Ayuda porfavor es para una tarea de mi capacitación de desarrollo microempresarial

a coil consists of 200 turns of copper wire and has a cross-sectional area of 0.8mm square . The mean length per turn is 80 cm a

Amanda [17]

Answer:

The picture below with the answer. Hope it helps, have a great day/night and stay safe! Length of the coil,

8 0

3 years ago

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

CS3733: Homework/Practice 05 Suppose we would like to write a program called monitor which allows two other programs to communic

valina [46]

Answer:

#include<stdio.h>

#include<stdlib.h>

#include<unistd.h>

#include<sys/types.h>

#include<string.h>

#include<pthread.h>

//#include<sys/wait.h>

int main(int argc, char** argv)

{

int fd1[2];

int fd2[2];

int fd3[2];

int fd4[2];

char message[] = "abcd";

char input_str[100];

pid_t p,q;

if (pipe(fd1)==-1)

{

fprintf(stderr, "Pipe Failed" );

return 1;

}

if (pipe(fd2)==-1)

{

fprintf(stderr, "Pipe Failed" );

return 1;

}

if (pipe(fd3)==-1)

{

fprintf(stderr, "Pipe Failed" );

return 1;

}

if (pipe(fd4)==-1)

{

fprintf(stderr, "Pipe Failed" );

return 1;

}

p = fork();

if (p < 0)

{

fprintf(stderr, "fork Failed" );

return 1;

}

// child process-1

else if (p == 0)

{

close(fd1[0]);// Close reading end of first pipe

char concat_str[100];

printf("\n\tEnter meaaage:"):

scanf("%s",concat_str);

write(fd1[1], concat_str, strlen(concat_str)+1);

// Concatenate a fixed string with it

int k = strlen(concat_str);

int i;

for (i=0; i<strlen(fixed_str); i++)

{

concat_str[k++] = fixed_str[i];

}

concat_str[k] = '\0';//string ends with '\0'

// Close both writting ends

close(fd1[1]);

wait(NULL);

//.......................................................................

close(fd2[1]);

read(fd2[0], concat_str, 100);

if(strcmp(concat_str,"invalid")==0)

{

printf("\n\tmessage not send");

}

else

{

printf("\n\tmessage send to prog_2(child_2).");

}

close(fd2[0]);//close reading end of pipe 2

exit(0);

}

else

{

close(fd1[1]);//Close writting end of first pipe

char concat_str[100];

read(fd1[0], concal_str, strlen(concat_str)+1);

close(fd1[0]);

close(fd2[0]);//Close writing end of second pipe

if(/*check if msg is valid or not*/)

{

//if not then

write(fd2[1], "invalid",sizeof(concat_str));

return 0;

}

else

{

//if yes then

write(fd2[1], "valid",sizeof(concat_str));

close(fd2[1]);

q=fork();//create chile process 2

if(q>0)

{

close(fd3[0]);/*close read head offd3[] */

write(fd3[1],concat_str,sizeof(concat_str);//write message by monitor(main process) using fd3[1]

close(fd3[1]);

wait(NULL);//wait till child_process_2 send ACK

//...........................................................

close(fd4[1]);

read(fd4[0],concat_str,100);

close(fd4[0]);

if(sctcmp(concat_str,"ack")==0)

{

printf("Messageof child process_1 is received by child process_2");

}

else

{

printf("Messageof child process_1 is not received by child process_2");

}

else

{

if(p<0)

{

printf("Chiile_Procrss_2 not cheated");

}

else

{

close(fd3[1]);//Close writing end of first pipe

char concat_str[100];

read(fd3[0], concal_str, strlen(concat_str)+1);

close(fd3[0]);

close(fd4[0]);//Close writing end of second pipe

write(fd4[1], "ack",sizeof(concat_str));

}

close(fd2[1]);

}

8 0

3 years ago

An automated transfer line is to be designed. Based on previous experience, the average downtime per occurrence = 5.0 min, and t

IRINA_888 [86]

Answer:

a) 28 stations

b) Rp = 21.43

E = 0.5

Explanation:

Given:

Average downtime per occurrence = 5.0 min

Probability that leads to downtime, d= 0.01

Total work time, Tc = 39.2 min

a) For the optimum number of stations on the line that will maximize production rate.

Maximizing Rp =minimizing Tp

Tp = Tc + Ftd

$= \frac{39.2}{n} + (n * 0.01 * 5.0)$

$= \frac{39.2}{n} + (n * 0.05)$

At minimum pt. = 0, we have:

dTp/dn = 0

$= \frac{-39.2}{n^2} + 0.05 = 0$

Solving for n²:

$n^2 = \frac{39.2}{0.05} = 784$

$n = \sqrt{784} = 28$

The optimum number of stations on the line that will maximize production rate is 28 stations.

b) $Tp = \frac{39.2}{28} + (28 * 0.01 * 5)$

Tp = 1.4 +1.4 = 2.8

The production rate, Rp =

$\frac{60min}{2.8} = 21.43$

The proportion uptime,

$E = \frac{1.4}{2.8} = 0.5$

3 0

3 years ago

How many 10" diameter circles can be cut from a semicircular shape that has a 20"

N76 [4]

9514 1404 393

Answer:

1

Explanation:

Only one such circle can be drawn. The diameter of the 10" circle will be a radius of the semicircle. In order for the 10" circle to be wholly contained, the flat side of the semicircle must be tangent to the 10" circle. There is only one position in the figure where that can happen. (see attached).

3 0

3 years ago

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