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telo118 [61]
3 years ago
12

Plot da(t) if the output voltage of the converter pole-a is vaN(t)=Vd/2+0.85 Vd/2 sin(Ï1t), where Ï1=2Ï x 60 rad/s

Engineering
1 answer:
maw [93]3 years ago
5 0

Answer:

Explanation:

given data :

output voltage ( Van(t) ) = (Vd /2) + (0.85 Vd/2 sin ( w1 t ) )

w1 = 2\pi60 rad/sec

find the value of da(t) by inputting the value of Van (t) into

da = Van(t) / Vd

hence: da(t) = 0.5 + 0.425 sin ((2\pi60)t)

attached below is the plot of the da(t) against time  

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Write a complete C++ program that is made of functions main() and rShift(). The rShift() function must be a function without ret
erma4kov [3.2K]

Answer:

Explanation:

attached is an uploaded picture to support the answer.

the program is written thus;

#include<iostream>

using namespace std;

// function declaration

void rShift(int&, int&, int&, int&, int&, double&);

int main()

{

   // declare the variables

   int a1, a2, a3, a4;

   int maximum = 0;

   double average = 0;

   // inputting the numbers

   cout << "Enter all the 4 integers seperated by space -> ";

   cin >> a1 >> a2 >> a3 >> a4;

   cout << endl << "Value before shift." << endl;

   cout << "a1 = " << a1 << ", a2 = " << a2 << ", a3 = "  

        << a3 << ", a4 = " << a4 << endl << endl;

   // calling rSift()

   // passing the actual parameters

   rShift(a1,a2,a3,a4,maximum,average);

   // printing the values

   cout << "Value after shift." << endl;

   cout << "a1 = " << a1 << ", a2 = " << a2 << ", a3 = "  

           << a3 << ", a4 = " << a4 << endl << endl;

   cout << "Maximum value is: " << maximum << endl;

   cout << "Average is: " << average << endl;

}

// function to right shift the parameters circularly

// and calculate max, average of the numbers

// and return it to main using pass by reference

void rShift(int &n1, int &n2, int &n3, int &n4, int &max, double &avg)

{

   // calculating the max

   max = n1;

   if(n2 > max)

     max = n2;

   if(n3 > max)

     max = n3;

   if(n4 > max)

     max = n4;

   // calculating the average

   avg = (double)(n1+n2+n3+n4)/4;

   // right shifting the numbers circulary

   int temp = n2;

   n2 = n1;

   n1 = n4;

   n4 = n3;

   n3 = temp;

}

8 0
3 years ago
100 kg of R-134a at 200 kPa are contained in a piston–cylinder device whose volume is 12.322 m3. The piston is now moved until t
LekaFEV [45]

Answer:

T=151 K, U=-1.848*10^6J

Explanation:

The given process occurs when the pressure is constant. Given gas follows the Ideal Gas Law:

 pV=nRT

For the given scenario, we operate with the amount of the gas- n- calculated in moles. To find n, we use molar mass: M=102 g/mol.  

Using the given mass m, molar mass M, we can get the following equation:  

 pV=mRT/M

To calculate change in the internal energy, we need to know initial and final temperatures. We can calculate both temperatures as:

T=pVM/(Rm); so initial T=302.61K and final T=151.289K

 

Now we can calculate change of U:

U=3/2 mRT/M using T- difference in temperatures

 U=-1.848*10^6 J

Note, that the energy was taken away from the system.  

5 0
3 years ago
Find the time-domain sinusoid for the following phasors:_________
sattari [20]

<u>Answer</u>:

a.  r(t) = 6.40 cos (ωt + 38.66°) units

b.  r(t) = 6.40 cos (ωt - 38.66°) units

c.  r(t) = 6.40 cos (ωt - 38.66°) units

d.  r(t) = 6.40 cos (ωt + 38.66°) units

<u>Explanation</u>:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = \sqrt{a^2 + b^2}

Ф = direction = tan⁻¹ (\frac{b}{a})

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

<em>(i) convert to polar form</em>

r = \sqrt{5^2 + 4^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{4}{5})

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

<em>(i) convert to polar form</em>

r = \sqrt{5^2 + (-4)^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{-4}{5})

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

<em>(i) convert to polar form</em>

r = \sqrt{(-5)^2 + 4^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{4}{-5})

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

<em>(i) convert to polar form</em>

r = \sqrt{(-5)^2 + (-4)^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{-4}{-5})

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

3 0
3 years ago
Technician A states that a brake lathe is used to make a used brake rotor surface "like new". Technician B states that a brake l
nikitadnepr [17]

Answer:

Both Technician A and B are correct.

Explanation: A brake lathe is a special tool used to improve or work on the surface of brake pads it helps to smoothen the surface.

Brake lathe has been found to be very effective in removing rusts in rotors and unevenness in the brake pad surfaces in order to ensure the efficiency and effectiveness of the brake system of a vehicle. Hence, a brake lathe helps to make brake rotor surface as smooth as possible.

7 0
3 years ago
The welding method that requires the operator to observe and only make corrections is
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Automatic manual semiautomatic
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