answers
Convert 1 second into solar day.
otez555 [7]

DIVIDE 1 BY 86400 TO CONVERT 1 SECOND INTO SOLAR DAY.

6 0
3 years ago
Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of
Alchen [17]

Answer:

\theta=10.20^{\circ}  

\Delta l=0.10 ft    

Explanation:

First of all, we analyze the system of blocks before starting to move.

\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0  

\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0

11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0

11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0

11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0  

16sin(\theta)-2.91cos(\theta)=0  

tan(\theta)=0.18  

\theta=arctan(0.18)  

\theta=10.20^{\circ}  

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.  

P_{A}sin(\theta)-F_{fA}-F_{spring}=0

Where:

F_{spring} = k\Delta l=2.1\Delta l

P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0

\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}

\Delta l=0.10 ft    

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

4 0
4 years ago
At the presentation ceremony, a championship bowler is presented a 1.64-kg trophy which he holds at arm's length, a distance of
solong [7]

To solve the problem it is necessary to take into account the concepts of the kinetic equations for the description of the torque at the rate of force and distance.

By definition the torque is given by,

\tau = F*d

where,

F= Force

d = Distance

For the problem in question the mass of the trophy is 1.64Kg and the distance of the tropeo to the board (the shoulder) is 0.655m

PART A) For part A, the torque with the given mass and the stipulated torque in the horizontal plane must be calculated as well,

\tau = F*d

For Newton's second law

\tau = mg*d

\tau = 1.64*9.81*0.655

\tau = 10.5Nm

PART B) For part B there is an angle of 26 degrees with respect to the horizontal, therefore to know the net torque it is necessary to know the horizontal component to the formed angle, that is,

\tau = F*dcos\theta

\tau = mgdcos\theta

\tau = 1.64*9.81*0.655*cos26

\tau = 9.471Nm

3 0
4 years ago
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