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iragen [17]
3 years ago
13

A MOSFET differs from a JFET mainly because

Engineering
1 answer:
Solnce55 [7]3 years ago
5 0

Answer:

The answer is option

C . the JFET has a PN junction

Explanation:

Not only is option C in the question a dissimilarity between the MOSFET and the JFET we can go on with some more dissimilarities.

1.MOSFET stands for Metal Oxide Silicon Field Effect Transistor or Metal          Oxide Semiconductor Field Effect Transistor.

  (JFET) stands for junction gate field-effect transistor (JFET)  

2. JFET is a three-terminal semiconductor device, whereas  MOFET a four-terminal semiconductor device.  

3. In terms of areas of application of  JFETs are used in low noise applications while MOSFETs,  are used for high noise applications

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Power tools are usually run on one of the three types of power: Compressed air, electricity, or combustion
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Shops should avoid purchasing any material sold in ____________.
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Answer: Aerosol Cans

Explanation: I just did the quiz

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3 years ago
A spur gearset has a module of 6 mm and a velocity ratio of 4. The pinion has 16 teeth. Find the number of teeth on the driven g
levacccp [35]

Answer:

NG=64 teeth

dG=384mm

dP=96mm

C=240mm

Explanation:

step one:

given data

module m=6mm

velocity ratio VR=4

number of teeth of pinion Np=16

<u>Step two:</u>

<u>Required</u>

1. Number of teeth on the driven gear

N_G=N_P*V_R\\\\N_G=16*4\\\\N_G=64

<em>The driven gear has 64 teeth</em>

2.  The pitch diameters

The driven gear diameter

d_G=N_G*m\\\\d_G=64*6\\\\d_G=384

<em>The driven gear diameter is 384mm</em>

The pinion diameter

<em />d_P=N_P*m\\\\d_P=16*6\\\\d_P=96<em />

Pinion diameter is 96mm

3. Theoretical center-to-center distance

C=\frac{d_G+d_P}{2} \\\\C=\frac{384+96}{2} \\\\C=\frac{480}{2}\\\\C=240

The theoretical center-to-center distance is 240mm

5 0
3 years ago
Explain biometric senser.​
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3 years ago
Read 2 more answers
Assign numMatches with the number of elements in userValues that equal matchValue. userValues has NUM_VALS elements. Ex: If user
erica [24]

Answer:

import java.util.Scanner;

public class FindMatchValue {

  public static void main (String [] args) {

     Scanner scnr = new Scanner(System.in);

     final int NUM_VALS = 4;

     int[] userValues = new int[NUM_VALS];

     int i;

     int matchValue;

     int numMatches = -99; // Assign numMatches with 0 before your for loop

     matchValue = scnr.nextInt();

     for (i = 0; i < userValues.length; ++i) {

        userValues[i] = scnr.nextInt();

     }

     /* Your solution goes here */

         numMatches = 0;

     for (i = 0; i < userValues.length; ++i) {

        if(userValues[i] == matchValue) {

                       numMatches++;

                }

     }

     System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);

  }

}

6 0
3 years ago
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