2. Describe the kinds of pressures that might lead<br> to unethical behavior.

A horizontal 2-m-diameter conduit is half filled with a liquid (SG=1.6 ) and is capped at both ends with plane vertical surfaces

luda_lava [24]

Answer:

Resultant force = 639 kN and it acts at 0.99m from the bottom of the conduit

Explanation:

The pressure is given as 200 KPa and the specific gravity of the liquid is 1.6.

The resultant force acting on the vertical plate, Ft, is equivalent to the sum of the resultant force as a result of pressurized air and resultant force due to oil, which will be taken as F1 and F2 respectively.

Therefore,

Ft = F1 + F2

According to Pascal's law which states that a change in pressure at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere, the air pressure will act on the whole cap surface.

To get F1,

F1 = p x A

= p x (πr²)

Substituting values,

F1 = 200 x π x 1²

F1 = 628.32 kN

This resultant force acts at the center of the plate.

To get F2,

F2 = Π x hc x A

F2 = Π x (4r/3π) x (πr²/2)

Π - weight density of oil,

A - area on which oil pressure is acting,

hc - the distance between the axis of the conduit and the centroid of the semicircular area

Π = Specific gravity x 9.81 x 1000

Therefore

F2 = 1.6 x 9.81 x 1000 x (4(1)/3π) x (π(1)²/2)

F2 = 10.464 kN

Ft = F1 + F2

Ft = 628.32 + 10.464

Ft = 638.784 kN

The resultant force on the surface is 639 kN

Taking moments of the forces F1 and F2 about the centre,

Mo = Ft x y

Ft x y = (F1 x r) + F2(1 - 4r/3π)

Making y the subject,

y = (628.32 + 10.464(1 - 4/3π)/ 638.784

y = 0.993m

7 0

3 years ago

Carbon dioxide at a temperature of 0oC and a pressure of 600 kPa (abs) flows through a horizontal 40-mm- diameter pipe with an a

rusak2 [61]

Answer:

f = 0.04042

Explanation:

temperature = 0°C = 273k

p = 600 Kpa

d = 40 millemeter

e = 10 m

change in P = 235 N/m²

μ = 2m/s

R = 188.9 Nm/kgk

we solve this using this formula;

P = ρcos*R*T

we put in the values into this equation

600x10³ = ρcos * 188.9 * 273

600000 = ρcos51569.7

ρcos = 600000/51569.7

=11.63

from here we find the head loss due to friction

Δp/pg = feμ²/2D

235/11.63 = f*10*4/2*40x10⁻³

20.21 = 40f/0.08

20.21*0.08 = 40f

1.6168 = 40f

divide through by 40

f = 0.04042

5 0

3 years ago

If you have a hole diameter of 0.250 with a tolerance of ±0.005, what are the limits of the hole size?

sleet_krkn [62]

Answer:

The limits of the hole size are;

The maximum limit of the hole diameter 0.255

The minimum limit of the hole diameter = 0.245

Explanation:

Tolerance is a standardized form of language that can be used to define the intended 'tightness' or 'clearance' degree between mating parts in a mechanical assembly process and in metal joining processes such as welding and brazing processes

In tolerancing, the size used in the description of a part is known as the nominal size while allowable variation of the nominal size that will still allow the part to function properly is known as the tolerance

A tolerance given in the form ±P is known as bilateral tolerancing, with the value being added to or subtracted from the nominal size to get the maximum and minimum allowable limits of the dimensions of the nominal size

Therefore;

The given nominal dimension of the hole diameter = 0.250

The bilateral tolerance of the dimension, = ±0.005

Therefore;

The maximum limit of the diameter of the hole = 0.250 + 0.005 = 0.255

The minimum limit of the diameter of the hole = 0.250 - 0.005 = 0.245

7 0

2 years ago

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

Pick a subjectarea/field/topic that you are interested in. For each of the following Bonham- Carver uses of GIS give an example

Vanyuwa [196]

Answer:

I hope following attachment will help you a lot!

Explanation:

3 0

3 years ago

2. Describe the kinds of pressures that might lead to unethical behavior.