1. An automobile travels along a straight road at 15.65 m/s through a 11.18 m/sspeed zone. A police car observed the automobile. At the instant that the two
vehicles are abreast of each other, the police car starts to pursue the automobile at
a constant acceleration of 1.96 m/s². The motorist noticed the police car in his rear
view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s². (Hint: The police will not go against the law.)
a) Find the total time required for the police car to overtake the automobile.
b) Find the total distance travelled by the police car while overtaking the
automobile.
c) Find the speed of the police car at the time it overtakes the automobile.
d) Find the speed of the automobile at the time it was overtaken by the police car.

Answers

Answer 1
Answer:

Answer:

a.) Time = 17.13 seconds

b.) 31.88 m

c.) V = 11.18 m/s

d.) V = 7.1 m/s

Explanation:

The initial velocity U of the automobile is 15.65 m/s.

 At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile with initial velocity U = 0 at a constant acceleration of 1.96 m/s². Because the police is starting from rest.

For the automobile, let us use first equation of motion

V = U - at.

Acceleration a is negative since it is decelerating with a = 3.05 m/s² . And

V = 0.

Substitute U and a into the formula

0 = 15.65 - 3.05t

15.65 = 3.05t

t = 15.65/3.05

t = 5.13 seconds

But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s².

The total time required for the police car to overtake the automobile will be

12 + 5.13 = 17.13 seconds.

b.) Using the third equation of motion formula for the police car at V = 11.18 m/s and a = 1.96 m/s²

V^2 = U^2 + 2aS

Where S = distance travelled.

Substitute V and a into the formula

11.18^2 = 0 + 2 × 1.96 ×S

124.99 = 3.92S

S = 124.99/3.92

S = 31.88 m

c.) The speed of the police car at the time it overtakes the automobile will be in line with the speed zone which is 11.18 m/s

d.) That will be the final velocity V of the automobile car.

We will use third equation of motion to solve that.

V^2 = U^2 + 2as

V^2 = 15.65^2 - 2 × 3.05 × 31.88

V^2 = 244.9225 - 194.468

V = sqrt( 50.4545)

V = 7.1 m/s


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_________ items are similar to the free issue items, but their access is limited. (CLO5) a)-Bin stock items free issue b)-Bin stock controlled issue c)-Critical or insurance spares d)-Rebuildable spares e)-consumables

Answers

Answer:

a)-Bin stock items free issue

Explanation:

Bin stock items free issue items are similar to the free issue items, but their access is limited.

Bin stock items free issue items are similar to the free issue items, but their access is limited.

With the aid of a labbled diagram describe the operation of a core type single phase transformer​

Answers

Answer:

A simple single-phase transformer has each winding being wound cylindrically on a soft iron limb separately to provide a necessary magnetic circuit, which is commonly referred to as “transformer core”. It offers a path for the flow of the magnetic field to induce voltage between two windings.

write an interface downloadable that has a method "geturl" that returns the url of a downloadable object

Answers

Answer:

I want to believe the program is to be written in java and i hope your question is complete. The code is in the explanation section below

Explanation:

import java.util.Date;

public interface Downloadable {

  //abstract methods

  public String getUrl();

  public Date getLastDownloadDate();

 

}

IN JAVA,Knapsack Problem

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

Answers

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

Propane (C3H8) burns completely with 150% of theoretical air entering at 74°F, 1 atm, 50% relative humidity. The dry air component can be modeled as 21% O2 and 79% N2 on a molar basis. The combustion products leave at 1 atm. Determine the mole fraction of water in the products, in lbmol(water)/lbmol(products).

Answers

Answer:

y=0.12 lbmol(water)/lbmol(products)

Explanation:

First we find the humidity of air. Using humidity tables and the temperature we find that is 0.01 g water/L air.

Now we set the equation assuming dry air:

C_(3)H_(8)+7.5(O_(2)+3.76N_(2)) \longrightarrow 3CO_(2)+4H_(2)O+2.5O_(2)+28.2N_(2)

With this equation we have almost all moles that exit the reactor, we are just missing the initial moles of water due to the humidity. So we proceed to calculate it with the ideal gas law:

PV=nRT

Vair=867.7L

With the volume and the fraction of water, we can calculate the mass of water:

0.01 * 867.7=8.677 g of water

Now we calculate the moles of water:

8.677 g / 18 g/mol = 0.48 moles of water

Now we can calculate the total moles of water in the exit of the reactor:

0.48 + 4 = 4.48 moles of water

And finally we just need to sum all moles at the exit of the reactor and divide:

3 moles of CO2 + 2.5 moles of O2+ 4.48 moles of H2O + 28.2 moles of N2

And we have 38.18 moles in total, then:

4.48/38.18=y=0.12 moles of water/moles of products

As the relation moles/moles is equal to lb-moles/lb-moles, we have our fina result:

y=0.12 lbmol(water)/lbmol(products)

Describe the difference between design guidelines or criteria and design performance. Explain the relationship between the use of guideline/criteria tools and performance tools during the design process

Answers

Answer:

PART A

Design guidelines are sets of procedures to be followed in order to enhance the designing of an object or other things.

Design Performance is the actual process of carrying out the design process of an object using the design guidelines or criteria.

PART B

(1) Design guidelines tools helps to enhance design Performance.

(2) Design guidelines tools helps the designing performance tools to be effective.

Explanation:Design guidelines are the various steps which has special tools used to guide the designer in order to enhance the designing performance tools and ensure that the design process is done devoid of errors.

Design Performance tools are tools which helps to enhance the actual design Activities.