Which of these cars traveled faster during time interval <br> please show solution

Write torsion equation and explain the importance of each components.<br>

Elanso [62]

The equations are based on the following assumptions

1) The bar is straight and of uniform section
2) The material of the bar is has uniform properties.
3) The only loading is the applied torque which is applied normal to the axis of the bar.
4) The bar is stressed within its elastic limit.

Nomenclature

T = torque (Nm)
l = length of bar (m)
J = Polar moment of inertia.(Circular Sections) ( m^4)
J' = Polar moment of inertia.(Non circluar sections) ( m^4 )
K = Factor replacing J for non-circular sections.( m^4)
r = radial distance of point from center of section (m)
ro = radius of section OD (m)
τ = shear stress (N/m^2)
G Modulus of rigidity (N/m^2)
θ = angle of twist (radians)

4 0

3 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

3 years ago

A 4-stroke Diesel engine with a displacement of Vd = 2.5x10^-3m^3 produces a mean effective pressure of 6.4 bar at the speed of

yKpoI14uk [10]

Answer:

The power developed by engine is 167.55 KW

Explanation:

Given that

$V_d=2.5\times 10^{-3} m^3$

Mean effective pressure = 6.4 bar

Speed = 2000 rpm

We know that power is the work done per second.

So

$P=6.4\times 100\times 2.5\times 10^{-3}\times \dfrac{2\pi \times2000}{120}$

We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.

P=167.55 KW

So the power developed by engine is 167.55 KW

4 0

3 years ago

What is the benefit of theories? they are easily disproven they are written by famous people they were applying every circumstan

OlgaM077 [116]

Answer:

A scientific theory is an explanation of an aspect of the natural world and universe that has been repeatedly tested and verified in accordance with the scientific method, using accepted protocols of observation, measurement, and evaluation of results. Where possible, theories are tested under controlled conditions in an experiment. In circumstances not amenable to experimental testing, theories are evaluated through principles of abductive reasoning. Established scientific theories have withstood rigorous scrutiny and embody scientific knowledge.

A scientific theory differs from a scientific fact or scientific law in that a theory explains "why" or "how": a fact is a simple, basic observation, whereas a law is a statement (often a mathematical equation) about a relationship between facts. For example, Newton’s Law of Gravity is a mathematical equation that can be used to predict the attraction between bodies, but it is not a theory to explain how gravity works. Stephen Jay Gould wrote that "...facts and theories are different things, not rungs in a hierarchy of increasing certainty. Facts are the world's data. Theories are structures of ideas that explain and interpret facts.

mark me as brainlist

3 0

2 years ago

Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an el

Allisa [31]

Answer:

theoretical fracture strength = 16919.98 MPa

Explanation:

given data

Length (L) = 0.28 mm = 0.28 × 10⁻³ m

radius of curvature (r) = 0.002 mm = 0.002 × 10⁻³ m

Stress (s₀) = 1430 MPa = 1430 × 10⁶ Pa

solution

we get here theoretical fracture strength s that is express as

theoretical fracture strength = $s_{0} \times \sqrt{\frac{L}{r} }$ .............................1

put here value and we get

theoretical fracture strength = $1430 \times 10^6\times \sqrt{\frac{0.28\times 10^{-3}}{0.002\times 10^{-3}} }$

theoretical fracture strength = $16919.98 \times 10^6$

theoretical fracture strength = 16919.98 MPa

3 0

3 years ago

Which of these cars traveled faster during time interval please show solution