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Kamila [148]
3 years ago
6

Pls hurry

Engineering
1 answer:
sergey [27]3 years ago
5 0

Answer:The answer is Potassium!

Explanation: This is true because each label should tell you about the available amount of a certain element. The standard order is Nitrogen-Phosphorus-Potassium. They are referred to by their standard abbreviations in the periodic table. One problem with fertilizer labels are that they are only required to disclose the amounts of macronutrients (or Nitrogen-Phosphorus-Potassium.)

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A single crystal of a metal that has the BCC crystal structure is oriented such that a tensile stress is applied in the [100] di
VARVARA [1.3K]

Answer:

For [1 1 0] and  [1 0 1] plane, σₓ = 6.05 MPa

For [0 1 1] plane, σ = 0; slip will not occur

Explanation:

compute the resolved shear stress in [111] direction on each of the [110], [011] and on the [101] plane.

Given;

Stress direction: [1 0 0] ⇒ A

Slip direction: [1 1 1]

Normal to slip direction: [1 1 1] ⇒ B

∅ is the angle between A & B

Step 1: cos∅ = A·B/|A| |B| = \frac{[100][111]}{\sqrt{1}.\sqrt{3}  } ⇒ cos∅ = 1/\sqrt{3}

σₓ = τ/cos ∅·cosλ

where τ is the critical resolved shear stress given as 2.47MPa

Step 2: Solve for the slip along each plane

(a) [1 1 0]

cosλ = [1 1 0]·[1 0 0]/(\sqrt{2}·\sqrt{1})        

note: cosλ = slip D·stress D/|slip D||stress D|

cosλ = 1/\sqrt{2}

∵ σₓ = τ/\frac{1}{\sqrt{2} } ·\frac{1}{\sqrt{3} } = \sqrt{6} * 2.47MPa = 6.05MPa

Hence, stress necessary to cause slip on [1 1 0] is 6.05MPa

(b) [0 1 1]

cosλ = [0 1 1]·[1 0 0]/(\sqrt{2}·\sqrt{1}) = 0

∵ σₓ = 2.47MPa/0, which is not defined

Hence, for stress along [1 0 0], slip will not occur along [0 1 1]

(c) [1 0 1]

cosλ = [0 1 1]·[1 0 0]/(\sqrt{2}·\sqrt{1})

cosλ = 1/\sqrt{2}

∵ σₓ = τ/\frac{1}{\sqrt{2} } ·\frac{1}{\sqrt{3} } = \sqrt{6} * 2.47MPa = 6.05MPa

See attachment for the space diagram

3 0
3 years ago
I NEED HELP ASAP WILL AWARD BRAINLIEST
CaHeK987 [17]

Answer:

It was a power supply of a circuit that was horizontal of the ladder where they controlled the circuit.

Explanation:

6 0
3 years ago
Read 2 more answers
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
Consider a situation where water is used as the coolant in a heat exchanger for a micro-electromechanical device (MEM device). T
Arturiano [62]

Answer:

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Explanation:

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Who was first credited with the discovery of the intensity of current within a circuit?
jeyben [28]

Answer:

Hans Christian Ørsted

One of the most important discoveries relating to current was made accidentally by Hans Christian Ørsted in 1820, when, while preparing a lecture, he witnessed the current in a wire disturbing the needle of a magnetic compass.

Explanation:

is that what you were looking for

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