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Olenka [21]
2 years ago
13

A furniture manufacturer purchases a drill press machine enabled with 5G and edge computing capabilities to keep the machine ope

rators safe when at work.How could these capabilities help improve safety of the operators?
Engineering
1 answer:
Andreas93 [3]2 years ago
8 0

A lot of manufacturer often uses 5G machines. How these capabilities could help improve safety of the operators is that it does includes an emergency switch for the operator so that one can manually shut off when needed.

<h3>Edge computing with 5G</h3>

  • The edge computing along with 5G network and IoT devices can help put together  different safety features and limitations and on can use them to known the unsafe action and also data can be communicated.

Edge computing when use with 5G produces good opportunities in all industry. It is known to help bring computation and data storage close to where data is been produced and it enable good data control, reduced costs, etc.

Learn more about 5G network from

brainly.com/question/24664177

You might be interested in
How might an operations manager alter operations to meet customer demand? Name at least two ways.
Citrus2011 [14]
One way is manager changes itself and the other one is the same thing i think.
4 0
2 years ago
A water-filled manometer is used to measure the pressure in an air-filled tank. One leg of the manometer is open to atmosphere.
ddd [48]

Answer:

P = 150.335\,kPa (Option B)

Explanation:

The absolute pressure of the air-filled tank is:

P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{kg}{m^{3}} \right)\cdot (5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)

P = 150.335\,kPa

4 0
3 years ago
Write a program that prompts for a line of text and then transforms the text based on chosen actions. Actions include reversing
nlexa [21]

Answer:

public class TextConverterDemo

{

//Method definition of action1337

public static String action1337(String current)

{

//Replace each L or l with a 1 (numeral one)

 current = current.replace('L', '1');

 current = current.replace('l', '1');

 

 //Replace each E or e with a 3 (numeral three)

 current = current.replace('E', '3');

 current = current.replace('e', '3');

 //Replace each T or t with a 7 (numeral seven)

 current = current.replace('T', '7');

 current = current.replace('t', '7');

 //Replace each O or o with a 0 (numeral zero)

 current = current.replace('O', '0');

 current = current.replace('o', '0');

 

//Replace each S or s with a $ (dollar sign)

 current = current.replace('S', '$');

 current = current.replace('s', '$');

 return current;

}

//Method definition of actionReverse

//This method is used to reverses the order of

//characters in the current string

public static String actionReverse(String current)

{

 //Create a StringBuilder's object

 StringBuilder originalStr = new StringBuilder();

 //Append the original string to the StribgBuilder's object

 originalStr.append(current);

 //Use reverse method to reverse the original string

 originalStr = originalStr.reverse();

 

 //return the string in reversed order

 return originalStr.toString();

}

//Method definition of main

public static void main(String[] args)

{

    //Declare variables

 String input, action;

 

 //Prompt the input message

 System.out.println("Welcome to the Text Converter.");

 System.out.println("Available Actions:");

 System.out.println("\t1337) convert to 1337-speak");

 System.out.println("\trev) reverse the string");

 System.out.print("Please enter a string: ");

   

 //Create a Scanner class's object

 Scanner scn = new Scanner(System.in);

 

 //Read input from the user

 input = scn.nextLine();

 do

 {

  /*Based on the action the user chooses, call the appropriate

   * action method. If an unrecognized action is entered then

   * the message "Unrecognized action." should be shown on a

   * line by itself and then the user is prompted again just

   * as they were when an action was performed.

   * */

  System.out.print("Action (1337, rev, quit): ");

  action = scn.nextLine();

  if (action.equals("1337"))

  {

   input = action1337(input);

   System.out.println(input);

  } else if (action.equals("rev"))

  {

   input = actionReverse(input);

   System.out.println(input);

  } else if (!action.equals("quit"))

  {

   System.out.println("Unrecognized action.");

  }

 } while (!action.equals("quit"));

 System.out.println("See you next time!");

 scn.close();

}

}

7 0
3 years ago
Please write the following code in Python 3. Also please show all output(s) and share your code.
maksim [4K]

Answer:

sum2 = 0

counter = 0

lst = [65, 78, 21, 33]

while counter < len(lst):

   sum2 = sum2 + lst[counter]

   counter += 1

Explanation:

The counter variable is initialized to control the while loop and access the numbers in <em>lst</em>

While there are numbers in the <em>lst</em>,  loop through <em>lst</em>

Add the numbers in <em>lst</em> to the sum2

Increment <em>counter</em> by 1 after each iteration

6 0
3 years ago
A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength i
Mama L [17]

Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter d_{o} = 12.8 mm

Final diameter d_{f} = 10.7

Engineering stress  \alpha _{E} = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4(d_{o} )²  ; Ag = π/4(d_{f} )²

% = π/4 [ ( (d_{f} )² - (d_{o} )²) / ( π/4  (d_{o} )²) ]

= ( (d_{f} )² - (d_{o} )²) / (d_{o} )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress  \alpha _{T} = \alpha _{E} ( 1 +  E_{E} )

E_{E}  is engineering strain

E_{E}  = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49  

E_{E} = 0.431

so we substitute the value of E_{E}  into our initial equation;

True stress  \alpha _{T} = 460 ( 1 +  0.431)

True stress  \alpha _{T} = 460 (1.431)

True stress  \alpha _{T} = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

6 0
2 years ago
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