Answer:

**Answer:**

The critical depth of the rectangular channel is approximately 1.790 meters.

The flow velocity in the rectangular channel is 4.190 meters per second.

**Explanation:**

From Open Channel Theory we know that critical depth of the rectangular channel (), measured in meters, is calculated by using this equation:

**(Eq. 1)**

Where:

- Volume flow rate, measured in cubic meters per second.

- Gravitational acceleration, measured in meters per square second.

- Channel width, measured in meters.

If we know that , and , then the critical depth is:

The critical depth of the rectangular channel is approximately 1.790 meters.

Lastly, the flow velocity (), measured in meters, is obtained from this formula:

**(Eq. 2)**

If we know that , and , then the flow velocity in the rectangular channel is:

The flow velocity in the rectangular channel is 4.190 meters per second.

Use Euler’s Method: ????????????????=???? ????????????????=−????????−????3+????cos(????) ????(0)=1.0 ????(0)=1.0 ????=0.4 ????=20.0 ℎ=0.01 ????=10000 ▪ Write the data (y1, y2) to a file named "LASTNAME_Prob1.dat" Example: If your name is John Doe – file name would be "DOE_Prob1.dat" ▪ Plot the result with lines using GNUPLOT (Hint: see lecture 08) ▪ Submit full code (copy and paste). Plot must be on a separate page. ▪ Run the code again and plot the result for: ????=0.1 ????=11.0

Which statement best describes how power and work are related?O A. Power is the ability to do more work with less force.O B. Power is a measure of how quickly work is done.O C. Power and work have the same unit of measurementO D. Power is the amount of work needed to overcome friction.Pls answer quick

The 50mm diameter cylinder is made from Am 1004-T61 magnesium (E = 44.7GPa, a = 26x10^-6/°C)and is placed in the clamp when the temperature is T1 = 15°C. If the two 304-stainless-steel (E =193GPa, a = 17x10^-6/°C) carriage bolts of the clamp each have a diameter of 10mm, and they holdthe cylinder snug with a negligible force against the rigid jaws, determine the temperature at whichthe average normal stress in either the magnesium or steel becomes 12 MPa.

A wind turbine system has the following specifications: Diameter:45 m Rated power 700 kW at the wind speed of 12 m/s Turbine speed 1500 rpm Determine the swept area of the wind turbine. a)- 1640 m^2 B)- 1690 m^2 c)- 1590 m^2 d)- 1540 m^2

Automobiles must be able to sustain a frontal impact. Specifically, the design must allow low speed impacts with little damage, while allowing the vehicle front end structure to deform and absorb impact energy at higher speeds. Consider a frontal impact test of a vehicle with a mass of 1000 kg. a. For a low speed test (v = 2.5 m/s), compute the energy in the vehicle just prior to impact. If the bumper is a pure elastic element, what is the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm? b. At a higher speed impact of v = 25 m/s, considerable deformation occurs. To absorb the energy, the front end of a vehicle is designed to deform while providing a nearly constant force. For this condition, what is the amount of energy that must be absorbed by the deformation [neglecting the energy stored in the elastic deformation in (a)? If it is desired to limit the deformation to 10 cm, what level of resistance force is required? What is the deacceleration of the vehicle in this condition?

Which statement best describes how power and work are related?O A. Power is the ability to do more work with less force.O B. Power is a measure of how quickly work is done.O C. Power and work have the same unit of measurementO D. Power is the amount of work needed to overcome friction.Pls answer quick

The 50mm diameter cylinder is made from Am 1004-T61 magnesium (E = 44.7GPa, a = 26x10^-6/°C)and is placed in the clamp when the temperature is T1 = 15°C. If the two 304-stainless-steel (E =193GPa, a = 17x10^-6/°C) carriage bolts of the clamp each have a diameter of 10mm, and they holdthe cylinder snug with a negligible force against the rigid jaws, determine the temperature at whichthe average normal stress in either the magnesium or steel becomes 12 MPa.

A wind turbine system has the following specifications: Diameter:45 m Rated power 700 kW at the wind speed of 12 m/s Turbine speed 1500 rpm Determine the swept area of the wind turbine. a)- 1640 m^2 B)- 1690 m^2 c)- 1590 m^2 d)- 1540 m^2

Automobiles must be able to sustain a frontal impact. Specifically, the design must allow low speed impacts with little damage, while allowing the vehicle front end structure to deform and absorb impact energy at higher speeds. Consider a frontal impact test of a vehicle with a mass of 1000 kg. a. For a low speed test (v = 2.5 m/s), compute the energy in the vehicle just prior to impact. If the bumper is a pure elastic element, what is the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm? b. At a higher speed impact of v = 25 m/s, considerable deformation occurs. To absorb the energy, the front end of a vehicle is designed to deform while providing a nearly constant force. For this condition, what is the amount of energy that must be absorbed by the deformation [neglecting the energy stored in the elastic deformation in (a)? If it is desired to limit the deformation to 10 cm, what level of resistance force is required? What is the deacceleration of the vehicle in this condition?

**Answer:**

Safety check is defined as rounding to make sure that the patients and the milieu (patients living quarters) is secured and free of harmful items that can be used to hurt someone.

By **Riemann sums**, the *boundary***work** done by a **gas** during an *expansion***process** based on the information given by the statement is approximately 0.243 joules.

A **process** is a consecution of **states** of a **system**. The **boundary work** (* W*), in kilojoules, is the

**(1)**

Where:

- Pressure, in kilopascals.**p**- Volume, in cubic meters.**V**

Now we proceed to calculate the boundary work:

W = 0.5 · [(300 kPa + 290 kPa) · (1.1 × 10⁻³ m³ - 1 × 10⁻³ m³) + (270 kPa + 290 kPa) · (1.2 × 10⁻³ m³ - 1.1 × 10⁻³ m³) + (250 kPa + 270 kPa) · (1.4 × 10⁻³ m³ - 1.2 × 10⁻³ m³) + (220 kPa + 250 kPa) · (1.7 × 10⁻³ m³ - 1.4 × 10⁻³ m³) + (200 kPa + 220 kPa) · (2 × 10⁻³ m³ - 1.7 × 10⁻³ m³)]

W = 0.243 kJ

By **Riemann sums**, the *boundary***work** done by a **gas** during an *expansion***process** based on the information given by the statement is approximately 0.243 joules.

To learn more on **boundary work**, we kindly invite to check this: brainly.com/question/17136485

#SPJ2

Answer:

attached below

Explanation:

**Answer:**

**Explanation:**

Adjusting the distance between the two electrodes is called gapping your spark plugs. You need a feeler gauge to gap your spark plugs properly If you're re-gapping a used plug, make sure that it's clean (gently scrub it with a wire brush)

hope this you

Answer:

Output:-

Enter the five digit lottery number

Enter the digit 1 : 23

Enter the digit 2 : 44

Enter the digit 3 : 43

Enter the digit 4 : 66

Enter the digit 5 : 33

YOU LOSS!!

Computer Generated Lottery Number :

|12|38|47|48|49|

Lottery Number Of user:

|23|33|43|44|66|

Number Of digit matched: 0

Code:-

import java.util.Arrays;

import java.util.Random;

import java.util.Scanner;

public class Lottery {

int[] lotteryNumbers = new int[5];

public int[] getLotteryNumbers() {

return lotteryNumbers;

}

Lottery() {

Random randomVal = new Random();

for (int i = 0; i < lotteryNumbers.length; i++) {

lotteryNumbers[i] = randomVal.nextInt((50 - 1) + 1);

}

}

int compare(int[] personLottery) {

int count = 0;

Arrays.sort(lotteryNumbers);

Arrays.sort(personLottery);

for (int i = 0; i < lotteryNumbers.length; i++) {

if (lotteryNumbers[i] == personLottery[i]) {

count++;

}

}

return count;

}

public static void main(String[] args) {

int[] personLotteryNum = new int[5];

int matchNum;

Lottery lnum = new Lottery();

Scanner input = new Scanner(System.in);

System.out.println("Enten the five digit lottery number");

for (int i = 0; i < personLotteryNum.length; i++) {

System.out.println("Enter the digit " + (i + 1) + " :");

personLotteryNum[i] = input.nextInt();

}

matchNum = lnum.compare(personLotteryNum);

if (matchNum == 5)

System.out.println("YOU WIN!!");

else

System.out.println("YOU LOSS!!");

System.out.println("Computer Generated Lottery Number :");

System.out.print("|");

for (int i = 0; i < lnum.getLotteryNumbers().length; i++) {

System.out.print(lnum.getLotteryNumbers()[i] + "|");

}

System.out.println("\n\nLottery Number Of user:");

System.out.print("|");

for (int i = 0; i < personLotteryNum.length; i++) {

System.out.print(personLotteryNum[i] + "|");

}

System.out.println();

System.out.println("Number Of digit matched: " + matchNum);

}

}

Explanation:

Who is right?

**Technician **A says that the **unitized structure **of a **hybrid vehicle **is considerably different when compared to the same **conventional model **is right.

**Hybrid vehicle **are defined as a powered by a **combustion engine **and/or a number of electric motors that draw power from batteries. A **gas**-**powered **car simply has a traditional gas engine, but a hybrid car also features an **electric motor**.

One important advantage of hybrid cars is their capacity to reduce the size of the main engine, which improves **fuel efficiency**. Many hybrid vehicles employ electric motors to accelerate slowly at first until they reach higher speeds. They then use **gasoline**-**powered engines **to increase fuel efficiency.

Thus, **technician **A says that the **unitized structure **of a **hybrid vehicle **is considerably different when compared to the same **conventional model **is right.

To learn more about **hybrid vehicle, **refer to the link below:

#SPJ5

A train was right get it

The file KnapsackData1.txt and KnapsackData2.txt are sample input files

for the following Knapsack Problem that you will solve.

KnapsackData1.txt contains a list of four prospective projects for the upcoming year for a particular

company:

Project0 6 30

Project1 3 14

Project2 4 16

Project3 2 9

Each line in the file provides three pieces of information:

1) String: The name of the project;

2) Integer: The amount of employee labor that will be demanded by the project, measured in work weeks;

3) Integer: The net profit that the company can expect from engaging in the project, measured in thousands

of dollars.

Your task is to write a program that:

1) Prompts the user for the number of work weeks available (integer);

2) Prompts the user for the name of the input file (string);

3) Prompts the user for the name of the output file (string);

4) Reads the available projects from the input file;

5) Dolves the corresponding knapsack problem, without repetition of items; and

6) Writes to the output file a summary of the results, including the expected profit and a list of the best

projects for the company to undertake.

Here is a sample session with the program:

Enter the number of available employee work weeks: 10

Enter the name of input file: KnapsackData1.txt

Enter the name of output file: Output1.txt

Number of projects = 4

Done

For the above example, here is the output that should be written to Output1.txt:

Number of projects available: 4

Available employee work weeks: 10

Number of projects chosen: 2

Number of projectsTotal profit: 46

Project0 6 30

Project2 4 16

The file KnapsackData2.txt, contains one thousand prospective projects. Your program should also be able to handle this larger problem as well. The corresponding output file,

WardOutput2.txt, is below.

With a thousand prospective projects to consider, it will be impossible for your program to finish in a

reasonable amount of time if it uses a "brute-force search" that explicitly considers every possible

combination of projects. You are required to use a dynamic programming approach to this problem.

WardOutput2.txt:

Number of projects available: 1000

Available employee work weeks: 100

Number of projects chosen: 66

Total profit: 16096

Project15 2 236

Project73 3 397

Project90 2 302

Project114 1 139

Project117 1 158

Project153 3 354

Project161 2 344

Project181 1 140

Project211 1 191

Project213 2 268

Project214 2 386

Project254 1 170

Project257 4 427

Project274 1 148

Project275 1 212

Project281 2 414

Project290 1 215

Project306 2 455

Project334 3 339

Project346 2 215

Project356 3 337

Project363 1 159

Project377 1 105

Project389 1 142

Project397 1 321

Project399 1 351

Project407 3 340

Project414 1 266

Project431 1 114

Project435 3 382

Project446 1 139

Project452 1 127

Project456 1 229

Project461 1 319

Project478 1 158

Project482 2 273

Project492 1 142

Project525 1 144

Project531 1 382

Project574 1 170

Project594 1 125

Project636 2 345

Project644 1 169

Project668 1 191

Project676 1 117

Project684 1 143

Project689 1 108

Project690 1 216

Project713 1 367

Project724 1 127

Project729 2 239

Project738 1 252

Project779 1 115

Project791 1 110

Project818 2 434

Project820 1 222

Project830 1 179

Project888 3 381

Project934 3 461

Project939 3 358

Project951 1 165

Project959 2 351

Project962 1 316

Project967 1 191

Project984 1 117

Project997 1 187

**Answer:**

**Explanation:**

Code:

import java.io.File;

import java.io.FileWriter;

import java.io.IOException;

import java.util.Scanner;

public class Knapsack {

public static void knapsack(int wk[], int pr[], int W, String ofile) throws IOException

{

int i, w;

int[][] Ksack = new int[wk.length + 1][W + 1];

for (i = 0; i <= wk.length; i++) {

for (w = 0; w <= W; w++) {

if (i == 0 || w == 0)

Ksack[i][w] = 0;

else if (wk[i - 1] <= w)

Ksack[i][w] = Math.max(pr[i - 1] + Ksack[i - 1][w - wk[i - 1]], Ksack[i - 1][w]);

else

Ksack[i][w] = Ksack[i - 1][w];

}

}

int maxProfit = Ksack[wk.length][W];

int tempProfit = maxProfit;

int count = 0;

w = W;

int[] projectIncluded = new int[1000];

for (i = wk.length; i > 0 && tempProfit > 0; i--) {

if (tempProfit == Ksack[i - 1][w])

continue;

else {

projectIncluded[count++] = i-1;

tempProfit = tempProfit - pr[i - 1];

w = w - wk[i - 1];

}

FileWriter f =new FileWriter("C:\\Users\\gshubhita\\Desktop\\"+ ofile);

f.write("Number of projects available: "+ wk.length+ "\r\n");

f.write("Available employee work weeks: "+ W + "\r\n");

f.write("Number of projects chosen: "+ count + "\r\n");

f.write("Total profit: "+ maxProfit + "\r\n");

for (int j = 0; j < count; j++)

f.write("\nProject"+ projectIncluded[j] +" " +wk[projectIncluded[j]]+ " "+ pr[projectIncluded[j]] + "\r\n");

f.close();

}

}

public static void main(String[] args) throws Exception

{

Scanner sc = new Scanner(System.in);

System.out.print("Enter the number of available employee work weeks: ");

int avbWeeks = sc.nextInt();

System.out.print("Enter the name of input file: ");

String inputFile = sc.next();

System.out.print("Enter the name of output file: ");

String outputFile = sc.next();

System.out.print("Number of projects = ");

int projects = sc.nextInt();

int[] workWeeks = new int[projects];

int[] profit = new int[projects];

File file = new File("C:\\Users\\gshubhita\\Desktop\\" + inputFile);

Scanner fl = new Scanner(file);

int count = 0;

while (fl.hasNextLine()){

String line = fl.nextLine();

String[] x = line.split(" ");

workWeeks[count] = Integer.parseInt(x[1]);

profit[count] = Integer.parseInt(x[2]);

count++;

}

knapsack(workWeeks, profit, avbWeeks, outputFile);

}

}

Console Output:

Enter the number of available employee work weeks: 10

Enter the name of input file: input.txt

Enter the name of output file: output.txt

Number of projects = 4

Output.txt:

Number of projects available: 4

Available employee work weeks: 10

Number of projects chosen: 2

Total profit: 46

Project2 4 16

Project0 6 30