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3 years ago

Ductility (increases/decreases/does not change) with temperature.

Engineering

1 answer:

PSYCHO15rus [73]3 years ago

8 0

Answer:

Increases

Explanation:

Ductility:

Ductility is the property of material to go permanent deformation due to tensile load.In other words the ability of material to deform in wire by the help of tensile load.

When temperature is increase then ductility will also increases.And when temperature decreases then the ductility will also decreases.As we know that at very low temperature material become brittle and this is know as ductile brittle transition.

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This is problem 4 from chapter 6 of the course text. Find the Thevenin equivalent seen at the terminal A-B. Select which answer

miv72 [106K]

Answer:

Option C is correct.

Vs = 9 V, Rth = 30 ohms

Explanation:

In the circuit diagram, there are two sources, so, using superposition, we'll find voltage across A and B, Vab, with respect to each of the sources.

Taking the voltage source as primary source.

We will open circuit the current source just like it is presented in the first drawing of the 2nd image I have added to this solution.

By open-circuiting the current source, the three resistors R₁, R₂ and R₃ are in series.

This means that the Vab is the voltage across the R₃ resistor.

Using voltage divider rule,

Vab = (R₃/R₁ + R₂ + R₃) Vs₁

Vab = [40/(60 + 60 + 40)] (12)

Vab = 3 V

Taking the current source as the primary source.

We will now short circuit the voltage source as shown in the 2nd drawing of the 2nd image I have attached to this solution.

As shown in the 2nd drawing of the 2nd file I attached to this solution, the R₂ is in parallel with R₁ and R₃, that is R₂//(R₁ + R₃)

Using the current divider rule

Current in the (R₁ + R₃) branch = [R₂/(R₂ + (R₁ + R₃))] Is₂ = (60/(60 + 60 + 40)) (0.4) = 0.15 A

Vab = voltage across the R₃ resistor = IR₃ = 0.15 × 40 = 6 V

Total voltage across A and B, Vab = Vab due to Vs₁ + Vab due to Is₂ = 3 + 6 = 9 V

And for the Rth, we open-circuit the current source and short-circuit the voltage source simultaneously and look at the resistance of the setup from the AB terminal, as shown in the first drawing of the 3rd file attached to this solution.

It is evident that R₃ is in a parallel combination with (R₁ + R₂)

Rth = (R₁ + R₂)//R₃ = (60 + 60)//40 = 120//40 = (120× 40)/(120 + 40) = 30 ohms

The image of the thevenin equivalent of the circuit as seen from terminals AB is presented in the 2nd drawing on the 3rd image attached.

5 0

3 years ago

Air flows steadily and isentropically from standard atmospheric conditions to a receiver pipe through a converging duct. The cro

Liula [17]

Answer:

The answer is "0.0728"

Explanation:

Given value:

$P_0= 14.696\ ps\\\\\ p _{0}= 0.00238 \frac{slue}{ft^{3}}\\\\\ A= 0.05 ft^2\\\\\ T_0= 59^{\circ}f = 518.67R\\\\\ air \ k= 1\\\\ \ cirtical \ pressure ( P^*)=P_0\times \frac{2}{k+1}^{\frac{k}{k-1}}\\$

$= 14.696\times (\frac{2}{1.4+1})^{\frac{1.4}{1.4-1}}\\\\=7.763 Psia\\\\$

if $P$ flow is chocked

if $P>P^{*} \to$ flow is not chocked

When P= 10 psia < $P^{*}$ $\to$ not chocked

match number:

$\ for \ P= \ 10\ G= \sqrt{\frac{2}{k-1}[(\frac{\ p_{0}}{p})^{\frac{k-1}{k}}-1]}$

$= \sqrt{\frac{2}{1.4-1}[(\frac{14.696}{10})^{\frac{1.4-1}{1.4}}-1]}$

$M_0=7.625$

$p=p_0(1+\frac{k-1}{2} M_0 r)^\frac{1}{1-k}$

$=0.00238(1+\frac{1.4-1}{2}0.7625`)^{\frac{1}{1-1.4}}\\\\\ p=0.001808 \frac{slug}{ft^3}$

$\ T= T_0(1+\frac{k-1}{2} Ma^r)^{-1}\\\\\ T=518.67(1+\frac{1.4-1}{2} 0.7625^2)^{-1}\\\\\ T=464.6R\\\\$

$\ velocity \ of \ sound \ (C)=\sqrt{KRT}\\\\$

$=\sqrt{1.4\times1716\times464.6}\\\\=1057 ft^3\\\\$

R= gas constant=1716

$m=PAV\\\\$

$=0.001808\times0.05\times(Ma.C)\\\\=0.001808\times0.05\times0.7625\times 1057\\\\=0.0728\frac{slug}{s}$

5 0

3 years ago

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

3 years ago

Two stepped bar is supported at both ends.At the join point of two segments,the force F is applied(downwards).Calculate reactive

nlexa [21]

Answer:

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Explanation:

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Answer:

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Explanation:

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