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borishaifa [10]
2 years ago
5

g Asbestos is a fibrous silicate mineral with remarkably high tensile strength. But is no longer used because airborne asbestos

particles can cause lung cancer. Grunerite, a type of asbestos, has a tensile strength of 3.5 × 10ଶ kg mmଶ ⁄ (meaning a strand of grunerite with a 1-mmଶ cross-sectional area can hold up to 3.5 × 10ଶ kg). The tensile strength of aluminum and Steel No. 5137 are 2.4 × 10ସ lb inଶ ⁄ and 5.0 × 10ସ lb inଶ ⁄ respectively. Calculate the cross section areas in mmଶ of wires of aluminum and of Steel No. 5137 that have the same tensile strength as a fiber of grunerite with a cross-sectional area of 1.0 μmଶ
Engineering
1 answer:
otez555 [7]2 years ago
8 0

Answer:

Aluminum cross sectional area = 1.99 * 10^-5 mm^2

Steel cross sectional area = 9.95* 10^-6 mm^2

Explanation:

Given data:

Tensile strength of Grunerite = 3.5 * 10^2 kg/mm^2 = 3.5 * 10^-4 kg/<em>u</em>m^2

Tensile strength of Aluminum = 2.5 × 10^4 lb/in2 =  2.5 × 10^4 * 703.07 kg/m^2

Tensile strength of Steel number 5137 = 5.0 × 10^4 lb/in2 = 5.0 × 10^4*703.07 kg/m^2

<u>Calculating the cross sectional area of the wires of aluminum and steel No5137</u>

first we will determine the cross sectional area of the aluminum wire ( A ) by equating tensile strength of aluminum with the tensile strength

of Grunerite

(2.5 × 10^4 * 703.07)  * A = 3.5 * 10^-4 kg

Hence ; A = 1.99 * 10^-5 mm^2

Next we calculate the cross sectional area of steel

5.0 × 10^4*703.07 kg/m^2 * A" = 3.5 * 10^-4 kg

Hence ; A" = 9.95* 10^-6 mm^2

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How do you extablish a chain of dimensions​
kap26 [50]

Answer:

Certamente você conhece três dimensões: comprimento, largura e profundidade. Além disso, quando se pensa um pouco fora da caixa também seria possível adicionar a dimensão do tempo.

Provavelmente, algumas pessoas viajam na maionese quando toca-se nesse assunto. Vem em suas mentes universos paralelos e até mesmo realidades alternativas. Mas também não se trata disso.

Explanation:

Basicamente as dimensões são as facetas do que nós percebemos a ser realidade. Existem muitos debates sobre dimensões na física. Um dos que mais chamam a atenção se chama Teoria das Cordas.

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5 0
3 years ago
Can someone tell me what car year and model this is please
Arlecino [84]

Answer:

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Explanation:

5 0
3 years ago
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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
Determine the nature of the following cycle (reversible, irreversible, or impossible): a refrigeration cycle draws heat from a c
vlabodo [156]

Answer:

Impossible.

Explanation:

The ideal Coefficient of Performance is:

COP_{i} = \frac{250\,K}{300\,K-250\,K}

COP_{i} = 5

The real Coefficient of Performance is:

COP_{r} = \frac{950\,kJ-70\,kJ}{70\,kJ}

COP_{r} = 12.571

Which leads to an absurds, since the real Coefficient of Performance must be equal to or lesser than ideal Coefficient of Performance. Then, the cycle is impossible, since it violates the Second Law of Thermodynamics.

6 0
3 years ago
Which system of linear inequalities is represented by the graph? y &gt; x – 2 and y x + 1 y x + 1 y &gt; x – 2 and y &lt; x + 1
KatRina [158]

Answer:

The graph representing the linear inequalities is attached below.

Explanation:

The inequalities given are :

y>x-2   and y<x+1

For tables for values of x and y and get coordinates to plot for both equation.

In the first equation;

y>x-2

y=x-2

y-x = -2

The table will be :

x    y

-2  -4

-1    -3

0     -2

1      -1

2      0

The coordinates to plot are : (-2,-4) , (-1,-3), (0,-2), (1,-1) ,(2,0)

Use a dotted line and shade the part right hand side of the line.

Do the same for the second inequality equation and plot then shade the part satisfying the inequality.

The graph attached shows results.

5 0
4 years ago
Read 2 more answers
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