What material are the rocker/valve cover gaskets made out of?

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 an air compressor the compression takes place at a constant internal energy and 50KJ of heat are rejected to the cooling wate

lozanna [386]

Answer:

work is 50 kj

Explanation:

Given data

heat (Q) = 50 kj

To find out

work input for the compression stroke per kilogram of air

Solution

we will apply here "first law of thermodynamics" i.e.

The First Law of Thermodynamics states that heat is a form of energy, subject to the principle of conservation of energy, that heat energy cannot be created or destroyed. It can be transferred from one location to another location. i.e.

ΔU = Q – W ................1

here ΔU is change in internal energy, Q is heat and W is work done

here U = 0 because air compressor the compression takes place at a constant internal energy in question

so that by equation 1

Q = W

and Q = 50

so work will be 50 kj

8 0

3 years ago

g Part 2: The features arrived the grammar Splitting categories and non-terminals gets out of hand fast. An alternative to the p

koban [17]

Answer:

The split is given by including spaces in both tabs

Explanation:

The bracket notation can be used to indicate the split. Here is an example:

String [ ] parts = s. split ( "[/]")

3 0

3 years ago

For a bronze alloy, the stress at which plastic deformation begins is 275 MPa, and the modulus of elasticity is 115 GPa. (a) Wha

Anton [14]

Answer:

89375 N

Explanation:

Rearrange the formula for normal stress for F:

$\sigma=\frac{F}{A}$

$F=\sigma*A$

Convert given values to base units:

275 MPa = $275*10^{6}$ Pa

325 $mm^{2}$ = 0.000325 $m^{2}$

Substituting in given values:

F = $(275*10^{6})*(0.000325)=89375$ N

3 0

3 years ago

g Four primary sedimentation basins are designed for total average flow of 1.2 m3/s. The TSS concentration in the primary treate

fiasKO [112]

Answer:

Capacity for the sludge pump = 0.217 m³/min

Explanation:

Detailed explanation is given in the attached document.

5 0

4 years ago