Answer every question of this quiz

How do I do this?<br> Blueprints, complete the missing view.

Ymorist [56]

Explanation:

Look at the drawings and decide which view is missing. Front? Side? Top? Then draw it

7 0

3 years ago

name the three exposure techniques in photolithography. what are the alternatives to photolithography in ic processing?

zhenek [66]

The three exposure techniques in photolithography are:

Contact
Proximity
Projection

Alternatives to photolithography in IC processing include;

X-ray
UV
Ion, and
Electron lithography

<h3>What is Photolithography?</h3>

Photolithography is a term in integrated circuit development that describes the patterned films that are formed when a beam of light falls on a substance.

This phenomenon protects the surface of sensitive materials such as glass during some operations like etching. UV and X-rays can be used for this purpose.

Learn more about photolithography here:

brainly.com/question/13650094

#SPJ11

6 0

2 years ago

Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical

dusya [7]

Answer:

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

Explanation:

For steel bolt

Stress = 210 MPa or 210 N/mm2

Pressure = Stress* Area

Pbolt = 210 N/mm2 * 16^2 *(pi)/4

Pbolt = 210 N/mm2 * 200.96 mm^2 = 42201.6 N

For Brass spacer

Pressure = 42201.6 N

Area of Brass spacer = Pressure/Stress

Area of Brass spacer = 42201.6 N/145 N/mm^2 = 291.044 mm^2

Area of Brass spacer = (pi) (d^2 - 16^2)/4 = 291.044 mm^2

d^2 - 16^2 = 291.044 mm^2* 4/(pi) = 370.758

d^2 = 370.758 + 16^2

d^2 = 626.758

d = 25.03 mm

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

5 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 silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur

anyanavicka [17]

Answer:

hello your question has some missing values below are the missing values

Mirror Radius (mm) Bending Failure Stress (MPa)

.603 225

.203 368

.162 442

answer : 191 mPa

Explanation:

<u>Determine the stress present at the time of fracture for the original plate</u>

Bending stress ∝ 1 / ( mirror radius )^n ------ ( 1 )

at 0.603 bending stress = 225

at 0.203 bending stress = 368

at 0.162 bending stress = 442

<u>applying equation 1 determine the value of n for several combinations</u>

( 225 / 368 ) = ( 0.203 / 0.603 )^n

hence : n = 0.452

also

( 368/442 ) = ( 0.162 / 0.203 ) ^n

hence : n = 0.821

also

( 225 / 442 ) = ( 0.162 / 0.603 ) ^n

hence : n = 0.514

Next determine the average value of n

n ( mean value ) = ( 0.452 + 0.821 + 0.514 ) / 3 = 0.596

Calculate estimated stress present at the time of fracture for the original plate

= bending stress at x = 0.796 / bending stress at x = 0.603

= x / 225 = ( 0.603 / 0.796 ) ^ 0.596

therefore X ( stress present at the time of fracture of original plate )

= 225 * 0.84747

<em>= 191 mPa </em>

3 0

3 years ago