6. Staples are the most common item used to secure and support cables in residential wiring.

Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr

son4ous [18]

Answer:

The entropy change of the air is $0.240kJ/kgK$

Explanation:

$T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa$

$T_{2}$ is unknown

we can apply the following expression to find $T_{2}$

$-w_{out} =mc_{v} (T_{2} -T_{1} )$

$T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }$

now substitute

$T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K$

To find entropy change of the air we can apply the ideal gas relationship

Δ $s_{air}$ $=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }$

Δ $s_{air} =$ $1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )$

Δ $s_{air} =0.240kJ/kgK$

4 0

3 years ago

The best saw for cutting miter joints is the

ZanzabumX [31]

Answer:

The best saw for cutting miter joints is the backsaw.

Add-on:

i hope this helped at all.

6 0

3 years ago

What are 4 activities that you do in your daily life that DO require energy?

SashulF [63]

Answer: Some activities that I do in my daily life that require energy are:

1. Doing ballet

2. Studying

3. Walking up and down stairs

4. Stretching

5. Running on the treadmill

Hope this helps! :)

Explanation:

4 0

3 years ago

For a project in C++ we are supposed toDesign a class named Month. The class should have the following private members:-name: a

mart [117]

Answer:

include <iostream>

using namespace std;

class Month

{

public:

Month (char firstLetter, char secondLetter, char thirdLetter);

Month (int monthNum);

.

Month();

void outputMonth_num();

void outputMonthLetters();

private:

int month;

};

int main ()

{

//

// Variable declarations

//

int monthNum;

char firstLetter, secondLetter, thirdLetter;

char testAgain;

do {

cout << endl;

cout << "Testing the default constructor ..." << endl;

Month defaultMonth;

defaultMonth.outputMonth_num();

defaultMonth.outputMonthLetters();

//

// Construct a month using the constructor with one integer argument

//

cout << endl;

cout << "Testing the constructor with one integer argument..." << endl;

cout << "Enter a month number: ";

cin >> monthNum;

Month testMonth1(monthNum);

testMonth1.outputMonth_num();

testMonth1.outputMonthLetters();

//

// Construct a month using the constructor with three letters as arguments

//

cout << endl;

cout << "Testing the constructor with 3 letters as arguments ..." << endl;

cout << "Enter the first three letters of a month (lowercase): ";

cin >> firstLetter >> secondLetter >> thirdLetter;

cout << endl;

Month testMonth2(firstLetter, secondLetter, thirdLetter);

testMonth2.outputMonth_num();

testMonth2.outputMonthLetters();

//

// See if user wants to try another month

//

cout << endl;

cout << "Do you want to test again? (y or n) ";

cin >> testAgain;

}

while (testAgain == 'y' || testAgain == 'Y');

return 0;

}

Month::Month(char firstLetter, char secondLetter, char thirdLetter)

{

if ((firstLetter == 'j')&&(secondLetter == 'a')&&(thirdLetter == 'n'))

outputMonth_num = 1;

if ((firstLetter == 'f')&&(secondLetter == 'e')&&(thirdLetter == 'b'))

outputMonth_num = 2;

if ((firstLetter == 'm')&&(secondLetter == 'a')&&(thirdLetter == 'r'))

outputMonth_num = 3;

if ((firstLetter = 'a')&&(secondLetter == 'p')&&(thirdLetter == 'r'))

outputMonth_num = 4;

if ((firstLetter == 'm')&&(secondLetter == 'a')&&(thirdLetter == 'y'))

outputMonth_num = 5;

if ((firstLetter == 'j')&&(secondLetter == 'u')&&(thirdLetter == 'n'))

outputMonth_num = 6;

if ((firstLetter == 'j')&&(secondLetter == 'u')&&(.thirdLetter == 'l'))

outputMonth_num = 7;

if ((firstLetter == 'a')&&(secondLetter == 'u')&&(thirdLetter == 'g'))

outputMonth_num = 8;

if ((firstLetter == 's')&&(secondLetter == 'e')&&(thirdLetter == 'p'))

outputMonth_num = 9;

if ((firstLetter == 'o')&&(secondLetter == 'c')&&(thirdLetter == 't'))

outputMonth_num = 10;

if ((firstLetter == 'n')&&(secondLetter == 'o')&&(thirdLetter == 'v'))

outputMonth_num = 11;

if ((firstLetter == 'd')&&(secondLetter == 'e')&&(thirdLetter == 'c'))

outputMonth_num = 12;

}

Month::inputMonthByNumber

{

if (Month_num > 12 && Month_num < 1)

cout << "Invalid number for Month, please choose 1-12)\n";

}

void Month::outputMonth_num()

{

if (month >= 1 && month <= 12)

cout ><< "Month: " << month << endl;

else

cout << "Error - The month is not a valid!" << endl;

}

void Month::outputMonthLetters()

{

switch (month)

{

case 1:

cout << "Jan" << endl;

break;

case 2:

cout << "Feb" << endl;

break;

case 3:

cout << "Mar" << endl;

break;

case 4:

cout << "Apr" << endl;

break;

case 5:

cout << "May" << endl;

break;

case 6:

cout << "Jun" << endl;

break;

case 7:

cout << "Jul" << endl;

break;

case 8:

cout << "Aug" << endl;

break;

case 9:

cout << "Sep" << endl;

break;

case 10:

cout << "Oct" << endl;

break;

case 11:

cout << "Nov" << endl;

break;

case 12:

cout << "Dec" << endl;

break;

default:

cout << "Error - the month is not a valid!" << endl;

}

7 0

3 years ago

In 1945, the United States tested the world’s first atomic bomb in what was called the Trinity test. Following the test, images

Zarrin [17]

Answer:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

$A= \frac{r^5 \rho}{t^2}$

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

Explanation:

Notation

In order to do the dimensional analysis we need to take in count that we need to conditions:

a) The energy A is released in a small place

b) The shock follows a spherical pattern

We can assume that the size of the explosion r is a function of the time t, and depends of A (energy), the time (t) and the density of the air is constant $\rho_{air}$ .

And now we can solve the dimensional problem. We assume that L is for the distance T for the time and M for the mass.

[r]=L with r representing the radius

[A]= $\frac{ML^2}{T^2}$ A represent the energy and is defined as the mass times the velocity square, and the velocity is defined as $\frac{L}{T}$

[t]=T represent the time

$[\rho]=\frac{M}{L^3}$ represent the density.

Solution to the problem

And if we analyze the function for r we got this:

$[r]=L=[A]^x [\rho]^y [t]^z$

And if we replpace the formulas for each on we got:

$[r]=L =(\frac{ML^2}{T^2})^x (\frac{M}{L^3})^y (T)^z$

And using algebra properties we can express this like that:

$[r]=L=M^{x+y} L^{2x-3y} T^{-2x+z}$

And on this case we can use the exponents to solve the values of x, y and z. We have the following system.

$x+y =0 , 2x-3y=1, -2x+z=0$

We can solve for x like this x=-y and replacing into quation 2 we got:

$2(-y)-3y = 1$

$-5y = 1$

$y= -\frac{1}{5}$

And then we can solve for x and we got:

$x = -y = -(-\frac{1}{5})=\frac{1}{5}$

And if we solve for z we got:

$z=2x =2 \frac{1}{5}=\frac{2}{5}$

And now we can express the radius in terms of the dimensional analysis like this:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

And K represent a constant in order to make the porportional relation and equality.

The problem says that we can assume the constant K=1.

And if we solve for the energy we got:

$A^{1/5}=\frac{r}{t^{2/5} \rho^{-1/5}}$

$A= \frac{r^5 \rho}{t^2}$

And now we can replace the values given. On this case t =0.025 s, the radius r =140 m, and the density is a constant assumed $\rho =1.2 kg/m^2$ , and replacing we got:

$A=\frac{140^5 1.2 kg/m^3}{(0.025 s)^2}=1.033x10^{14} \frac{kg m^2}{s^2}$

And we can convert this into ergs we got:

$A= 1.033x10^{14} \frac{kgm^2}{s^2} * \frac{1 x10^7 egrs}{1 \frac{kgm^2}{s^2}}=1.033x10^{21} ergs$

And then we know that 1 g of TNT have $4x10^4 erg$

And we got:

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

3 0

3 years ago

6. Staples are the most common item used to secure and support cables in residential wiring.​

6. Staples are the most common item used to secure and support cables in residential wiring.