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Harlamova29_29 [7]
3 years ago
7

(a) For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary e

nergy? Why? (b) The grain boundary energy of a small-angle grain boundary is less than for a high-angle one. Why is this so?
Engineering
1 answer:
aksik [14]3 years ago
5 0

Answer:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

Explanation:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

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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:

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3 years ago
Why does teachers grade things that are not due yet​
pav-90 [236]
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In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by
zvonat [6]

Answer:

R= 1.25

Explanation:

As given the local heat transfer,

Nu_x = 0.035 Re^{0.8}_x Pr^{1/3}

But we know as well that,

Nu=\frac{hx}{k}\\h=\frac{Nuk}{x}

Replacing the values

h_x=Nu_x \frac{k}{x}\\h_x= 0.035Re^{0.8}_xPr^{1/3} \frac{k}{x}

Reynolds number is define as,

Re_x = \frac{Vx}{\upsilon}

Where V is the velocity of the fluid and \upsilon is the Kinematic viscosity

Then replacing we have

h_x=0.035(\frac{Vx}{\upsilon})^{0.8}Pr^{1/3}kx^{-1}

h_x=0.035(\frac{V}{\upsilon})^{0.8}Pr^{1/3}kx^{0.8-1}

h_x=Ax^{-0.2}

<em>*Note that A is just a 'summary' of all of that constat there.</em>

<em>That is A=0.035(\frac{V}{\upsilon})^{0.8}Pr^{1/3}k</em>

Therefore at x=L the local convection heat transfer coefficient is

h_{x=L}=AL^{-0.2}

Definen that we need to find the average convection heat transfer coefficient in the entire plate lenght, so

h=\frac{1}{L}\int\limit^L_0 h_x dx\\h=\frac{1}{L}\int\limit^L_0 AL^{-0.2}dx\\h=\frac{A}{0.8L}L^{0.8}\\h=1.25AL^{-0.2}

The ratio of the average heat transfer coefficient over the entire plate  to the local convection heat transfer coefficient is

R = \frac{h}{h_L}\\R= \frac{1.25Al^{-0.2}}{AL^{-0.2}}\\R= 1.25

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BabaBlast [244]

Properties of Carpenter's hammer possess

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2.The hammer head should also be very hard, so that it does not deform while driving or eradicate any nails in wood.

3.Carpenter's hammer is used to impact smaller areas of an object.It can drive nails in the wood,can crush  the rock and shape the metal.It is not suitable for heavy work.

How hammer head is manufactured :

1.Hammer head is produced by metal forging process.

2.In this process metal is heated and this molten metal is placed in the cavities said to be dies.

3.One die is fixed and another die is movable.Ram forces the two dies under the forces which gives the metal desired shape.

4.The third process is repeated for several times.

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A short framing member that fills the space between the rough sill and the soleplate is a cripple stud.

(HAVE A GOOD DAY!!!)

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