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AVprozaik [17]
3 years ago
9

Joe, a technician, is attempting to connect two hubs to add a new segment to his local network. He uses one of his CAT5 patch ca

bles to connect them; however, he is unable to reach the new network segment from his workstation. He can only connect to it from a workstation within that segment. Which of the following is MOST likely the problem?
A. One of the hubs is defective.
B. The new hub is powered down.
C. The patch cable needs to be a CAT6 patch cable.
D. The technician used a straight-through cable.
Engineering
1 answer:
mart [117]3 years ago
3 0

Answer:

Option D. is correct

Explanation:

Joe uses one of his CAT5 patch cables to connect two hubs to add a new segment to his local network. As he can only connect to it from a workstation within that segment,  he is not able to reach the new network segment from his workstation.

The most problem is that the technician used a straight-through cable.

Option D. is correct.

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Along with refining craft skills another way to increase the odds for career advancement is to
Xelga [282]

The acquisition of additional certifications with a personal refined craft skills can increase the odds for career advancemen.

<h3>What is a career advancement?</h3>

An advancement is achieved in a career if a professional use their skill sets, determination or perserverance to achieve new career height.

An example of a career advancement is when an employee progresses from entry-level position to management and transits from an occupation to another.

Therefore, the Option A is correct.

Read more about career advancement

<em>brainly.com/question/7053706</em>

7 0
2 years ago
A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri
babymother [125]

Answer:

Tension in cable BE= 196.2 N

Reactions A and D both are  73.575 N

Explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

T_{BE}-W=0 hence

T_{BE}=W=20*9.81=196.2 N

Therefore, tension in the cable, T_{BE}=196.2 N

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0

24.525-39.24+0.2D_x=0

D_x=73.575 N

Similarly,

A_x-D_y=0

A_x=73.575 N

Therefore, both reactions at A and D are 73.575 N

7 0
3 years ago
What material property would still cause strain in a strain gauge that is positionedperpendicular to the direction of force if i
svetlana [45]

Answer:

oof

Explanation:

I don't know but please don't report me

I am trying to do a challenge

Thank you-

If you don't report me!

5 0
3 years ago
two pints of ethyl ether evaporate over a period of 1,5 hours. what is the air flow necessary to remain at 10% or less of ethyl
lesya692 [45]

The air flow necessary to remain at the lower explosive level is 4515. 04cfm

<h3>How to solve for the rate of air flow</h3>

First we have to find the rate of emission. This is solved as

2pints/1.5 x 1min

= 2/1.5x60

We have the following details

SG = 0.71

LEL = 1.9%

B = 10% = 0.1 a constant

The molecular weight is given as 74.12

Then we would have Q as

403*100*0.2222 / 74.12 * 0.71 * 0.1

= Q = 4515. 04

Hence we can conclude that the air flow necessary to remain at the lower explosive level is 4515. 04cfm

Read more on the rate of air flow on brainly.com/question/13289839

#SPJ1

7 0
2 years ago
java Write a program that simulates tossing a coin. Prompt the user for how many times to toss the coin. Code a method with no p
max2010maxim [7]

Answer:

The solution code is written in Java.

  1. public class Main {
  2.    public static void main(String[] args) {
  3.        Scanner inNum = new Scanner(System.in);
  4.        System.out.print("Enter number of toss: ");
  5.        int num = inNum.nextInt();
  6.        for(int i=0; i < num; i++){
  7.            System.out.println(toss());
  8.        }
  9.    }
  10.    public static String toss(){
  11.        String option[] = {"heads", "tails"};
  12.        Random rand = new Random();
  13.        return option[rand.nextInt(2)];
  14.    }
  15. }

Explanation:

Firstly, we create a function <em>toss()</em> with no parameter but will return a string (Line 14). Within the function body, create an option array with two elements, "heads" and "tails" (Line 15). Next create a Random object (Line 16) and use <em>nextInt()</em> method to get random value either 0 or 1. Please note we need to pass the value of 2 into <em>nextInx() </em>method to ensure the random value generated is either 0 or 1.  We use this generate random value as an index of <em>option </em>array and return either "heads" or "tails" as output (Line 17).

In the main program, we create Scanner object and use it to prompt user to input an number for how many times to toss the coin (Line 6 - 7). Next, we use the input num to control how many times a for loop should run (Line 9). In each round of the loop, call the function <em>toss() </em>and print the output to terminal (Line 10).  

4 0
4 years ago
Read 2 more answers
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