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Hitman42 [59]
3 years ago
8

Which of the following is MOST likely to be true about a service manager?

Engineering
1 answer:
forsale [732]3 years ago
7 0

Answer:

a.they are responsible for the function of the service department

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Storm sewer backup causes your basement to flood at the steady rate of 1 in. of depth per hour. The basement floor area is 2600
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Answer:

attached below

Explanation:

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3 years ago
State five applications of thermochromic materials
rusak2 [61]

Explanation:

The end-use industries of thermochromic materials include packaging, printing & coating, medical, textile, industrial, and others. Printing & coating is the fastest-growing end-use industry of thermochromic materials owing to a significant increase in the demand for thermal paper for POS systems. The use of thermochromic materials is gaining momentum for interactive packaging that encourages consumers to take a product off the shelf and use it.

8 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
4 years ago
In order to impress your neighbors and improve your vision in traffic jams, you decide to mount a cylindrical periscope 2.0 m hi
kondaur [170]
Follow @richard.gbe on Instagram for the answer
5 0
4 years ago
Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale
artcher [175]

The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

  • HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

<h3>How can percentage conversion give the contents of the product stream?</h3>

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ \longrightarrow 2Cl₂ + 2H₂O

Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \  of \ limiting \ reactant \ supplied \ in \ the \, feed}}

Which gives;

80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}

Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h

Which gives;

Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h

Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Similarly;

Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h

The composition of the production stream is therefore;

  • <u>HCl: 0.4 kmol/h</u>
  • <u>Cl₂: 1.6 kmol/h</u>
  • <u>H₂O: 1.6 kmol/h</u>
  • <u>O₂: 0.5 kmol/h</u>

Learn more about theoretical and actual yield here:

brainly.com/question/14668990

brainly.com/question/82989

7 0
3 years ago
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