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wolverine [178]
3 years ago
7

Tech A says that in some cases, the electronic brake control module can be programmed with a new tire size to restore proper ele

ctronic brake control performance. Tech B says that there could be an electronic brake control module available for installation that matches a new tire size. Who is correct?A. Tech AB. Tech BC. Both Techs A and BD. Neither Tech A nor B
Engineering
1 answer:
Vlad [161]3 years ago
3 0

Answer:

Both Techs A and B

Explanation:

Electronic braking systems are controlled by the electronic brake control module. It is a microprocessor that processes information from wheel-speed sensors and the hydraulic brake system to determine when to release braking pressure at a wheel that's about to lock up and start skidding  and activates the anti lock braking system or traction system when it detects it is necessary.

Some electronic brake control modules can be programmed to the size of the vehicle's new tires to restore proper electronic brake control performance. While others may require replacing the module to match the module's programming to the installed tire size. So, both technicians A and B are correct.

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What are some advantages of making electronic components like transistors increasingly smaller?
Stells [14]

Answer:

Dr. Engelbart, who would later help develop the computer mouse and other personal computing technologies, theorized that as electronic circuits were made smaller, their components would get faster, require less power and become cheaper to produce — all at an accelerating pace

4 0
2 years ago
Explain the working of McLeod gauge for measurement of very low pressure.
AfilCa [17]

Explanation:

McLeod gauge:

  It is used to measure very low pressure of gas.It measure gas pressure by the help of mercury(Hg).it measure absolute pressure of gas.

McLeod gauge on the principle of Boyle's law.Boyle's law state that ,at constant temperature the pressure of gas is directly proportional to its  volume.From Boyle's law

P_1V_1=P_2V_2

A sample of gas is taken from vacuum then its pressure is measured by help  Hg.

4 0
3 years ago
Write a GUI-based program that plays a guess-the-number game in which the roles of the computer and the user are the reverse of
AURORKA [14]

Answer:

import javax.swing.*;

import java.awt.*;

import java.util.Random;

import java.awt.event.*;

public class Guess extends JFrame

{

   private static final long serialVersionUID = 1L;

   private JButton newGame;

   private JButton enter;

   private JButton exit;

   private JTextField guess;

   private JLabel initialTextLabel;

   private JLabel enterLabel;

   private JLabel userMessageLabel;

   private int randNum;

   private int userInput;

   private int maxtries = 0;

   public Guess()

   {

       super("Guessing Game");

       newGame = new JButton("New Game");

       exit = new JButton("Exit Game");

       enter = new JButton("Enter");

       guess = new JTextField(4);

       initialTextLabel = new JLabel("I'm thinking of a number between 1 and 100. Guess it!");

       enterLabel = new JLabel("Enter your guess.");

       userMessageLabel = new JLabel("");

       randNum = new Random().nextInt(100) + 1;

       setLayout(new FlowLayout());

       add(initialTextLabel);

       add(enterLabel);

       add(guess);

       add(newGame);

       add(enter);

       add(exit);

       add(userMessageLabel);

   

       setSize(500, 300);

       addWindowListener(new WindowAdapter()

       {

           public void windowClosing(WindowEvent e)

           {

               System.exit(0);

           }

       });

       newGameButtonHandler nghandler = new newGameButtonHandler();

       newGame.addActionListener(nghandler);

       ExitButtonHandler exithandler = new ExitButtonHandler();

       exit.addActionListener(exithandler);

       enterButtonHandler enterhandler = new enterButtonHandler();

       enter.addActionListener(enterhandler);

   }

   class newGameButtonHandler implements ActionListener

   {

       public void actionPerformed(ActionEvent e)

       {

           setBackground(Color.ORANGE);

           guess.setEnabled(true);

           guess.setText("");

           enter.setEnabled(true);

           maxtries = 0;

           userMessageLabel.setText("");

           randNum = new Random().nextInt(100) + 1;

       }

   }

   class ExitButtonHandler implements ActionListener

   {

       public void actionPerformed(ActionEvent e)

       {

           System.exit(0);

       }

   }

   class enterButtonHandler implements ActionListener

   {

       public void actionPerformed(ActionEvent e)

       {

           userInput = Integer.parseInt(guess.getText());

           checkGuess(randNum);

      if(userInput > 100 )

          {

                               userMessageLabel.setText("invalid entry");

          }

       }

   }

   public void checkGuess(int randomNumber)

   {

       maxtries++;

     if(maxtries==10){

           userMessageLabel.setText("You Lose!!");

           guess.setEnabled(false);

           enter.setEnabled(false);

         

       }else if (userInput == randomNumber)

           {

               userMessageLabel.setText("Correct !");

           }

       else if (userInput > randomNumber)

           {

               userMessageLabel.setText("Too high");

           }

       else if (userInput < randomNumber)

           {

               userMessageLabel.setText("Too Low");

           }

   }

   public static void main(String[] args)

   {

       Guess game = new Guess();

       game.setVisible(true);

   }

}

8 0
3 years ago
Define Electromechanical systems.
dedylja [7]

Answer:

Electromechanical systems or devices are systems or devices that involves the interaction between electrical systems and mechanical systems in which the motion of mechanical parts is converted to electrical energy or made to interact with energy or in which electrical energy is converted to mechanical energy or interacts with a moving mechanical system

Therefore;

Electromechanical systems convert <u>electrical energy</u> input into a <u>mechanical energy</u> output

Explanation:

3 0
3 years ago
A disk is rotating around an axis located at its center. The angular velocity is 0.5 rad/s. The radius of the disk is 0.4 m. Wha
dimaraw [331]

Answer:

0.2 m/s

Explanation:

The velocity of a point on the edge of a disk rotating disk can be calculated as:

v=\omega*r

Where \omega is the angular velocity and r the radius of the disk. This leads to:

v=0.5\,rad/s\,*\,0.4\,m=0.2\,m/s

4 0
3 years ago
Read 2 more answers
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