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3 years ago

Refectories are one of the types of ceramics that have low melting temperature. a)-True b)-False

Engineering

1 answer:

MaRussiya [10]3 years ago

3 0

Answer:

b). False

Explanation:

A refractory material is a type of material that can withstand high temperatures without loosing its strength. They are used in reactors, furnaces, kilns, etc.

Refractory materials are certain super alloys and ceramics materials.

Properties of refractory materials :

1. Refractory materials have high melting point.

2.They acts barriers between high heat zone and low heat zone.

3. The specific heat of refractory material is very low.

4. Refractories that have high bulk densities are better in quality.

Hence, Refractory materials have a very high melting temperature.

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A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is

-BARSIC- [3]

Answer:

$\mathbf{C_{10} = 137.611 \ kN}$

Explanation:

From the information given:

Life requirement = 40 kh = 40 $40 \times 10^{3} \ h$

Speed (N) = 520 rev/min

Reliability goal $(R_D)$ = 0.9

Radial load $(F_D)$ = 2600 lbf

To find C10 value by using the formula:

$C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}$

where;

$x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}$

$\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}$

The Weibull parameters include:

$x_o = 0.02$

$(\theta - x_o) = 4.439$

$b= 1.483$

∴

Using the above formula:

$C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}$

$C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}$

$C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}$

$C_{10} = 30962.449 \ lbf$

Recall that:

1 kN = 225 lbf

∴

$C_{10} = \dfrac{30962.449}{225}$

$\mathbf{C_{10} = 137.611 \ kN}$

7 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

About ceramics: Only can be optically opaque or semi-transparent. a) True b)-False

julia-pushkina [17]

Answer: True

Explanation: Ceramics have the property that when the band gap present between the atoms are larger than the light energy then the tend to become opaque because the light scattering is caused . They also show the property of being translucent when there are chances of the light to get a path through the surface of ceramic so they get the light at some parts e.g.porcelain .Therefore the statement given is true that ceramics can be optically opaque or semi-transparent(translucent).

6 0

3 years ago

A two-dimensional flow field described by

Oduvanchick [21]

Answer:

the answer is

Explanation:

<h2> We now focus on purely two-dimensional flows, in which the velocity takes the form </h2><h2>u(x, y, t) = u(x, y, t)i + v(x, y, t)j. (2.1) </h2><h2>With the velocity given by (2.1), the vorticity takes the form </h2><h2>ω = ∇ × u = </h2><h2> </h2><h2>∂v </h2><h2>∂x − </h2><h2>∂u </h2><h2>∂y </h2><h2>k. (2.2) </h2><h2>We assume throughout that the flow is irrotational, i.e. that ∇ × u ≡ 0 and hence </h2><h2>∂v </h2><h2>∂x − </h2><h2>∂u </h2><h2>∂y = 0. (2.3) </h2><h2>We have already shown in Section 1 that this condition implies the existence of a velocity </h2><h2>potential φ such that u ≡ ∇φ, that is </h2><h2>u = </h2><h2>∂φ </h2><h2>∂x, v = </h2><h2>∂φ </h2><h2>∂y . (2.4) </h2><h2>We also recall the definition of φ as </h2><h2>φ(x, y, t) = φ0(t) + Z x </h2><h2>0 </h2><h2>u · dx = φ0(t) + Z x </h2><h2>0 </h2><h2>(u dx + v dy), (2.5) </h2><h2>where the scalar function φ0(t) is arbitrary, and the value of φ(x, y, t) is independent </h2><h2>of the integration path chosen to join the origin 0 to the point x = (x, y). This fact is </h2><h2>even easier to establish when we restrict our attention to two dimensions. If we consider </h2><h2>two alternative paths, whose union forms a simple closed contour C in the (x, y)-plane, </h2><h2>Green’s Theorem implies that </h2><h2> </h2><h2> </h2><h2> </h2><h2> </h2><h2> </h2><h2></h2><h2></h2>

5 0

3 years ago

1. What affects the weight of an airplane?<br> a. Mass<br> b. Matter<br> c. Materials<br> d. Speed

BARSIC [14]

Mass affects the weight of an airplane because everything has mass

5 0

2 years ago

Read 2 more answers