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Y_Kistochka [10]
3 years ago
5

What are cars manufactured with today that allows for quick stopping without the brakes locking up?

Engineering
1 answer:
stiks02 [169]3 years ago
5 0

Answer:

ABS

Explanation:

an Anti Lock Breaking system will not lock up your breaks even when used in a tight situation. This helps inexperienced drivers with no breaking technique break proprly.

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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
Explain the purpose of wrench plzzz
Rus_ich [418]

Answer:

Wrenches are made in various shapes and sizes and are used for gripping, fastening, turning, tightening and loosening things like pipes, pipe fittings, nuts and bolts. There are basically two major kinds of wrenches: Pipe wrenches used in plumbing for gripping round (cylindrical) things.

4 0
2 years ago
Read 2 more answers
The diffusion coefficients for species A in metal B are given at two temperatures:
Kruka [31]

Answer:

a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s

Explanation:

Diffusion is governed by Arrhenius equation

D = D_0e^{\frac{-Q_d}{RT} }

I will be using R in the equation instead of k_b as the problem asks for molar activation energy

I will be using

R = 8.314\ J/mol*K

and

°C + 273 = K

here, adjust your precision as neccessary

Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm

So:

ln(6.69*10^{-17})=ln(D_0) -\frac{Q_d}{R*(1030+273)}

and

ln(6.56*10^{-16}) = ln(D_0) -\frac{Q_d}{R*(1290+273)}

You might notice that these equations have the form of  

d=y-ax

You can solve this equation system easily using calculator, and you will eventually get

D_0 =6.11*10^{-11}\ m^2/s\\ Q_d=1.49 *10^3\ J/mol

After you got those 2 parameters, the rest is easy, you can just plug them all   including the given temperature of 1180°C into the Arrhenius equation

6.11*10^{-11}e^{\frac{149\ 000}{8.143*(1180+273)}

And you should get D = 2.76*10^-16 m^/s as an answer for c)

5 0
3 years ago
An electric power plant uses solid waste for fuel in the production of electricity. The cost Y in dollars per hour to produce el
andrew-mc [135]

Answer:

15.64 MW

Explanation:

The computation of value of X that gives maximum profit is shown below:-

Profit = Revenue - Cost

= 15x - 0.2x 2 - 12 - 0.3x - 0.27x 2

= 14.7x - .47x^2 - 12

After solving the above equation we will get maximum differentiate  for profit that is

14.7 - 0.94x = 0

So,

x = 15.64 MW

Therefore for computing the value of X that gives maximum profit we simply solve the above equation.

8 0
3 years ago
What is the maximal coefficient of performance of a refrigerator which cools down 10 kg of water (and then ice) to -6C. Upper he
inysia [295]

Given:

Temperature of water, T_{1} = -6^{\circ}C =273 +(-6) =267 K

Temperature surrounding refrigerator, T_{2} = 21^{\circ}C =273 + 21 =294 K

Specific heat given for water, C_{w} = 4.19 KJ/kg/K

Specific heat given for ice, C_{ice} = 2.1 KJ/kg/K

Latent heat of fusion,  L_{fusion} = 335KJ/kg

Solution:

Coefficient of Performance (COP) for refrigerator is given by:

Max COP_{refrigerator} = \frac{T_{2}}{T_{2} - T_{1}}

= \frac{267}{294 - 267} = 9.89

Coefficient of Performance (COP) for heat pump is given by:

Max COP_{heat pump} = \frac{T_{1}}{T_{2} - T_{1}}\frac{294}{294 - 267} = 10.89

6 0
3 years ago
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