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Lostsunrise [7]
3 years ago
6

How to walk a dog dududududududududesssss

Engineering
2 answers:
bagirrra123 [75]3 years ago
6 0
Pick up a leash put the dog on the leash open the door/cage take the dog out and walk it
rusak2 [61]3 years ago
4 0
Putting them on a leash and picking a route and walking it
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A gas flows through a one-inlet, one-exit control volume operating at steady state. Considering an adiabatic control volume with
Hunter-Best [27]

Answer:

b. equal to the specific entropy of the gas at the inlet.

Explanation:

Isentropic process is the process in which the entropy of the system remains unchanged. The word isentropic is formed from the combination of the prefix "iso" which means "equal" and the word entropy.

If a process is completely reversible, without the need to provide energy in the form of heat, then the process is isentropic.

3 0
3 years ago
Compute L, T, M, LC, and R and stations of the BC and EC for the circular curve with the given data of: I (delta) = 22°15′00" an
Mars2501 [29]

Answer:

L = 475.718

T = 240.89 ft

M = 23.0195

LC = 472.728

R = 1225 ft

Explanation:

See the attached file for the calculation.

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
3 years ago
A 1000 kg turbine has a rotating unbalance of 0.1 kg.m. The turbine operates at a speed between 500 to 750 rpm. What is the maxi
raketka [301]

Answer:

maximum isolator stiffness k =1764 kN-m

Explanation:

mean speed of rotation =\frac{N_1 +N_2}{2}

Nm = \frac{500+750}{2} = 625 rpm

w =\frac{2\pi Nm}{60}

  =65.44 rad/sec

F_T = mw^2 e

F_T = mew^2

       = 0.1*(65.44)^2

F_T =428.36 N

Transmission ratio =\frac{300}{428.36} = 0.7

also

transmission ratio = \frac{1}{[\frac{w}{w_n}]^{2} -1}

0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}

SOLVING FOR Wn

Wn = 42 rad/sec

Wn = \sqrt {\frac{k}{m}

k = m*W^2_n

k = 1000*42^2 = 1764 kN-m

k =1764 kN-m

3 0
3 years ago
A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane-strain fracture tough
jeyben [28]

Complete question:

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.

Answer:

Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection

so that critical flow is subject to detection  

Explanation:

We are given:

Plane strain fracture toughness K = 98.9 MPa \sqrt{m}

Yield strength Y = 860 MPa

Flaw detection apparatus = 3.0mm (12in)

y = 1.0

Let's use the expression:

oc = \frac{K}{Y \sqrt{pi * a}}

We already know

K= design

a = length of surface creak

Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.

Therefore

a= \frac{1}{pi} * [\frac{k}{y*a}]^2

Substituting figures in the expression above, we have:

= \frac{1}{pi} * [\frac{98.9 MPa \sqrt{m}} {10 * \frac{860MPa}{2}}]^2

= 0.0168 m

= 17mm

Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection  

3 0
3 years ago
Read 2 more answers
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