According to mcclelland human motivation theory, individuals primarily motivated by achievement performance to work in teams

The following electrical characteristics have been determined for both intrinsic and p-type extrinsic gallium antimonide (GaSb)

xxTIMURxx [149]

Answer:

0.5m^2/Vs and 0.14m^2/Vs

Explanation:

To calculate the mobility of electron and mobility of hole for gallium antimonide we have,

$\sigma = n|e|\mu_e+p|e|\mu_h$ (S)

Where

e= charge of electron

n= number of electrons

p= number of holes

$\mu_e=$ mobility of electron

$\mu_h=$ mobility of holes

$\sigma =$ electrical conductivity

Making the substitution in (S)

Mobility of electron

$8.9*10^4=(8.7*10^{23}*(-1.602*10^{-19})*\mu_e)+(8.7*10^{23}*(-1.602*10^{-19})*\mu_h)$

$0.639=\mu_e+\mu_h$

Mobility of hole in (S)

$2.3*10^5 = (7.6*10^{22}*(-1.602*10^{-19})*\mu_e)+(1*10^{25}*(-1.602*10^{-19}*\mu_h))$

$0.1436 = 7.6*10^{-3}\mu_e+\mu_h$

Then, solving the equation:

$0.639=\mu_e+\mu_h$ (1)

$0.1436 = 7.6*10^{-3}\mu_e+\mu_h$ (2)

We have,

Mobility of electron $\mu_e = 0.5m^2/V.s$

Mobility of hole is $\mu_h = 0.14m^2/V.s$

6 0

3 years ago

You have a motor such that if you give it 12 Volt, it will eventually reach a steady state speed of 200 rad/s. If it starts from

Aleksandr [31]

Answer:

a) $\frac{Ws}{Es} = \frac{200}{1+1.2s}$

b) attached below

c) type zero system

d) k > $\frac{g}{200}$

e) The gain K increases above % error as the steady state speed increases

Explanation:

Given data:

Motor voltage = 12 v

steady state speed = 200 rad/s

time taken to reach 63.2% = 1.2 seconds

<u>a) The transfer function of the motor from voltage to speed</u>

let ; $\frac{K1}{1+St}$ be the transfer function of a motor

when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec

hence the transfer function of the motor from voltage to speed

= $\frac{Ws}{Es} = \frac{200}{1+1.2s}$

<u>b) draw the block diagram of the system with plant controller and the feedback path </u>

attached below is the remaining part of the detailed solution

c) The system is a type-zero system because the pole at the origin is zero

d) ) k > $\frac{g}{200}$

7 0

3 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

An escalator handles a steady load of 26 people per minute in elevating them from the first to the second floor through a vertic

photoshop1234 [79]

Answer:

$\eta = 70.711\,\%$

Explanation:

The power needed to make the escalator working is obtained by means of the Work-Energy Theorem:

$\dot W = \dot U_{g}$

$\dot W = \dot n \cdot m_{p}\cdot g \cdot \Delta y$

$\dot W = \left(26\,\frac{persons}{min}\right)\cdot (124\,lbm)\cdot \left(32.174\,\frac{ft}{s^{2}}\right)\cdot \left(\frac{1\,lbf}{32.174\,\frac{lbm\cdot ft}{s^{2}} } \right)\cdot (27.5\,ft)$

$\dot W = 88660\,\frac{lbf\cdot ft}{min}\,\left(2.687\,hp\right)$

The mechanical efficiency of the escalator is:

$\eta = \frac{2.687\,hp}{3.8\,hp}\times 100\,\%$

$\eta = 70.711\,\%$

3 0

3 years ago

How do you use the brakes in an airplane?

Paraphin [41]

Answer:

When a pilot pushes the top of the right pedal, it activates the brakes on the right main wheel/wheels, and when the pilot pushes the top of the left rudder pedal, it activates the brake on the left main wheel/wheels. The brakes work in a rather simple way: they convert the kinetic energy of motion into heat energy.

3 0

3 years ago