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3241004551 [841]
3 years ago
15

Consider the following program:

Engineering
1 answer:
EleoNora [17]3 years ago
5 0

Answer:

The solution of this question is given below in explanation section.

Explanation:

The correct answer to this question is A i.e. 112002 .

the correct code of this question is given below

#include  <stdio.h>

#include  <string.h>

#include  <sys/types.h>

//#include "csapp.h"

void end(void)

{

printf("2");

}

int main()

{

if (fork() == 0)

atexit(end);

if (fork() == 0)

printf("0");

else

printf("1");

exit(0);

}

/* $end forkprob2 */

when you run this program, fork function print different values.

However, it is noted that when you run the program repeatedly, it will show you different values. But the most possible value that this program will show is A.

The picture is attached of the running program to get the idea of code.

You might be interested in
The time factor for a doubly drained clay layer
Margarita [4]

Answer with Explanation:

Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is

T_v=\frac{\pi }{4}(\frac{U}{100})^2

Solving for 'U' we get

\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%

Since our assumption is correct thus we conclude that degree of consolidation is 50.46%

The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2

i)\frac{z}{H}=0.25=U=0.71 = 71% consolidation

ii)\frac{z}{H}=0.5=U=0.45 = 45% consolidation

iii)\frac{z}{H}=0.75U=0.3 = 30% consolidation

Part b)

The degree of consolidation is given by

\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm

Thus a settlement of 50.46 centimeters has occurred

For time factor 0.7, U is given by

T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59

thus consolidation of 85.59 % has occured if time factor is 0.7

The degree of consolidation is given by

\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm

5 0
3 years ago
A driver traveling in her 16-foot SUV at the speed limit of 30 mph was arrested for running a red light at 15th and Main, an int
Lynna [10]

Answer:

(a) Yes

(b) 102.8 ft

Explanation:

(a)First let convert mile per hour to feet per second

30 mph = 30 * 5280 / 3600 = 44 ft/s

The time it takes for this driver to decelerate comfortably to 0 speed is

t = v / a = 44 / 10 = 4.4 (s)

given that it also takes 1.5 seconds for the driver reaction, the total time she would need is 5.9 seconds. Therefore, if the yellow light was on for 4 seconds, that's not enough time and the dilemma zone would exist.

(b) At this rate the distance covered by the driver is

s = v_0t + \frac{at^2}{2}

s =44*1.5 + 44(4.4) - \frac{10*4.4^2}{2} = 162.8 (ft)

Since the intersection is only 60 feet wide, the dilemma zone must be

162.8 - 60 = 102.8 ft

4 0
3 years ago
C++ question <br> Thank you
Salsk061 [2.6K]
You're welcome lol! !!!!!!!!!!!!!!

:):):):)
5 0
3 years ago
Discuss the importance of dust and fluff removal from spinning mills and how it is realised in air conditioning plants
Alina [70]

Answer:

Removal of dust and fluff from spinning mill is important as it has adverse and detrimental effects on the health of the workers in these industries. Tiny and microscopic particles of various substances present in the surrounding air is transferred from one place to another and these causes various respiratory diseases and pose health hazards for the workers and make work environment unhealthy and hazardous thus affecting the over all efficiency and productivity.

Cotton dust , the major pollutant, when breathed in affetcs the lungs badly and workers experience symptoms such as respiratory problems, coughing, tightness in chest, etc.  Thus to ensure proper health of the workers spinning mills have been provided with powerful air conditioning to ensure purity of air, to maintain proper moisture levels and to ensure dust and fluff removal.

The dust and fluff laiden air is humidified, purified and then recirculated. Optimization of number of air changes/hour to clean air stream and prevent any health risk of the workers.

8 0
3 years ago
A tank contains initially 2500 liters of 50% solution. Water enters the tank at the rate of 25 iters per minute and the solution
sergejj [24]

Answer:

Percentage of solution=32. 96%

Explanation:

Given that initially tank have 50% solution.It means that amount of solution=0.5 x 2500 =1250 lts

Lets take the amount of solution at time t = A

So

\dfrac{dA}{dt}=rate\ in\ solution -rate\ out\ solution

\dfrac{dA}{dt}=concentration\times rate\ in\ of\ water -concentration\times rate\ out\ of\ water

\dfrac{dA}{dt}=0\times 25-\dfrac{A}{2500}\times 50

\dfrac{dA}{dt}=-\dfrac{A}{50}

Now by integration

\ln A=-\dfrac{1}{50}t+C

Where C is the constant

Given that at t=0 ,A=1250

So C=\ln 1250

\ln A=-\dfrac{1}{50}t+\ln 1250

When t=20 min

\ln A=-\dfrac{1}{50}\times 20+\ln 1250

A=837.90

So percentage of solution after 20 min

=\dfrac{1250-837.9}{1250}\times 100

Percentage of solution=32. 96%

4 0
3 years ago
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