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Nonamiya [84]
3 years ago
11

A capillary tube is immersed vertically in a water container. Knowing that water starts to evaporate when the pressure drops bel

ow 2 kPa, determine the maximum capillary rise and tube diameter for this maximum-rise case. Take the contact angle at the inner wall of the tube to be 6° and the surface tension to be 1.00 N/m.
Engineering
1 answer:
REY [17]3 years ago
8 0

Answer: 10.12m, 20micro meter

Explanation:

You might be interested in
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
Which of the following is an example of a hardwood? A maple B spruce C pine D fir
bearhunter [10]

Answer:

A. Maple

Explanation:

Maple is a hardwood.

Hope that helps!

7 0
2 years ago
An automobile having a mass of 1100 kg initially moves along a level highway at 120 km/h relative to the highway. It then climbs
soldier1979 [14.2K]

Answer:

-6111.11\ \text{kJ}

863.28\ \text{kJ}

Explanation:

m = Mass of automobile = 1100 kg

v = Velocity of car = 120 km/h = \dfrac{120}{3.6}\ \text{m/s}

h = Height of hill = 80 m

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Change in kinetic energy

KE=\dfrac{1}{2}m(u^2-v^2)\\\Rightarrow KE=\dfrac{1}{2}\times 1100\times (0-(\dfrac{120}{3.6})^2)\\\Rightarrow KE=-611111.11\ \text{J}

Change in kinetic energy is -6111.11\ \text{kJ}

Change in potential energy is given by

PE=mgh\\\Rightarrow PE=1100\times 9.81\times 80\\\Rightarrow PE=863280\ \text{J}

The change in potential energy is 863.28\ \text{kJ}.

8 0
3 years ago
in verification of ohms law the mass is 100g, initial length is 31, final length is 31.3 what is the extension?​
MissTica

Answer:

24

Explanation:

6 0
3 years ago
A thick casting with a thermal diffusivity of 5 x 10-6 m2/s is initially at a uniform temperature of 150oC. One surface of the c
fredd [130]

Answer:

T_o = 141.81 ^0C

Explanation:

Given that;

Thermal diffusivity \alpha = 5 \times 10 ^{-6} m^2/s

Thermal conductivity k = 20 \ W/m.K

Heat transfer coefficient h = ( we are to assume the imposed surface temperature ) = 20 W/m².K

Initial temperature = 150 ° C = (150+273) K = 423 K

Then coolant temperature with which the casting is exposed to = 20° C = (20+273)K = 293 K

Time = 40 seconds

Length = 20mm = 0.02 m

The objective is to determine the  temperature at the surface  at a depth of 20 mm after 40 seconds.

Bi = \dfrac{hL}{k}

Bi = \dfrac{20*0.02}{20}

Bi == 0.02

\tau = \dfrac{\alpha t}{L^2}

\tau=  \dfrac{5*10^{-6 }* 40}{0.020^2}

\tau = 0.5

For a wall at 0.2 Bi

A_1 = 1.0311

\lambda _1 = 0.4328

Therefore;

\dfrac{T_o - T_{\infty}}{T_i - T_{\infty}}= A_1 e ^{-( \lambda_1^2 \ \tau)

\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.438^2 \times 0.5 )

\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.0959 )

\dfrac{T_o - 293 }{130}= 1.0311 \times 0.9085

\dfrac{T_o - 293 }{130}= 0.937

T_o - 293= 0.937 \times 130

T_o - 293= 121.81

T_o = 121.81+ 293

T_o = 414.81 \ K

T_o = (414.81 - 273)^0C

T_o = 141.81 ^0C

4 0
4 years ago
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