Why does an aeroplane smoke in the air

Basile [38]

Answer:

Divide the difference in tax by the amount of income from the investment, and you'll get the economic marginal tax rate from investing. Most people refer to marginal tax rates as being identical to tax brackets.

hope this helps

have a good day :)

Explanation:

8 0

2 years ago

g For this project you are required to perform Matrix operations (Addition, Subtraction and Multiplication). For each of the ope

Kruka [31]

Answer:

C++ code is explained below

Explanation:

#include<iostream>

using namespace std;

//Function Declarations

void add();

void sub();

void mul();

//Main Code Displays Menu And Take User Input

int main()

{

int choice;

cout << "\nMenu";

cout << "\nChoice 1:addition";

cout << "\nChoice 2:subtraction";

cout << "\nChoice 3:multiplication";

cout << "\nChoice 0:exit";

cout << "\n\nEnter your choice: ";

cin >> choice;

cout << "\n";

switch(choice)

{

case 1: add();

break;

case 2: sub();

break;

case 3: mul();

break;

case 0: cout << "Exited";

exit(1);

default: cout << "Invalid";

}

main();

}

//Addition Of Matrix

void add()

{

int rows1,cols1,i,j,rows2,cols2;

cout << "\nmatrix1 # of rows: ";

cin >> rows1;

cout << "\nmatrix1 # of columns: ";

cin >> cols1;

int m1[rows1][cols1];

//Taking First Matrix

for(i=0;i<rows1;i++)

for(j=0;j<cols1;j++)

{

cout << "\nEnter element (" << i << "," << j << "): ";

cin >> m1[i][j];

cout << "\n";

}

//Printing 1st Matrix

for(i=0;i<rows1;i++)

{

for(j=0;j<cols1;j++)

cout << m1[i][j] << " ";

cout << "\n";

}

cout << "\nmatrix2 # of rows: ";

cin >> rows2;

cout << "\nmatrix2 # of columns: ";

cin >> cols2;

int m2[rows2][cols2];

//Taking Second Matrix

for(i=0;i<rows2;i++)

for(j=0;j<cols2;j++)

{

cout << "\nEnter element (" << i << "," << j << "): ";

cin >> m2[i][j];

cout << "\n";

}

//Displaying second Matrix

cout << "\n";

for(i=0;i<rows2;i++)

{

for(j=0;j<cols2;j++)

cout << m2[i][j] << " ";

cout << "\n";

}

//Displaying Sum of m1 & m2

if(rows1 == rows2 && cols1 == cols2)

{

cout << "\n";

for(i=0;i<rows1;i++)

{

for(j=0;j<cols1;j++)

cout << m1[i][j]+m2[i][j] << " ";

cout << "\n";

}

else

cout << "operation is not supported";

main();

}

void sub()

{

int rows1,cols1,i,j,k,rows2,cols2;

cout << "\nmatrix1 # of rows: ";

cin >> rows1;

cout << "\nmatrix1 # of columns: ";

cin >> cols1;

int m1[rows1][cols1];

for(i=0;i<rows1;i++)

for(j=0;j<cols1;j++)

{

cout << "\nEnter element (" << i << "," << j << "): ";

cin >> m1[i][j];

cout << "\n";

}

for(i=0;i<rows1;i++)

{

for(j=0;j<cols1;j++)

cout << m1[i][j] << " ";

cout << "\n";

}

cout << "\nmatrix2 # of rows: ";

cin >> rows2;

cout << "\nmatrix2 # of columns: ";

cin >> cols2;

int m2[rows2][cols2];

for(i=0;i<rows2;i++)

for(j=0;j<cols2;j++)

{

cout << "\nEnter element (" << i << "," << j << "): ";

cin >> m2[i][j];

cout << "\n";

}

for(i=0;i<rows2;i++)

{

for(j=0;j<cols2;j++)

cout << m1[i][j] << " ";

cout << "\n";

}

cout << "\n";

//Displaying Subtraction of m1 & m2

if(rows1 == rows2 && cols1 == cols2)

{

for(i=0;i<rows1;i++)

{

for(j=0;j<cols1;j++)

cout << m1[i][j]-m2[i][j] << " ";

cout << "\n";

}

else

cout << "operation is not supported";

main();

}

void mul()

{

int rows1,cols1,i,j,k,rows2,cols2,mul[10][10];

cout << "\nmatrix1 # of rows: ";

cin >> rows1;

cout << "\nmatrix1 # of columns: ";

cin >> cols1;

int m1[rows1][cols1];

for(i=0;i<rows1;i++)

for(j=0;j<cols1;j++)

{

cout << "\nEnter element (" << i << "," << j << "): ";

cin >> m1[i][j];

cout << "\n";

}

cout << "\n";

for(i=0;i<rows1;i++)

{

for(j=0;j<cols1;j++)

cout << m1[i][j] << " ";

cout << "\n";

}

cout << "\nmatrix2 # of rows: ";

cin >> rows2;

cout << "\nmatrix2 # of columns: ";

cin >> cols2;

int m2[rows2][cols2];

for(i=0;i<rows2;i++)

for(j=0;j<cols2;j++)

{

cout << "\nEnter element (" << i << "," << j << "): ";

cin >> m2[i][j];

cout << "\n";

}

cout << "\n";

//Displaying Matrix 2

for(i=0;i<rows2;i++)

{

for(j=0;j<cols2;j++)

cout << m2[i][j] << " ";

cout << "\n";

}

if(cols1!=rows2)

cout << "operation is not supported";

else

{

//Initializing results as 0

for(i = 0; i < rows1; ++i)

for(j = 0; j < cols2; ++j)

mul[i][j]=0;

// Multiplying matrix m1 and m2 and storing in array mul.

for(i = 0; i < rows1; i++)

for(j = 0; j < cols2; j++)

for(k = 0; k < cols1; k++)

mul[i][j] += m1[i][k] * m2[k][j];

// Displaying the result.

cout << "\n";

for(i = 0; i < rows1; ++i)

for(j = 0; j < cols2; ++j)

{

cout << " " << mul[i][j];

if(j == cols2-1)

cout << endl;

}

main();

}

5 0

2 years ago

An aircraft is flying at 300 mph true airspeed has a 50 mph tailwind. What is its ground speed?

Free_Kalibri [48]

Answer:

304.13 mph

Explanation:

Data provided in the question :

The Speed of the flying aircraft = 300 mph

Tailwind of the true airspeed = 50 mph

Now,

The ground speed will be calculated as:

ground speed = $\sqrt{300^2+50^2}$

or

The ground speed = $\sqrt{92500}$

or

The ground speed = 304.13 mph

Hence, the ground speed is 304.13 mph

8 0

3 years ago

The heat required to raise the temperature of m (kg) of a liquid from T1 to T2 at constant pressure is Z T2CpT dT (1) In high sc

a_sh-v [17]

Answer:

(a)

dQ = mdq

dq = $C_p$ dT

$q = \int\limits^{T_2}_{T_1} {C_p} \, dT$ = $C_p$ (T₂ - T₁)

From the above equations, the underlying assumption is that $C_p$ remains constant with change in temperature.

(b)

Given;

V = 2L

T₁ = 300 K

Q₁ = 16.73 KJ , Q₂ = 6.14 KJ

ΔT = 3.10 K , ΔT₂ = 3.10 K for calorimeter

Let $C_{cal}$ be heat constant of calorimeter

Q₂ = $C_{cal}$ ΔT

Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂

Q₁ - Q₂ = m $C_p$ ΔT

number of moles of n-C₆H₁₄, n = m/M

ρ = 650 kg/m³ at 300 K

M = 86.178 g/mol

m = ρv = 650 (2x10⁻³) = 1.3 kg

n = m/M => 1.3 / 0.086178 = 15.085 moles

Q₁ - Q₂ = m $C_p$ ' ΔT

$C_p$ = (16.73 - 6.14) / (15.085 x 3.10)

$C_p$ = 0.22646 KJ mol⁻¹ k⁻¹

6 0

2 years ago

Write the heat equation for each of the following cases:

jok3333 [9.3K]

Answer:

Explanation:

a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:

$\dfrac{\partial^2T}{\partial x^2}= \ 0 \ ; \ if \ T = f(x) \\ \\ \dfrac{\partial^2T}{\partial y^2}= \ 0 \ ; \ if \ T = f(y) \\ \\ \dfrac{\partial^2T}{\partial z^2}= \ 0 \ ; \ if \ T = f(z)$

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

$\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}$

where;

$Q_g$ = heat generated per unit volume

$\alpha$ = Thermal diffusivity

c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

$\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0$

where;

The radial directional term = $\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r})$ and the axial directional term is $\dfrac{\partial^2 T}{\partial z^2 }$

d) The heat equation for a wire going through a furnace is:

$\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]$

since;

the steady-state is zero, Then:

$\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]$ '

e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

$\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}$

4 0

3 years ago

Why does an aeroplane smoke in the air​

Why does an aeroplane smoke in the air