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2 years ago

Do not use C++ pointers (this is not a topic of this course).

Engineering

1 answer:

noname [10]2 years ago

4 0

The program is an illustration of functions in C++

C++ functions are collection of named program statements that are executed when called

<h3>The C++ program</h3>

The program in C++, where comments are used to explain each action is as follows:

#include <iostream>

#include <cmath>

using namespace std;

//This defines the boolean method

bool isoscelesTriangleHeightAndArea(double base, double hypotenuse, double height, double area) {

//This checks for invalid base and height

if(base < 0 || hypotenuse < 0){

return false;

}

//This returns true if base and height are valid

return true;

}

//The main begins here

int main(){

//This initializes the variables

double base, hypotenuse, height, area;

//This gets input for base and hypotenuse

cout<<"Base: "; cin>>base;

cout<<"Hypotenuse: "; cin>>hypotenuse;

//The calls the boolean method

bool success = isoscelesTriangleHeightAndArea(base, hypotenuse, height, area);

//If success is True

while (success){

//This calculates height

height = sqrt(hypotenuse*hypotenuse - (base/2) * (base/2));

//This calculates area

area = 0.5 * base * height;

//This prints height

cout << "Triangle height is " << height << endl;

//This prints area

cout << "Triangle area is " << area << endl;

//This gets input for base and hypotenuse

cout<<"Base: "; cin>>base;

cout<<"Hypotenuse: "; cin>>hypotenuse;

bool success = isoscelesTriangleHeightAndArea(base, hypotenuse, height, area);

}

cout << "Invalid Input!" << endl;

}

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Consider a Carnot heat-engine cycle executed in a closed system using 0.028 kg of steam as the working fluid. It is known that t

Paha777 [63]

Answer:

T=138 °C

Explanation:

Given that

m = 0.028 kg

Net work output W= 60 KJ

T₂=2T₁

As we know that efficiency of Carnot heat engine given as

$\eta=1-\dfrac{T_1}{T_2}$

$\eta=1-\dfrac{T_1}{2T_1}$

η = 0.5

We know that

$\eta=\dfrac{W}{Q_a}$

Qa=heat addition

W= net work output

$\eta=\dfrac{W}{Q_a}$

$0.5=\dfrac{60}{Q_a}$

Qa= 120 KJ

From first law

Qa= W+ Qr

Qr= 120 - 60

Qr= 60 KJ

Qr Is the heat rejection.

Heat rejection per unit mass

Qr=60 / 0.028 = 2142.85 KJ/kg

Qr= 2142.85 KJ/kg

Temperature at which latent heat of steam is 2142.85 KJ/kg will be our answer.

T=138 °C

The temperature corresponding to 2142.85 KJ/kg will be 138 °C.

T=138 °C

8 0

3 years ago

Air in a large tank at 300C and 400kPa, flows through a converging diverging nozzle with throat diameter 2cm. It exits smoothly

-Dominant- [34]

Answer:

The answer is " $3.74 \ cm\ \ and \ \ 0.186 \frac{kg}{s}$ "

Explanation:

Given data:

Initial temperature of tank $T_1 = 300^{\circ}\ C= 573 K$

Initial pressure of tank $P_1= 400 \ kPa$

Diameter of throat $d* = 2 \ cm$

Mach number at exit $M = 2.8$

In point a:

calculating the throat area:

$A*=\frac{\pi}{4} \times d^2$

$=\frac{\pi}{4} \times 2^2\\\\=\frac{\pi}{4} \times 4\\\\=3.14 \ cm^2$

Since, the Mach number at throat is approximately half the Mach number at exit.

Calculate the Mach number at throat.

$M*=\frac{M}{2}\\\\=\frac{2.8}{2}\\\\=1.4$

Calculate the exit area using isentropic flow equation.

$\frac{A}{A*}= (\frac{\gamma -1}{2})^{\frac{\gamma +1}{2(\gamma -1)}} (\frac{1+\frac{\gamma -1}{2} M*^2}{M*})^{\frac{\gamma +1}{2(\gamma -1)}}$

Here: $\gamma$ is the specific heat ratio. Substitute the values in above equation.

$\frac{A}{3.14}= (\frac{1.4-1}{2})^{-\frac{1.4+1}{2(1.4 -1)}} (\frac{1+\frac{1.4-1}{2} (1.4)^2}{1.4})^{\frac{1.4+1}{2(1.4-1)}} \\\\A=\frac{\pi}{4}d^2 \\\\10.99=\frac{\pi}{4}d^2 \\\\d = 3.74 \ cm$

exit diameter is 3.74 cm

In point b:

Calculate the temperature at throat.

$\frac{T*}{T}=(1+\frac{\Gamma-1}{2} M*^2)^{-1}\\\\\frac{T*}{573}=(1+\frac{1.4-1}{2} (1.4)^2)^{-1}\\\\T*=411.41 \ K$

Calculate the velocity at exit.

$V*=M*\sqrt{ \gamma R T*}$

Here: R is the gas constant.

$V*=1.4 \times \sqrt{1.4 \times 287 \times 411.41}\\\\=569.21 \ \frac{m}{s}$

Calculate the density of air at inlet

$\rho_1 =\frac{P_1}{RT_1}\\\\=\frac{400}{ 0.287 \times 573}\\\\=2.43\ \frac{kg}{m^3}$

Calculate the density of air at throat using isentropic flow equation.

$\frac{\rho}{\rho_1}=(1+\frac{\Gamma -1}{2} M*^2)^{-\frac{1}{\Gamma -1}} \\\\\frac{\rho *}{2.43}=(1+\frac{1.4-1}{2} (1.4)*^2)^{-\frac{1}{1.4-1}} \\\\\rho*= 1.045 \ \frac{kg}{m^3}$

Calculate the mass flow rate.

$m= \rho* \times A* \times V*\\\\= 1.045 \times 3.14 times 10^{-4} \times 569.21\\\\= 0.186 \frac{kg}{s}$

5 0

3 years ago

Ethylene glycol, the ingredient in antifreeze, does not cause health problems because it is a clear liquid.

netineya [11]

B. false

Explanation: because it is

5 0

3 years ago

Please help this is due today!!!!!

White raven [17]

Answer:

1:c 2:False

Explanation:

7 0

3 years ago

In c the square root of a number N can be approximated by repeated calculation using the formula NG = 0.5(LG + N/LG) where NG st

DanielleElmas [232]

Answer:

Following are the program to the given question:

#include <stdio.h>//header file

double square_root(double N, double initialGuess)//defining a method square_root that takes two variable in parameters

{

double NG, LG = initialGuess,diff;//defining double variable

while(1)//use loop to calculate square root value

{

NG = 0.5 * (LG + N / LG);//using given formula

diff = NG - LG;//calculating difference

if(diff < 0)//use if to check difference is less than 0

diff = -diff;//decreaing difference

if(diff < 0.005)//use if that check difference is less than 0.005

break;//using break keyword

else//defining else block

{

LG = NG;//holding value

}

return NG;//return value

}

int main()//defining main method

{

double ans, n,initialguess = 1.0;//defining double variable

n = 4;//use n to hold value

ans = square_root(n, initialguess);//calculating the square root value and print its value