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melisa1 [442]
3 years ago
15

Imagine that you have been troubleshooting a network cable (either copper or fiber optic), and you determine that there is a bre

ak in the cable. What would you use to identify exactly where the break exists?
Computers and Technology
1 answer:
Gnoma [55]3 years ago
8 0

Answer: Time-domain reflectometer for copper cabling                      Optical time domain reflectometer for fiber-optic cabling

Explanation:

Time-domain reflectometer(TDR) is defined as the device that is utilized for locating and identifying any breakage and faults in circuit board, wires,connectors,metal cables or any path of electricity.

  • It locates the discontinuation in metal wires such as coaxial cables with the help of time domain reflectometry.

Optical Time Domain Reflectometer (OTDR) is the electronic device that searches and locates faults and break in fiber optic cable.

  • This specifically works for identifying disruption, loss and measurement of cable.
  • Thus, for identifying the exact location of breakage in copper cable,time domain reflectometer can be used and for fiber optical cable, optical time domain can be used while troubleshooting.
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Answer is in the provided screenshot!

Explanation:

Steps required to solve this problem:

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Design pseudocode for a program that will permit a user to store exactly 50 numbers in an array. Create an array big enough to h
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Answer:

#include<stdio.h>

//declare a named constant

#define MAX 50

int main()

{

   //declare the array

   int a[MAX],i;

   //for loop to access the elements from user

   for(i=0;i<MAX;i++)

   {

      printf("\n Enter a number to a[%d]",i+1);

      scanf("%d",&a[i]);

  }

  //display the input elements

  printf("\n The array elements are :");

  for(i=0;i<=MAX;i++)

  printf(" %d ",a[i]);

}

Explanation:

PSEUDOCODE INPUTARRAY(A[MAX])

REPEAT FOR I EQUALS TO 1 TO MAX

PRINT “Enter a number to A[I]”

READ A[I]

[END OF LOOP]

REPEAT FOR I EQUALS TO 1 TO MAX

PRINT A[I]

[END OF LOOP]

RETURN

ALGORITHM

ALGORITHM PRINTARRAY(A[MAX])

REPEAT FOR I<=1 TO MAX

PRINT “Enter a number”

INPUT A[I]

[END OF LOOP]

REPEAT FOR I<=1 TO MAX

PRINT A[I]

[END OF LOOP]

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