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nadezda [96]
3 years ago
15

Complete the grading of fine aggregate table given below. Plot grading curve and calculate

Engineering
2 answers:
abruzzese [7]3 years ago
4 0

Answer:

I don't know

Explanation:

I don't know it buddy some smarty pant would surely come

Oksi-84 [34.3K]3 years ago
3 0

Answer:

Is there supposed to be an attachment?

Explanation:

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When handling chemicals and solvents, technicians are recommended to
Luda [366]

Answer:

  1. To wear PPE
  2. Have prior knowledge of explosive levels and elemental properties
  3. Know procedure to eliminate any threat
7 0
3 years ago
Read 2 more answers
An open glass tube is inserted into a pan of fresh water at 20 °C. What tube diameter is needed to make the height of capillary
kiruha [24]

Answer:

The tube diameter is 2.71 mm.

Explanation:

Given:

Open glass tube is inserted into a pan of fresh water at 20°C.

Height of capillary raise is four times tube diameter.

h = 4d

Assumption:

Take water as pure water as the water is fresh enough. So, the angle of contact is 0 degree.

Take surface tension of water at 20°C as 72.53\times 10^{-3} N/m.

Take density of water as 100 kg/m3.

Calculation:

Step1

Expression for height of capillary rise is gives as follows:

h=\frac{4\sigma\cos\theta}{dg\rho}

Step2

Substitute the value of height h, surface tension, density of water, acceleration due to gravity and contact angle in the above equation as follows:

4d=\frac{4\times72.53\times10^{-3}\cos0^{\circ}}{d\times9.81\times1000}

d^{2}=7.39\times10^{-6}

d=2.719\times10^{-3} m.

Or

d=(2.719\times10^{-3}m)(\frac{1000mm}{1m})

d=2.719 mm

Thus, the tube diameter is 2.719 mm.

 

5 0
4 years ago
IN JAVA,
Citrus2011 [14]

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

8 0
3 years ago
Typically each development platform consists of the following components, except:Select one:a.Operating systemb.System softwarec
damaskus [11]

Typically each development platform consists of the following components except compilers and assemblers

  • The platform development simply means the development of the fundamental software which is vital in making hardware work.

  • Operating system: This refers to the low-level software that communicates with the hardware so that other programs can be able to run.

  • System software: This is the software that's designed in order to provide a platform for the other software. Examples include search engines, Microsoft Windows, etc.

  • Compilers and assemblers: Compliers are sued in converting source code to a machine-level language. Assembler is used in converting assembly code to machine code.

  • Hardware platform: This is a set of hardware where the software applications are run.

In conclusion, the correct option is Compilers and assemblers.

Read related link on:

brainly.com/question/21650058

4 0
3 years ago
A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3. The propane undergoes a process t
iren2701 [21]

Work done = -19.7 KJ

Heat transferred = 17.4 KJ

Explanation:

Given-

Temperature, T = 27°C

Volume, V = 0.2 m³

Pressure, P_{1}= 1 bar

v_{2} = 4 bar

pV¹°¹ = constant

From superheated propane table, at  P_{1}= 1 bar andT_{1}  = 27⁰C

v_{1} = 0.557 m³/kg

v_{2} = 473.73 KJ/kg

(a) Work = ?

We know,

V1¹°¹ = p2V2¹°¹

V2 = (\frac{P1}{P2})^\frac{1}{1.1} * V1  \\\\V2 = \frac{1}{4}^\frac{1}{1.1} * 0.557  \\\\V2 = 0.158 m^3/kg

At  = 4 bar and v = 0.158 m³/kg

u2 = 548.45K J/kg

To find work done in the process:

W = \frac{P2V2 - P1V1}{1-n} \\\\W = \frac{m(P2V2 - P1V1)}{1-n} \\\\W = \frac{v}{u} * \frac{P2V2 - P1V1}{1-n}\\  \\W = \frac{0.2}{0.5571} * \frac{4 X 0.158 - 1 X 0.577}{1-1.1} X 10^5 \frac{Pa}{Bar} \frac{1KJ}{10^3Nm} \\   \\W = -19.75KJ

(b) Heat transfer = ?

Q = m(u2 - u1) + W\\\\Q = \frac{0.2}{0.5571} * (548.45 - 473.73) + (-19.7)\\\\Q = 17.4KJ

8 0
4 years ago
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