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.
Answer:
a) 0.39795 kJ/K
b) 79.589.37 kJ
Explanation:
m = Mass of air = 2 kg
Temperature = 200 K
P₁ = Initial pressure = 300 kPa
P₂ = Final pressure = 600 kPa
R = mass-specific gas constant for air = 287.058 J/kgK
a) For isentropic process
∴ Entropy is generated in the process is 0.39795 kJ/K
b)
∴ Amount of lost work is 79.589.37 kJ
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
In the given circuit, V(t)=12cos(2000t+45)V, R1=R2=2Ω, L1=L2=L3=3mH and C1=250μF. You are required to find the Thevenin equivalent of this circuit using phasors.
a. If you write the Thevenin voltage, Vth, in phasor form, what is the magnitude of this phasor? Put your answer in the box below without units.
b. What is the value of the angle associated with the phasor Vth, in degrees?
c. Now, calculate the Thevenin impedance, Zth. What is the magnitude of this phasor?
d. What is the angle associated with the phasor Zth, in degrees?
Answer:
Vth = 6 < 45° V
Zth = 1.414 < 45°
a. The magnitude of the Thevenin voltage is 6 V
b. The phase angle of the Thevenin voltage is 45°
c. The magnitude of the Thevenin impedance is 1.414 V
d. The phase angle of the Thevenin impedance is 45°
Explanation:
The given voltage is
V(t)=12cos(2000t+45)
In phasor form,
V(t) = 12 < 45° V
So the magnitude of voltage is 12 V and the phase angle is 45°
Also the frequency ω = 2000
then the inductance is
L₁ = L₂ = L₃ = jωL = j×2000×0.003 = j6 Ω
and the capacitance is
C₁ = 1/jωC = 1/(j×2000×250x10⁻⁶) = -j2 Ω
and the resistance is
R₁ = R₂ = 2 Ω
Thevenin voltage:
The Thevenin voltage is the voltage that appears across the open-circuited terminals a-b (after removing L₃)
The Thevenin voltage is given by
Vth = V(t) × [ (R₂ + L₂) / (R₁ + L₁) + (R₂ + L₂) ]
Please note that there is no current flow in the capacitor due to open-circuited terminal a-b
Vth = 12 < 45° × [ (2 + j6) / (2 + j6) + (2 + j6) ]
Vth = 12 < 45° × [ (2 + j6) / (4 + j12) ]
Vth = 4.24264 + j4.24264 V
In phasor form,
Vth = 6 < 45° V
a. The magnitude of the Thevenin voltage is 6 V
b. The phase angle of the Thevenin voltage is 45°
Thevenin Impedance:
The Thevenin Impedance is the impedance of the circuit calculated when looking from the terminal a-b
Zth = [ (R₁ + L₁) × (R₂ + L₂) / (R₁ + L₁) + (R₂ + L₂) ] + (-j2)
Zth = [ (2 + j6) × (2 + j6) / (2 + j6) + (2 + j6) ] - j2
Zth = [ (2 + j6) × (2 + j6) / (2 + j6) + (2 + j6) ] - j2
Zth = [ (-32 + j24) / (4 + j12) ] -j2
Zth = [ (1 +j3) ] - j2
Zth = 1 + j Ω
In phasor form,
Zth = 1.414 < 45°
c. The magnitude of the Thevenin impedance is 1.414 V
d. The phase angle of the Thevenin impedance is 45°
Answer:
a) the directivity of the antenna is evaluated to be = 1.761db
b) radiation efficiency is evaluated to be = 98.67%
c) reflection efficiency gain is evaluated to be = 68%
Explanation:
kindly check the attachment for step to step explanations for better understanding.
Answer:
The power produced by the turbine is 23309.1856 kW
Explanation:
h₁ = 3755.39
s₁ = 7.0955
s₂ = sf + x₂sfg =
Interpolating fot the pressure at 3.25 bar gives;
570.935 +(3.25 - 3.2)/(3.3 - 3.2)*(575.500 - 570.935) = 573.2175
2156.92 +(3.25 - 3.2)/(3.3 - 3.2)*(2153.77- 2156.92) = 2155.345
h₂ = 573.2175 + 0.94*2155.345 = 2599.2418 kJ/kg
Power output of the turbine formula =
Which gives;
= -8*((2599.2418 - 3755.39)+(15^2 - 60^2)/2 ) = -22749.1856
= -22749.1856 - 560 = -23309.1856 kJ
= 23309.1856 kJ
Power produced by the turbine = Work done per second = 23309.1856 kW.