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rosijanka [135]
1 year ago
7

What is the correct procedure for mounting the m240 on the m122a1 tripod after the pintle is attached to the receiver?

Engineering
1 answer:
cluponka [151]1 year ago
8 0

Using the front sight adjusting tool, loosen (turn counterclockwise) the adjusting screw on the front sight assembly the desired amount. Then tighten (turn clockwise) the opposite side screw on the left exactly the same number of clicks.

<h3>What is a Machine gun ?</h3>

The term "machine gun" refers to a rifled, autoloading, fully automatic weapon intended for continuous direct fire using rifle rounds. Other automatic weapons, such as automatic rifles, are often intended to fire in brief bursts rather than continuously, and are not regarded as real machine guns because of this.

  • Machine guns and other automatic weapons vary in that they fire rounds continually until the shooter lets off of the trigger after pulling it once. Fully automatic guns are uncommon compared to semi-automatic rifles.

Learn more about Machine gun here:

brainly.com/question/1358898

#SPJ4

You might be interested in
It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
denis23 [38]

The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

brainly.com/question/16699941

brainly.com/question/10121263

6 0
2 years ago
Some Tiny College staff employees i s are information technology (IT) personnel. Some IT personnel provide technology support fo
Westkost [7]

Answer:

solution in the picture attached

Explanation:

3 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
3.3 Equation (2) for VCPP is rather difficult to prove at this time. Take it as a challenge to derive it as you learn increasing
podryga [215]

Answer:

For an RC integrator circuit, the input signal is applied to the resistance with the output taken across the capacitor, then VOUT equals VC. As the capacitor is a frequency dependant element, the amount of charge that is established across the plates is equal to the time domain integral of the current. That is it takes a certain amount of time for the capacitor to fully charge as the capacitor can not charge instantaneously only charge exponentially.

Therefore the capacitor current can be written as:

 

his basic equation above of iC = C(dVc/dt) can also be expressed as the instantaneous rate of change of charge, Q with respect to time giving us the following standard equation of: iC = dQ/dt where the charge Q = C x Vc, that is capacitance times voltage.

The rate at which the capacitor charges (or discharges) is directly proportional to the amount of the resistance and capacitance giving the time constant of the circuit. Thus the time constant of a RC integrator circuit is the time interval that equals the product of R and C.

Since capacitance is equal to Q/Vc where electrical charge, Q is the flow of a current (i) over time (t), that is the product of i x t in coulombs, and from Ohms law we know that voltage (V) is equal to i x R, substituting these into the equation for the RC time constant gives:

We have seen here that the RC integrator is basically a series RC low-pass filter circuit which when a step voltage pulse is applied to its input produces an output that is proportional to the integral of its input. This produces a standard equation of: Vo = ∫Vidt where Vi is the signal fed to the integrator and Vo is the integrated output signal.

The integration of the input step function produces an output that resembles a triangular ramp function with an amplitude smaller than that of the original pulse input with the amount of attenuation being determined by the time constant. Thus the shape of the output waveform depends on the relationship between the time constant of the circuit and the frequency (period) of the input pulse.

By connecting two RC integrator circuits together in parallel has the effect of a double integration on the input pulse. The result of this double integration is that the first integrator circuit converts the step voltage pulse into a triangular waveform and the second integrator circuit converts the triangular waveform shape by rounding off the points of the triangular waveform producing a sine wave output waveform with a greatly reduced amplitude.

RC Differentiator

For a passive RC differentiator circuit, the input is connected to a capacitor while the output voltage is taken from across a resistance being the exact opposite to the RC Integrator Circuit.

A passive RC differentiator is nothing more than a capacitance in series with a resistance, that is a frequency dependentTherefore the capacitor current can be written as:

 

 

device which has reactance in series with a fixed resistance (the opposite to an integrator). Just like the integrator circuit, the output voltage depends on the circuits RC time constant and input frequency.

Thus at low input frequencies the reactance, XC of the capacitor is high blocking any d.c. voltage or slowly varying input signals. While at high input frequencies the capacitors reactance is low allowing rapidly varying pulses to pass directly from the input to the output.

This is because the ratio of the capacitive reactance (XC) to resistance (R) is different for different frequencies and the lower the frequency the less output. So for a given time constant, as the frequency of the input pulses increases, the output pulses more and more resemble the input pulses in shape.

We saw this effect in our tutorial about Passive High Pass Filters and if the input signal is a sine wave, an rc differentiator will simply act as a simple high pass filter (HPF) with a cut-off or corner frequency that corresponds to the RC time constant (tau, τ) of the series network.

Thus when fed with a pure sine wave an RC differentiator circuit acts as a simple passive high pass filter due to the standard capacitive reactance formula of XC = 1/(2πƒC).

But a simple RC network can also be configured to perform differentiation of the input signal. We know from previous tutorials that the current through a capacitor is a complex exponential given by: iC = C(dVc/dt). The rate at which the capacitor charges (or discharges) is directly proportional to the amount of resistance and capacitance giving the time constant of the circuit. Thus the time constant of a RC differentiator circuit is the time interval that equals the product of R and C. Consider the basic RC series circuit below.

Explanation:

3 0
3 years ago
Guess my birthday it’s in September you should be good with it
Paraphin [41]

Answer:

spetember 7th

Explanation:

5 0
3 years ago
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