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andrew11 [14]
2 years ago
10

Elements of parallel computing

Engineering
1 answer:
Leya [2.2K]2 years ago
6 0

\huge{\orange}\fcolorbox{purple}{cyan}{\bf{\underline{\green{\color{pink}Answer}}}}

<h3><u>E</u><u>lements of parallel </u><u>computing:</u></h3>

  • Computer systems organization.
  • Computing methodologies.
  • General and reference.
  • Networks.
  • Software and its engineering.
  • Theory of computation.

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Briefly explain how each of the following influences the tensile modulus of a semicrystalline polymer and why:(a) molecular weig
marin [14]

Answer:

(a) Increases

(b) Increases

(c) Increases

(d) Increases

(e) Decreases

Explanation:

The tensile modulus of a semi-crystalline polymer depends on the given factors as:

(a) Molecular Weight:

It increases with the increase in the molecular weight of the polymer.

(b) Degree of crystallinity:

Tensile strength of the semi-crystalline polymer increases with the increase in the degree of crystallinity of the polymer.

(c) Deformation by drawing:

The deformation by drawing in the polymer results in the finely oriented chain structure of the polymer with the greater inter chain secondary bonding structure resulting in the increase in the tensile strength of the polymer.

(d) Annealing of an undeformed material:

This also results in an increase in the tensile strength of the material.

(e) Annealing of  a drawn material:

A semi crystalline material which is drawn when annealed results in the decreased tensile strength of the material.

5 0
3 years ago
A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.
sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c)Stress\ ratio =-0.65

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}

2\sigma_m={\sigma_1+\sigma_2}

2 x 46.2 =  σ₁ +  σ₂

 σ₁ +  σ₂ = 92.4 MPa    --------1

The amplitude stress given as

\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}

2\sigma_a={\sigma_1-\sigma_2}

2 x 219 =  σ₁ -  σ₂

 σ₁ -  σ₂ = 438 MPa    --------2

By adding the above equation

2  σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}

Stress\ ratio =\dfrac{-172.8}{265.2}

Stress\ ratio =-0.65

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

8 0
3 years ago
The AGC control voltage: ___________
lyudmila [28]

Answer:

The AGC circuit operates with an input voltage range of 60 dB (5 mV p-p to 5 V p-p), with a fixed output voltage of 250 mV p-p.

Explanation:

3 0
3 years ago
What type of models can be communicated in more than one way.
guajiro [1.7K]
A 3-D model can be communicated, and can also be a visual model.
4 0
3 years ago
Write a program that prompts the user to enter time in 12-hour notation. The program then outputs the time in 24-hour notation.
Juliette [100K]

Answer:

THE CODE FOR THE PROGRAM IS GIVEN BELOW:

#include <iostream>

#include "ConvertTimeHeader.h"

using namespace std;

int main()

{

convertTime convert;

int hr, mn, sc = 0;

 

cout << "Please input hours in 12 hr notation: ";

cin >> hr;

cout << "Please input minutes: ";

cin >> mn;

cout << "Please input seconds: ";

cin >> sc;

 

convert.invalidHr(hr);

convert.invalidMin(mn);

convert.invalidSec(sc);

convert.printMilTime();

 

system("Pause");

 

return 0;  

 

}

#include <iostream>

#include "ConvertTimeHeader.h"

using namespace std;

int convertTime::invalidHr (int hour)

{

try{

 if (hour < 13 && hour > 0)

  {hour = hour + 12;

  return hour;}

 else{

 

  cin.clear();

  cin.ignore();

  cout << "Invalid input! Please input hour again in correct 12 hour format: ";

  cin >> hour;

  invalidHr(hour);

  throw 10;

 }

   

}

catch (int c) { cout << "Invalid hour input!";}

}

int convertTime::invalidMin (int min)

{

try{

 if (min < 60 && min > 0)

  {return min;}

 else{

 

  cin.clear();

  cin.ignore();

  cout << "Invalid input! Please input minutes again in correct 12 hour format: ";

  cin >> min;

  invalidMin(min);

  throw 20;

  return 0;

 }

   

}

catch (int e) { cout << "Invalid minute input!" << endl;}

}

int convertTime::invalidSec(int sec)

{

try{

 if (sec < 60 && sec > 0)

  {return sec;}

 else{

 

  cin.clear();

  cin.ignore();

  cout << "Invalid input! Please input seconds again in correct 12 hour format: ";

  cin >> sec;

  invalidSec(sec);

  throw 30;

  return 0;

 }

   

}

catch (int t) { cout << "Invalid second input!" << endl;}

}

void convertTime::printMilTime()

{

cout << "Your time converted: " << hour << ":" << min << ":" << sec;

}

Explanation:

4 0
3 years ago
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