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irga5000 [103]
3 years ago
8

A rectangular workpiece has the following original dimensions: 2a=100mm, h=25mm, and width=20mm. The metal has a strengh coeffic

ient of 400MPa and strain-hardening exponent of 0.5. It is being forged in plane strain with mu=0.2. Calculate the force required when the height is reduced by 20 percent. Do not use average pressure formulas.
Engineering
1 answer:
Elena-2011 [213]3 years ago
4 0

Answer:

See attachment for detailed answer.

Explanation:

Download pdf
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Swing arm restraints are intended to prevent a vehicle from falling off a lift.
castortr0y [4]

The question is asking whether that statement is true or false. Options are;

A) True

B) False

This is about usage of Swing arm restraints.

<em><u>B) False</u></em>

There are different safety features that people employ when a vehicle is lifted. However, for this question, we will only talk about swing arm restraints.

  • Swing arm restraints are lifting restraint devices that are used to prevent a cars arms from shifting or going out of position after that car has been lifted and mounted.

  • This swing arm restraint does not prevent a vehicle from falling off a lift as it just helps to ensure that the swing arms that are unloaded basically maintain their position.

Read more at; brainly.com/question/17972874

4 0
3 years ago
C++A palindrome is a string such as "madam", "radar", "Dad", and "I", that reads the same forwards and backwards. The empty stri
jeka57 [31]

Answer:

See explaination

Explanation:

#include <iostream>

#include<string.h>

using namespace std;

bool isPalindrome(string str, int lower, int upper){

if(str.length() == 0 || lower>=upper){

return true;

}

else{

if(str.at(lower) == str.at(upper)){

return isPalindrome(str,lower+1,upper-1);

}

else{

return false;

}

}

}

int main(){

string input;

cout<<"Enter string: ";

cin>>input;

if(isPalindrome(input,0,input.length()-1)){

cout<<input<<" is a palindrome"<<endl;

}

else{

cout<<input<<" is NOT a palindrome"<<endl;

}

return 0;

}

5 0
3 years ago
Derive the following conversion factors: (a) Convert a pressure of 1 psi to kPa (b) Convert a vol ume of 1 liter to gallons (c)
kirill [66]

Answer:

a)6.8 KPa

b)0.264 gallon

c)47.84 Pa.s

Explanation:

We know that

1 lbf=  4.48 N

1 ft =0.30 m

a)

Given that

P= 1 psi

psi is called pound force per square inch.

We know that 1 psi = 6.8 KPa.

b)

Given that

Volume = 1 liter

We know that 1000 liter = 1 cubic meter.

1 liter =0.264 gallon.

c)

1\ \frac{lb.s}{ft^2}=47.84\ \frac{Pa.s}{ft^2}

4 0
3 years ago
What separates the work of technology transfer research from implementation of the products of such research?
tatuchka [14]

Answer:

The thing that separates the work of technology transfer research from implementation of the products of such research is:

Technology transfer research looks at how technology may be transferred but does not actually make the transfer.

Explanation:

This suggests that technology transfer research is different from the technology transfer (implementation) itself.  The first stops at making scientific investigations into technology transfer activities while the next step performs the actual transfer or implementation.  In other words, technology transfer conveys the results of scientific and technological research to the marketplace and to the wider society.  It is the bridge-builder between the research and the implementation.

4 0
3 years ago
Write 3 classes with three levels of hierarchy: Base, Derive (child of base), and D1 (child of Derive class). Within each class,
PIT_PIT [208]

Answer:

class Base

{

void m1()

{

System.out.println("Origin: Base Class");

}

}

class Derive extends Base

{

void m1()

{

System.out.println("Origin: Derived Class");

}

}

class D1 extends Derive

{

void m1()

{

System.out.println("Origin: D1 - Child of Derive Class");

}

}

class TestDynamicBinding

{

public static void main(String args[])

{

Base base = new Base(); // object of Base class

Derive derive = new Derive(); // object of Derive class

D1 d1 = new D1(); // object of D1 class

 

Base reference; // Reference of type Base

reference = base; // reference referring to the object of Base class

reference.m1();   //call made to Base Class m1 method

 

reference = derive;   // reference referring to the object of Derive class

reference.m1(); //call made to Derive Class m1 method

 

reference = d1;    // reference referring to the object of D1 class

reference.m1(); //call made to D1 Class m1 method

}

}

Explanation:

The solution demonstrates dynamic binding behavior because the linking procedure used calls overwritten method m1() is made at run time rather than doing it at the compile time. The code to be executed for this specific procedural call is also known at run time only.

6 0
3 years ago
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