Anyone want to play among us with me the code is MMJSUF

During January, at a location in Alaska winds at -20°C can be observed, However, several meters below ground the temperature rem

maw [93]

Answer:

a) $\eta_{th} = 10.910\,\%$ , b) Yes.

Explanation:

a) The maximum thermal efficiency is given by the Carnot's Cycle, whose formula is:

$\eta_{th} =\left(1-\frac{253.15\,K}{284.15\,K} \right) \times 100\,\%$

$\eta_{th} = 10.910\,\%$

b) The claim of the inventor is possible since real efficiency is lower than maximum thermal efficiency.

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

A thick steel slab (rho= 7800 kg/m3 , cp= 480 J/kg K, k= 50 W/m K) is initially at 300 °C and is cooled by water jets impinging

dimaraw [331]

Answer:

t = 2244.3 sec

Explanation:

calculate the thermal diffusivity

$\alpha = \frac{k}{\rho c}$

$= \frac{50}{7800\times 480} = 1.34 \times 10^{-5} m^2/s$

Temperature at 28 mm distance after t time = = 50 degree C

we know that

$\frac[ T_{28} - T_s}{T_i -T_s} = erf(\frac{x}{2\sqrt{at}})$

$\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]$

$0.909 = erf{\frac{3.8245}{\sqrt{t}}}$

from gaussian error function table , similarity variable w calculated as

erf w = 0.909

it is lie between erf w = 0.9008 and erf w = 0.11246 so by interpolation we have

w = 0.08073

$erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]$

$0.08073 = \frac{3.8245}{\sqrt{t}}$

solving fot t we get

t = 2244.3 sec

3 0

3 years ago

A manometer is used to measure the air pressure in a tanlc The fluid used has a specific gravity of 1.25, and the differentialhe

Alenkasestr [34]

Answer:

(a) 11.437 psia

(b) 13.963 psia

Explanation:

The pressure exerted by a fluid can be estimated by multiplying the density of the fluid, acceleration due to gravity and the depth of the fluid. To determine the fluid density, we have:

fluid density = specific gravity * density of water = 1.25 * 62.4 lbm/ft^3 = 78 lbm/ft^3

height = 28 in * (1 ft/12 in) = 2.33 ft

acceleration due to gravity = 32.174 ft/s^2

The change in pressure = fluid density*acceleration due to gravity*height = 78*32.174*(28/12) = 5855.668 lbm*ft/(s^2 * ft^2) = 5855.668 lbf/ft^2

The we convert from lbf/ft^2 to psi:

(5855.668/32.174)*0.00694 psi = 1.263 psi

(a) pressure = atmospheric pressure - change in pressure = 12.7 - 1.263 = 11.437 psia

(b) pressure = atmospheric pressure + change in pressure = 12.7 + 1.263 = 13.963 psia

8 0

3 years ago

Match the words with their definitions

Pavlova-9 [17]

Wooden 2 X 8's, 10's, or 12's that run parallel to one another and support a floor or ceiling, and supported in turn by larger beams, girders, or bearing walls- Joist

Explanation:

Wooden 2 X 8's, 10's, or 12's that run parallel to one another and support a floor or ceiling, and supported in turn by larger beams, girders, or bearing walls- Joist
A point where a bearing/structural weight is concentrated and transferred to the foundation-Point Load
Includes all exterior walls and any interior wall that is aligned above a support beam or girder-Load Bearing Wall
Plywood substitute made of coarse sawdust that is mixed with resin and pressed into sheets-Particleboard
The replacement of excavated earth into a trench around or against a basement/crawl space foundation wall- Backfill
Continuous 8m" or 10" thick concrete pad installed before and supports the foundation wall or monopost-Footing
Ribbed steel bars installed in foundation concrete walls, footers and poured in place concrete structures designed to strengthen concrete-Rebar
The end, upper, triangular area of a home, beneath the roof-Gable
A manufactured 4X8 wood panel made out of 1-2 wood chips and glue. Often used as a substitute for plywood-OSB
The incline slope of a roof or the ratio of the total width of a house-Pitch
A manufactured structual wood beam. It is constructed of pressure and adhesive wood strands of wood-Microlam

5 0

3 years ago