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4 years ago

In modern construction, studs are usually cut to exact length at the mill. They are designated by the letters

Engineering

1 answer:

ehidna [41]4 years ago

3 0

Answer:

P.E.T

Explanation:

Stud is a short form of stud wall which are typically vertical framing members in a wall of a building. Wall studs are available in standard lengths but since the height of ceiling vary, sometimes one has to cut the stud walls on site to fit. Therefore, there are pre-cut studs which are more efficient in implementation. These are usually referred as Precision-End Trimmed studs, (P.E.T). Therefore, stud walls which are cut to exact length at the mill for modern construction are designated by letters P.E.T

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Peter the postman became bored one night and, to break the monotony of the night shift, he carried out the following experiment

aleksandrvk [35]

Answer:

//This Program is written in C++

// Comments are used for explanatory purpose

#include <iostream>

using namespace std;

enum mailbox{open, close};

int box[149];

void closeAllBoxes();

void OpenClose();

void printAll();

int main()

{

closeAllBoxes();

OpenClose();

printAll();

return 0;

}

void closeAllBoxes()

{

for (int i = 0; i < 150; i++) //Iterate through from 0 to 149 which literarily means 1 to 150

{

box[i] = close; //Close all boxes

}

void OpenClose()

{

for(int i = 2; i < 150; i++) {

for(int j = i; j < 150; j += i) {

if (box[j] == close) //Open box if box is closed

box[j] = open;

else

box[j] = close; // Close box if box is opened

}

// At the end of this test, all boxes would be closed

}

void printAll()

{

for (int x = 0; x < 150; x++) //use this to test

{

if (box[x] = 1)

{

cout << "Mailbox #" << x+1 << " is closed" << endl;

// Print all close boxes

}

Explanation:

7 0

3 years ago

An ideal gas initially occupying 0.020 m3 at 1.0 MPa is quasistatically expanded inside a piston-cylinder device at a constant p

egoroff_w [7]

An ideal gas initially occupying 0.020 m3 at 1.0 MPa is quasistatically expanded inside a piston-cylinder device at a constant pressure until its volume doubles. Next the expansion is continued at constant volume till the pressure reaches half of the initial pressure. Finally it is brought back to the initial state in a polytropic process with exponent n=1.6

a. Draw the processes on a P-v diagram and calculate the total work.
b. Calculate the total heat transfered, what is the difference between the initial and final temperature?an answer is to present a question of how you are not able to join the world and how you can help please answer

8 0

3 years ago

Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the

myrzilka [38]

Answer:

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

$\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}$

The head associated with the pump is cleared:

$h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})$

Inlet and outlet velocities are found:

$v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }$

$v_{1} \approx 4.278\,\frac{m}{s}$

$v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }$

$v_{2} \approx 11.573\,\frac{m}{s}$

Now, the head associated with the pump is finally computed:

$h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m$