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NikAS [45]
3 years ago
12

Ignition for heavy fuel oil?

Engineering
2 answers:
ipn [44]3 years ago
7 0

Answer:

What heavy fuel oil?

Heavy Fuel Oil (HFO) is a category of fuel oils of a tar-like consistency. Also known as bunker fuel, or residual fuel oil, HFO is the result or remnant from the distillation and cracking process of petroleum.

Explanation:

lianna [129]3 years ago
3 0
I mean…. Fire? Matches?
You might be interested in
if both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will
egoroff_w [7]

If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will: a) increase during a climb.

<h3>What is a ram air input?</h3>

A ram air input can be defined as an air intake system which is designed and developed to use the dynamic air pressure that is created due to vehicular motion, or ram pressure, in order to increase the static air pressure within the intake manifold of an internal combustion engine of an automobile.

This ultimately implies that, a ram air input allows a greater mass-flow of air through the engine of an automobile, thereby, increasing the engine's power.

In conclusion, indicated airspeed will increase during a climb when both the ram air input and drain hole of the pitot system become blocked.

Read more on pilots here: brainly.com/question/10381526

#SPJ1

Complete Question:

If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will

a) increase during a climb

b) decrease during a climb

c) remain constant regardless of altitude change

6 0
2 years ago
g For this project you are required to perform Matrix operations (Addition, Subtraction and Multiplication). For each of the ope
Kruka [31]

Answer:

C++ code is explained below

Explanation:

#include<iostream>

using namespace std;

//Function Declarations

void add();

void sub();

void mul();

//Main Code Displays Menu And Take User Input

int main()

{

  int choice;

  cout << "\nMenu";

  cout << "\nChoice 1:addition";

  cout << "\nChoice 2:subtraction";

  cout << "\nChoice 3:multiplication";

  cout << "\nChoice 0:exit";

 

  cout << "\n\nEnter your choice: ";

 

  cin >> choice;

 

  cout << "\n";

 

  switch(choice)

  {

      case 1: add();

              break;

             

      case 2: sub();

              break;

             

      case 3: mul();

              break;

     

      case 0: cout << "Exited";

              exit(1);

     

      default: cout << "Invalid";      

  }

  main();  

}

//Addition Of Matrix

void add()

{

  int rows1,cols1,i,j,rows2,cols2;

 

  cout << "\nmatrix1 # of rows: ";

  cin >> rows1;

 

  cout << "\nmatrix1 # of columns: ";

  cin >> cols1;

 

   int m1[rows1][cols1];

 

  //Taking First Matrix

  for(i=0;i<rows1;i++)

      for(j=0;j<cols1;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m1[i][j];

          cout << "\n";

      }

  //Printing 1st Matrix

  for(i=0;i<rows1;i++)

  {

      for(j=0;j<cols1;j++)

          cout << m1[i][j] << " ";

      cout << "\n";

  }

     

  cout << "\nmatrix2 # of rows: ";

  cin >> rows2;

 

  cout << "\nmatrix2 # of columns: ";

  cin >> cols2;

 

  int m2[rows2][cols2];

  //Taking Second Matrix

  for(i=0;i<rows2;i++)

      for(j=0;j<cols2;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m2[i][j];

          cout << "\n";

      }

  //Displaying second Matrix

  cout << "\n";

  for(i=0;i<rows2;i++)

  {

      for(j=0;j<cols2;j++)

          cout << m2[i][j] << " ";

      cout << "\n";

  }

  //Displaying Sum of m1 & m2

  if(rows1 == rows2 && cols1 == cols2)

  {

      cout << "\n";

      for(i=0;i<rows1;i++)

      {

          for(j=0;j<cols1;j++)

              cout << m1[i][j]+m2[i][j] << " ";

          cout << "\n";  

      }

  }

  else

      cout << "operation is not supported";

     

  main();

 

}

void sub()

{

  int rows1,cols1,i,j,k,rows2,cols2;

  cout << "\nmatrix1 # of rows: ";

  cin >> rows1;

 

  cout << "\nmatrix1 # of columns: ";

  cin >> cols1;

 

   int m1[rows1][cols1];

 

  for(i=0;i<rows1;i++)

      for(j=0;j<cols1;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m1[i][j];

          cout << "\n";

      }

 

  for(i=0;i<rows1;i++)

  {

      for(j=0;j<cols1;j++)

          cout << m1[i][j] << " ";

      cout << "\n";

  }

     

  cout << "\nmatrix2 # of rows: ";

  cin >> rows2;

 

  cout << "\nmatrix2 # of columns: ";

  cin >> cols2;

 

  int m2[rows2][cols2];

 

  for(i=0;i<rows2;i++)

      for(j=0;j<cols2;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m2[i][j];

          cout << "\n";

      }

 

  for(i=0;i<rows2;i++)

  {

      for(j=0;j<cols2;j++)

          cout << m1[i][j] << " ";

      cout << "\n";

  }

  cout << "\n";

  //Displaying Subtraction of m1 & m2

  if(rows1 == rows2 && cols1 == cols2)

  {

      for(i=0;i<rows1;i++)

      {

          for(j=0;j<cols1;j++)

              cout << m1[i][j]-m2[i][j] << " ";

          cout << "\n";  

      }

  }

  else

      cout << "operation is not supported";

     

  main();

 

}

void mul()

{

  int rows1,cols1,i,j,k,rows2,cols2,mul[10][10];

  cout << "\nmatrix1 # of rows: ";

  cin >> rows1;

 

  cout << "\nmatrix1 # of columns: ";

  cin >> cols1;

 

   int m1[rows1][cols1];

 

  for(i=0;i<rows1;i++)

      for(j=0;j<cols1;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m1[i][j];

          cout << "\n";

      }

  cout << "\n";

  for(i=0;i<rows1;i++)

  {

      for(j=0;j<cols1;j++)

          cout << m1[i][j] << " ";

      cout << "\n";

  }

     

  cout << "\nmatrix2 # of rows: ";

  cin >> rows2;

 

  cout << "\nmatrix2 # of columns: ";

  cin >> cols2;

 

  int m2[rows2][cols2];

 

  for(i=0;i<rows2;i++)

      for(j=0;j<cols2;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m2[i][j];

          cout << "\n";

      }

  cout << "\n";

  //Displaying Matrix 2

  for(i=0;i<rows2;i++)

  {

      for(j=0;j<cols2;j++)

          cout << m2[i][j] << " ";

      cout << "\n";

  }

     

  if(cols1!=rows2)

      cout << "operation is not supported";

  else

  {

      //Initializing results as 0

      for(i = 0; i < rows1; ++i)

  for(j = 0; j < cols2; ++j)

  mul[i][j]=0;

// Multiplying matrix m1 and m2 and storing in array mul.

  for(i = 0; i < rows1; i++)

  for(j = 0; j < cols2; j++)

  for(k = 0; k < cols1; k++)

  mul[i][j] += m1[i][k] * m2[k][j];

// Displaying the result.

  cout << "\n";

  for(i = 0; i < rows1; ++i)

      for(j = 0; j < cols2; ++j)

      {

      cout << " " << mul[i][j];

      if(j == cols2-1)

      cout << endl;

      }

      }  

  main();

 }

5 0
3 years ago
Water at 200C flows through a pipe of 10 mm diameter pipe at 1 m/s. Is the flow Turbulent ? a. Yes b. No
Degger [83]

Answer:

Yes, the flow is turbulent.

Explanation:

Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.

Given:

Diameter of pipe is 10mm.

Velocity of the pipe is 1m/s.

Temperature of water is 200°C.

The kinematic viscosity at temperature 200°C is 1.557\times10^{-7}m2/s.

Calculation:

Step1

Expression for Reynolds number is given as follows:

Re=\frac{vd}{\nu}

Here, v is velocity, \nu is kinematic viscosity, d is diameter and Re is Reynolds number.

Substitute the values in the above equation as follows:

Re=\frac{vd}{\nu}

Re=\frac{1\times(10mm)(\frac{1m}{1000mm})}{1.557\times10^{-7}}

Re=64226.07579

Thus, the Reynolds number is 64226.07579. This is greater than 2000.

Hence, the given flow is turbulent flow.

5 0
3 years ago
Select all that apply: Contaminated sharps should<br><br> not be<br><br> ----
PIT_PIT [208]

Answer:

Contaminated sharps should not be bent, recapped or removed.

Explanation:

Contaminated sharps are defined as "any contaminated object that can penetrate the skin including, but not limited to, needles, scalpels, broken glass, broken capillary tubes and exposed ends of dental wires".

4 0
3 years ago
A European car manufacturer reports that the fuel efficiency of the new MicroCar is 48.5 km/L highway and 42.0 km/L city. What a
statuscvo [17]

Answer:

Fuel efficiency for highway = 114.08 miles/gallon

Fuel efficiency for city = 98.79 miles/gallon

Explanation:

1 gallon = 3.7854 litres

1 mile = 1.6093 km

Let's first convert the efficiency to km/gallon:

48.5 km/litre = (48.5 * 3.7854) km/gallon

48.5 km/litre =  183.5919 km/gallon (highway)

42.0 km/litre = (42.0 * 3.7854) km/gallon

42.0 km/litre = 158.9868 km/gallon (city)

Next, we convert these to miles/gallon:

183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon

183.5919 km/gallon = 114.08 miles/gallon (highway)

158.9868 km/gallon = (158.9868 /1.6093) miles/gallon

158.9868 km/gallon = 98.79 miles/gallon (city)

3 0
3 years ago
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