What phenomenon allows water to reach the top of a building?

An aluminium alloy tube has a length of 750 mm at a temperature of 223°C. What will be its length at 23°C if its coefficient of

uranmaximum [27]

Answer:

Final length= 746.175 mm

Explanation:

Given that Length of aluminium at 223 C is 750 mm.As we know that when temperature of material increases or decreases then dimensions of material also increases or decreases respectively with temperature.

Here temperature of aluminium decreases so the final length of aluminium decreases .

As we know that

$\Delta L=L\alpha\Delta T$

Now by putting the values

$\Delta L=750\times \25.5\times 10^{-6}\times 200$

ΔL=3.82 mm

So final length =750-3.82 mm

Final length= 746.175 mm

3 0

3 years ago

what is an example of an innovative solution to an engineering problem? Explain briefly why you chose this answer.

Leviafan [203]

Answer:

robotic technology

Explanation:

Innovation is nothing but the use of various things such as ideas, products, people to build up a solution for the benefit of the human. It can be any product or any solution which is new and can solve people's problems.

Innovation solution makes use of technology to provide and dispatch new solutions or services which is a combination of both technology and ideas.

One such example of an innovative solution we can see is the use of "Robots" in medical science or in any military operations or rescue operation.

Sometimes it is difficult for humans to do everything or go to everywhere. Thus scientist and engineers have developed many advance robots or machines using new ideas and technology to find solutions to these problems.

Using innovations and technologies, one can find solutions to many problems which is difficult for the peoples. Robots can be used in any surveillance operation or in places of radioactive surrounding where there is a danger of humans to get exposed to such threats. They are also used in medical sciences to operate and support the patient.

3 0

3 years ago

2.) A fluid moves in a steady manner between two sections in a flow

Talja [164]

Answer:

$250\ \text{lbm/min}$

$625\ \text{ft/min}$

Explanation:

$A_1$ = Area of section 1 = $10\ \text{ft}^2$

$V_1$ = Velocity of water at section 1 = 100 ft/min

$v_1$ = Specific volume at section 1 = $4\ \text{ft}^3/\text{lbm}$

$\rho$ = Density of fluid = $0.2\ \text{lb/ft}^3$

$A_2$ = Area of section 2 = $2\ \text{ft}^2$

Mass flow rate is given by

$m=\rho A_1V_1=\dfrac{A_1V_1}{v_1}\\\Rightarrow m=\dfrac{10\times 100}{4}\\\Rightarrow m=250\ \text{lbm/min}$

The mass flow rate through the pipe is $250\ \text{lbm/min}$

As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

$m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}$

The speed at section 2 is $625\ \text{ft/min}$ .

3 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

The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with di

jeka57 [31]

Answer:

a) at T = 5800 k

band emission = 0.2261

at T = 2900 k

band emission = 0.0442

b) daylight (d) = 0.50 μm

Incandescent ( i ) = 1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

Values are gotten from the table named: blackbody radiation functions

a) Calculate the band emission fractions for the visible region

at T = 5800 k

band emission = 0.2261

at T = 2900 k

band emission = 0.0442

attached below is a detailed solution to the problem

b)calculate wavelength corresponding to the maximum spectral intensity

For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

3 0

2 years ago