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3 years ago

How many parts (screws and bolts included) does the average car have?

Engineering

2 answers:

Vladimir [108]3 years ago

7 0

Answer:

Explanation:

30,000 parts

A single car has about 30,000 parts, counting every part down to the smallest screws.

Leokris [45]3 years ago

5 0

A single car has about 30,000 parts, counting every part down to the smallest screws

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. Consider the single-engine light plane described in Prob. 2. If the specific fuel consumption is 0.42 lb of fuel per horsepowe

Trava [24]

Answer:

Hence the Range and Endurance of single engine plane is given by

650.644 miles and 5.3528 hrs at standard sea level.

Explanation:

Given :

A single engine light plane with ,

Specific fuel consumption 0.42lb/hr/hp.

Fuel capacity =44 gal.

Gross weight =3400 lb.

To find :

Range and Endurance of the plane.

Solution:

Consider all standard measures of standard single engine propeller plane

Wing span =35.8 fts.

Wing swing area=174 sq ft

parasite drag coefficient =Cd.o.=0.025

Oswald's eff. factor= 0.8

ρ=0.002377= corresponds to standard sea level constant.

Now

Formula for Range is given by, Breguent formula.

R=(η/c) *(Cl/Cd)*ln(W1/W0)

here η is Oswald's constant,

Now calculating lift(Cl) and drag coefficient (Cd)

Cl=W/(1/2*ρ*v^2*S)

W=Gross weight

ρ=0.002377

Assume v=200 ft/sec normally,

S=174 Sq .ft.

CI=3400/(1/2*0.002377*200*200*174)

=6800/16543.9

=0.4110

Now calculating drag constant,

AR=(wing span)^2/wing swing area

=(35.8)^2/174

=7.37

Now

Drag Coefficient

Cd=Cd.o.+ (Cl^2)/(pie*e*AR)

=0.025+(0.4110)^2/(3.142*0.8*7.36)

=0.0342

Given that 44 gal fuel capacity and in Aviation weight of fuel is 5.64 lb/gal

hence weight of fuel=W1=3400- (44*5.64)

=3151.84

Now

for specific fuel consumption=0.42 lb/hp/hr

=0.42 lb*(1/550 ft)*(1/3600)sec

=2.12 *10^-7 lb/ft/sec

Now further calculating range

R=(η/c) *(Cl/Cd)*ln(W1/W0)

={0.8/(2.12*10^-7)}*(0.4110/0.0342)*ln(3151.84/3400)

=0.024908/0.072504

=0.34354*10^7

=3.4353 *10^6 fts.

1mi =5280 ft

=(3.4353/5280)*10^6

=650.644 miles

Now

For Endurance

E=(η/c)*{(Cl^3/2)/Cd}*(2*ρ*S)^1/2*[1/(W1)^1/2 -1/(W0)^1/2].

=(0.8/2.12*10^-7)*{(0.4110^3/2)/0.0342}*(2*0.002377*174)^1/2*[1/(3151.84)^1/2 -1/(3400)^1/2]

=3.7735*10^6*7.7043*0.8272*0.0006629

=0.01927*10^6

=1.927*10^4 sec

here 1hr =3600 sec

E=(1.927/3600)*10^4

=5.3528 hrs

7 0

3 years ago

An alloy has a yield strength of 818 MPa and an elastic modulus of 104 GPa. Calculate the modulus of resilience for this alloy [

crimeas [40]

Answer:

Modulus of resilience will be $3216942.308j/m^3$

Explanation:

We have given yield strength $\sigma _y=818MPa$

Elastic modulus E = 104 GPa

We have to find the modulus

Modulus of resilience is given by

Modulus of resilience $=\frac{\sigma _y^2}{2E}$ , here $\sigma _y$ is yield strength and E is elastic modulus

Modulus of resilience $=\frac{(818\times 10^6)^2}{2\times 104\times 10^9}=3216942.308j/m^3$

5 0

3 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

A complete mix of an activated sludge system without primary clarification is used for treatment of municipal wastewater with a

Hitman42 [59]

Answer:

sorry di ko alam

Explanation:

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2 years ago

Examine a process whereby air at 300 K, 100 kPa is compressed in a piston/cylinder arrangement to 600 kPa. Assume the process is

professor190 [17]

Answer:

See attachment and explanation.

Explanation:

- The following question can be solved better with the help of a MATLAB program as follows. The code is given in the attachment.

- The plot of the graph is given in attachment.

- The code covers the entire spectrum of the poly-tropic range ( 1.2 - 1.6 ) and 20 steps ( cases ) have been plotted and compared in the attached plot.

3 0

3 years ago