Mr.Haussman has 17 students in his class

In this assignment, you will write a user interface for your calculator using JavaFX. Your graphical user interface (GUI) should

Zolol [24]

Answer:

Kindly note that, you're to replace "at" with shift 2 as the brainly text editor can't take the symbol

Explanation:

import javafx.application.Application;

import javafx.stage.Stage;

import javafx.scene.Group;

import javafx.scene.Scene;

import javafx.scene.layout.VBox;

import javafx.scene.layout.HBox;

import javafx.scene.control.TextField;

import javafx.scene.control.Button;

public class Calculator extends Application {

public static void main(String[] args) {

// TODO Auto-generated method stub

launch(args);

}

"at"Override

public void start(Stage primaryStage) throws Exception {

// TODO Auto-generated method stub

Group root = new Group();

VBox mainBox = new VBox();

HBox inpBox = new HBox();

TextField txtInput = new TextField ();

txtInput.setEditable(false);

txtInput.setStyle("-fx-font: 20 mono-spaced;");

txtInput.setText("0.0");

txtInput.setMinHeight(20);

txtInput.setMinWidth(200);

inpBox.getChildren().add(txtInput);

Scene scene = new Scene(root, 200, 294);

mainBox.getChildren().add(inpBox);

HBox rowOne = new HBox();

Button btn7 = new Button("7");

btn7.setMinWidth(50);

btn7.setMinHeight(50);

Button btn8 = new Button("8");

btn8.setMinWidth(50);

btn8.setMinHeight(50);

Button btn9 = new Button("9");

btn9.setMinWidth(50);

btn9.setMinHeight(50);

Button btnDiv = new Button("/");

btnDiv.setMinWidth(50);

btnDiv.setMinHeight(50);

rowOne.getChildren().addAll(btn7,btn8,btn9,btnDiv);

mainBox.getChildren().add(rowOne);

HBox rowTwo = new HBox();

Button btn4 = new Button("4");

btn4.setMinWidth(50);

btn4.setMinHeight(50);

Button btn5 = new Button("5");

btn5.setMinWidth(50);

btn5.setMinHeight(50);

Button btn6 = new Button("6");

btn6.setMinWidth(50);

btn6.setMinHeight(50);

Button btnMul = new Button("*");

btnMul.setMinWidth(50);

btnMul.setMinHeight(50);

rowTwo.getChildren().addAll(btn4,btn5,btn6,btnMul);

mainBox.getChildren().add(rowTwo);

HBox rowThree = new HBox();

Button btn1 = new Button("1");

btn1.setMinWidth(50);

btn1.setMinHeight(50);

Button btn2 = new Button("2");

btn2.setMinWidth(50);

btn2.setMinHeight(50);

Button btn3 = new Button("3");

btn3.setMinWidth(50);

btn3.setMinHeight(50);

Button btnSub = new Button("-");

btnSub.setMinWidth(50);

btnSub.setMinHeight(50);

rowThree.getChildren().addAll(btn1,btn2,btn3,btnSub);

mainBox.getChildren().add(rowThree);

HBox rowFour = new HBox();

Button btnC = new Button("C");

btnC.setMinWidth(50);

btnC.setMinHeight(50);

Button btn0 = new Button("0");

btn0.setMinWidth(50);

btn0.setMinHeight(50);

Button btnDot = new Button(".");

btnDot.setMinWidth(50);

btnDot.setMinHeight(50);

Button btnAdd = new Button("+");

btnAdd.setMinWidth(50);

btnAdd.setMinHeight(50);

rowFour.getChildren().addAll(btnC,btn0,btnDot,btnAdd);

mainBox.getChildren().add(rowFour);

HBox rowFive = new HBox();

Button btnEq = new Button("=");

btnEq.setMinWidth(200);

btnEq.setMinHeight(50);

rowFive.getChildren().add(btnEq);

mainBox.getChildren().add(rowFive);

root.getChildren().add(mainBox);

primaryStage.setScene(scene);

primaryStage.setTitle("GUI Calculator");

primaryStage.show();

}

4 0

3 years ago

Does anyone know what this is

sammy [17]

Answer:

Looks like mold that got frosted over

Explanation:

4 0

2 years ago

Read 2 more answers

The input and output signals of a system is related by the following equation: fraction numerator d squared y over denominator d

Colt1911 [192]

Answer:

Explanation:

The given equation is :

$\frac{d^{2}y }{dx^{2} } + sin(3y) \frac{dy}{dt} + y = t\frac{df}{dt} + f$

5 0

2 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

c)

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

2 years ago

Which of the following best describes the role of engineers

Fantom [35]

Problem Solvers

Explanation:

Engineers find problems in the world, and then they find solutions for them.

8 0

2 years ago