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
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.
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.
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();
}
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
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:
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)
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.