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kolezko [41]
3 years ago
13

What is the operation of an earth leakage tester

Computers and Technology
1 answer:
Vladimir [108]3 years ago
5 0

In any electrical installation, some current will flow through the protective ground conductor to ground. This is usually called leakage current. Leakage current most commonly flows in the insulation surrounding conductors and in the filters protecting electronic equipment around the home or office. So what's the problem? On circuits protected by GFCIs (Ground Fault Current Interrupters), leakage current can cause unnecessary and intermittent tripping. In extreme cases, it can cause a rise in voltage on accessible conductive parts. 


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A method a. may have zero or more parameters b. never has parameter variables c. must have at least two parameter variables d. m
Lynna [10]

Answer:

The answer is "Option a"

Explanation:

A method is a technique, that associated with both a message and an object. It includes information, behavior, and actions of an interface defining, how the object can be used, and wrong choices can be described as follows:

  • In option b, It includes parameters in the method.
  • In option c, It contains a parameter, that may be one or more.
  • In option d, It contains one parameter also.

5 0
3 years ago
How do you find the exterior angle measure of any regular polygon? (The degree in which to turn the turtle to draw a shape)
Diano4ka-milaya [45]

Answer:

360/number of sides

Explanation:

all the exterior angles add up to 360°

so to find each angle mesure, divide the 360 by the number of sides the figure has.

7 0
3 years ago
Write a Java test program, named TestGuitar, to create 3different Guitars representing each representing a unique test case and
alina1380 [7]

Answer:

Explanation:

//Guitar.java

import java.awt.Color;

import java.lang.reflect.Field;

import java.util.Random;

public class Guitar

     /**

     * these two fields are used to generate random note and duration

     */

     static char[] validNotes = 'A', 'B', 'C', 'D', 'E', 'F', 'G' ;

     static double[] validDuration = 0.25, 0.5, 1, 2, 4 ;

     /**

     * basic guitar attributes

     */

     private int numStrings;

     private double guitarLength;

     private String guitarManufacturer;

     private Color guitarColor;

     public Guitar()

           /**

           * default constructor

           */

           numStrings = 6;

           guitarLength = 28.2;

           guitarManufacturer = Gibson;

           guitarColor = Color.RED;

     

     public Guitar(int numStrings, double guitarLength,

                 String guitarManufacturer, Color guitarColor)

           /**

           * parameterized constructor

           */

           this.numStrings = numStrings;

           this.guitarLength = guitarLength;

           this.guitarManufacturer = guitarManufacturer;

           this.guitarColor = guitarColor;

     

     /**

     * required getters and setters

     */

     public static char[] getValidNotes()

           return validNotes;

     public static void setValidNotes(char[] validNotes)

           Guitar.validNotes = validNotes;      

     public static double[] getValidDuration()

           return validDuration;

     public

4 0
3 years ago
Kevin is working in the Tasks folder of his Outlook account. Part of his computer screen is shown below.
klasskru [66]

Answer:

D

Explanation:

7 0
3 years ago
Given two 2x3 (2 rows, 3 columns) arrays of integer , x1 and x2, both already initialized , two integer variables , i and j, and
Gnesinka [82]
JAVA programming was employed...

What we have so far:
* Two 2x3 (2 rows and 3 columns) arrays. x1[i][j] (first 2x3 array) and x2[i][j] (second 2x3 array) .
* Let i = row and j = coulumn.
* A boolean vaiable, x1rules

Solution:

for(int i=0; i<2; i++)
{
   for(int j=0; j<3; j++)
   {
      x1[i][j] = num.nextInt();
   }
}// End of Array 1, x1.

for(int i=0; i<2; i++)
{
   for(int j=0; j<3; j++)
   {
      x2[i][j] = num.nextInt();
   }
}//End of Array 2, x2
This should check if all the elements in x1 is greater than x2:

x1rules = false;
if(x1[0][0]>x2[0][0] && x1[0][1]>x2[0][1] && x1[0][2]>x2[0][2] && x1[1][0]>x2[1][0] && x1[1][1]>x2[1][1] && x1[1][2]>x2[1][2])
{
   x1rules = true;
   system.out.print(x1rules);
}
else
{
   system.out.print(x1rules);
}//Conditional Statement
7 0
3 years ago
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