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2 years ago

6

Consider a half-wave rectifier circuit with a triangular-wave input of LaTeX: 6V6 V peak-to-peak amplitude and zero average valu

e. For LaTeX: R=1kΩR = 1 k Ω, and assuming that the diode can be represented by a constant voltage drop model with LaTeX: V_D=0.7VV D = 0.7 V.

1 answer:

Varvara68 [4.7K]2 years ago

8 0

The question is not complete. We are supposed to find the average value of v_o.

Answer:

v_o,avg = 0.441V

Explanation:

Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;

3/(T/4) = 0.7/t1

So, 12/T = 0.7/t1

So, t1 = 0.7T/12

t1 = 0.0583 T

Also, from symmetry of triangles,

t2 = T/2 - t1

So, t2 = T/2 - 0.0583 T

t2 = 0.4417T

Average of voltage output is;

v_o,avg = (1/T) x Area under small triangle

v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)

v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)

v_o,avg = (1/T) x 2.3 x 0.1917T

T will cancel out to give;

v_o,avg = 0.441V

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Your Java program will be reading input from a file name strInput.txt. Each record contains String firstname String lastName Str

stiks02 [169]

Answer:

The program requires that you have the specified input files and it reads from each file at a time and processes salary in digits, states the city, state and bonus with respective first and last name as requested in the question. Note that you must have access to the mentioned output files for the program to work properly. Below is the java version of the program.

import java.io.File;

import java.io.FileNotFoundException;

import java.io.PrintWriter;

import java.util.Scanner;

class Driver

{

public static void main(String[] args) throws FileNotFoundException

{

Scanner sc = new Scanner(new File("strInput.txt"));

PrintWriter pd = new PrintWriter(new File("strOutputD"));

PrintWriter prf = new PrintWriter(new File("strOutputRF"));

String firstname = "", lastname = "", strSalary = "", status = "", cityState = "", city = "", state = "";

double salary = 0, bonus = 0;

int incorrectRecords = 0;

int dRecords = 0;

int fRecords = 0;

while(sc.hasNextLine())

{

firstname = sc.next();

lastname = sc.next();

strSalary = sc.next();

status = sc.next();

cityState = sc.next();

if(!status.equals("D") && !status.equals("F"))

{

System.out.println("Records is neither D nor F. Skipping this...");

incorrectRecords++;

continue;

}

else if(status.equals("D") || status.equals("F"))

{

char c = ' ';

int i = 0;

for(i=0; i<strSalary.length() && c != '.'; i++)

{

c = strSalary.charAt(i);

if(!Character.isDigit(c))

{

System.out.println("Char at position " + (i+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

}

if(c == '.')

{

if(i+1 == strSalary.length()-1)

{

if(!Character.isDigit(strSalary.charAt(i)))

{

System.out.println("Char at position " + (i+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

if(!Character.isDigit(strSalary.charAt(i+1)))

{

System.out.println("Char at position " + (i+1+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

}

else

{

System.out.println("Period is in the wrong position. Expected at " + (strSalary.length()-3) + " but found at " + (i+1));

continue;

}

}

city = cityState.split(",")[0];

state = cityState.split(",")[1];

salary = Double.parseDouble(strSalary);

if(status.equals("D"))

{

bonus = salary * 0.125;

dRecords++;

pd.write(firstname + " " + lastname + " " + status + " " + salary + " " + bonus + " " + city + " " + state);

}

else

{

bonus = salary * 0.18;

fRecords++;

prf.write(firstname + " " + lastname + " " + status + " " + salary + " " + bonus + " " + city + " " + state);

}

}

}

System.out.println("No of D records : " + dRecords);

System.out.println("No of F records : " + fRecords);

System.out.println("No of incorrect records : " + incorrectRecords);

}

}

6 0

2 years ago

A steel plate has a hole drilled through it. The plate is put into a furnace and heated. What happens to the size of the inside

djverab [1.8K]

Answer:

The diameter increases

Explanation:

The expansion in the metal is uniform in every dimension

4 0

2 years ago

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t

saveliy_v [14]

Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is $=21.29mm$

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as $\sigma_T$ .

True Strain

A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as $\epsilon_T$ .

The mathematical relation between stress to strain on the plastic region of deformation is

$\sigma _T =K\epsilon^n_T$

Where K is a constant

n is known as the strain hardening exponent

This constant K can be obtained as follows

$K = \frac{\sigma_T}{(\epsilon_T)^n}$

No substituting $345MPa \ for \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and \ 0.22 \ for \ n$ from the question we have

$K = \frac{345}{(0.02)^{0.22}}$

$= 815.82MPa$

Making $\epsilon_T$ the subject from the equation above

$\epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }$

$Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K \ and \ 0.22 \ for \ n$

$\epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }$

$=0.0443$

From the definition we mentioned instantaneous length and this can be obtained mathematically as follows

$l_i = l_o e^{\epsilon_T}$

Where

$l_i$ is the instantaneous length

$l_o$ is the original length

$Substituting \ 470mm \ for \ l_o \ and \ 0.0443 \ for \ \epsilon_T$

$l_i = 470 * e^{0.0443}$

$=491.28mm$

We can also obtain the elongated length mathematically as follows

$Elongated \ Length =l_i - l_o$

$Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i$

$Elongated \ Length = 491.28 - 470$

$=21.29mm$

4 0

3 years ago

HELP PLEASE!!!!!!!!!!!

MAXImum [283]

C is your answers!!!!!$3&2)//

7 0

3 years ago

Read 2 more answers

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and

k0ka [10]

Answer:

A.) 0.3088

B.) 0.0017

C.) part A

Explanation:

A.)

$z1=$ $\frac{\left(150-137\right)}{27.7}$ $=0.4693$

$z2=\frac{\left(201-137\right)}{27.7}=2.3105$

$P(0.4693$

B.)

$z1=\frac{150-137}{27.7/ \sqrt{39}} =2.9309\\z2=\frac{201-137}{27.7/ \sqrt{39}}=14.4289$

$\\P(2.9309$

C.) Since the seat performance for an individual pilot is more important than 39 different pilots.

3 0

3 years ago

Read 2 more answers