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yan [13]
3 years ago
12

A negative normal strain can be considered to increase or decrease volume depending on the coordinate system used. a)True b)- Fa

lse
Engineering
1 answer:
geniusboy [140]3 years ago
6 0

Answer:

The given statement "A negative normal strain can be considered to increase or decrease volume depending on coordinate system used" is

b) False

Explanation:

Normal strain refers to the strain due to normal stress which is when the applied stress is perpendicular to the surface.

Negative normal strain results in compression or contraction further leading to a decrease in volume while a positive normal strain results in elongation thus giving rise to an increase in volume.

You might be interested in
Sketch the velocity profile for laminar and turbulent flow.
Margarita [4]

Answer:

The laminar flow is generally given in high viscosity fluids such as honey or oil, it has the characteristic of flowing in an orderly manner, the walls of the tube have a zero speed while in the center it has a maximum speed.

turbulent flow is characterized by fluid velocity vectors presenting themselves in a disorderly manner and in all directions.

I attached the drawings for the velocity profile in laminar and turbulent flow.

4 0
3 years ago
A hollow aluminum sphere, with an electrical heater in the center, is used in tests to determine the thermal conductivity of ins
Stella [2.4K]

Answer:

K_{ins}=\frac {0.157892}{2.854263}=0.055318 W/m.K

Explanation:

Generally, thermal resistance for conduction heat transfer in a sphere.

R_{cond} = \frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi K}}  

Where R_{cond} is the thermal resistance for conduction, K is the thermal conductivity of the material, r_{i} is the inner radius of the sphere, and r_{o} is the outer radius of the sphere.

The surface area of sphere, A_{s} is given by

A_{s}=4\pi {r^2}

For aluminum sphere, the thermal resistance for conductive heat transfer is given by

Calculate the thermal resistance for conductive heat transfer through the aluminum sphere.

R_{cond,s{\rm{ - 1}}} = \frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}}

Where K_{Al} is aluminum’s thermal conductivity at T_{s}

Thermal resistance for conductive heat transfer through the insulation.

R_{cond,1{\rm{ - 2}}} = \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}}

Thermal resistance for convection is given by

R_{conv} = \frac{1}{{hA}}

Where h is convective heat transfer coefficient, R_{conv} is thermal resistance for convection and A is the cross-sectional area normal to the direction of flow of heat energy

Thermal resistance for convective heat transfer in-between the outer surface of the insulation and the ambient air.

R_{conv,2{\rm{ - }}\infty } = \frac{1}{{h{A_s}}}

Where h represents convective heat transfer coefficient at the outer surface of the insulation. Since A_{s} is already defined, substituting it into the above formula yields

R_{conv,2{\rm{ - }}\infty } = \frac{1}{{h\left( {4\pi {r^2}} \right)}}

To obtain radial distance of the outer surface of the insulation from the center of the sphere.

r = r_{o} + t where t is thickness of insulation

r=0.21+0.15=0.36m

Total thermal resistance

R_{eq} = {R_{cond,s{\rm{ - 1}}}} + {R_{cond,1{\rm{ - 2}}}} +{R_{conv,2{\rm{ - }}\infty }}

Where R_{eq} is total thermal resistance

R_{eq} = \frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}} + \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}} + \frac{1}{{h\left( {4\pi {r^2}} \right)}}

Consider the thermal conductivity of aluminum at temperature T_{s} as 234W/m.K

Rate of heat transfer for the given process

\dot Q_{s - \infty } = \frac{{{T_s} - {T_\infty }}}{{{R_{eq}}}}

Where \dot Q_{s - \infty }} is the steady state heat transfer rate in-between the inner surface of the sphere and the ambient air.

Substituting \left( {\frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}} + \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}} + \frac{1}{{h\left( {4\pi {r^2}} \right)}}} \right) for R_{eq} we obtain

\dot Q_{s - \infty } = \frac{{{T_s} - {T_\infty }}}{{\left( {\frac{{\left( {1/{r_i}} \right) - \left( {1/{r_o}} \right)}}{{4\pi {K_{Al}}}} + \frac{{\left( {1/{r_o}} \right) - \left( {1/r} \right)}}{{4\pi {K_{ins}}}} + \frac{1}{{h\left( {4\pi {r^2}} \right)}}} \right)}}

\begin{array}{l}\\80{\rm{ W}} = \frac{{250{\rm{ }}^\circ {\rm{C}} - 20{\rm{ }}^\circ {\rm{C}}}}{{\left( {\frac{{\left( {\frac{1}{{0.18{\rm{ m}}}}} \right) - \left( {\frac{1}{{0.21{\rm{ m}}}}} \right)}}{{4\pi \left( {234{\rm{ W/m}} \cdot {\rm{K}}} \right)}} + \frac{1}{{30{\rm{ W/}}{{\rm{m}}^2} \cdot {\rm{K}}\left( {4\pi {{\left( {0.36{\rm{ m}}} \right)}^2}} \right)}}\frac{{\left( {\frac{1}{{0.21{\rm{ m}}}}} \right) - \left( {\frac{1}{{0.36{\rm{ m}}}}} \right)}}{{4\pi {K_{ins}}}} + } \right)}}\\\\80{\rm{ W}}\left( {{\rm{0}}{\rm{0.020737 K/W}} + \frac{{{\rm{0}}{\rm{0.157892/m}}}}{{{K_{ins}}}}} \right) = 230{\rm{ K}}\\\\\frac{{{\rm{0}}{\rm{0.157892/m}}}}{{{K_{ins}}}} = \frac{{230{\rm{ K}}}}{{80{\rm{ W}}}} - {\rm{0}}{\rm{0.020737 K/W}}\\\\\frac{{{\rm{0}}{\rm{0.157892/m}}}}{{{K_{ins}}}} = {\rm{2}}{\rm{.854263 K/W}}\\\end{array}

K_{ins}=\frac {0.157892}{2.854263}=0.055318 W/m.K

7 0
3 years ago
Implement a quick sort algorithm that will accept an integer array of size n and in random order. Develop or research three diff
Nostrana [21]

Answer:

#include <cstdlib>  

#include <iostream>  

#include <array>  

using namespace std;  

const string APP_NAME = "Quick Sort Algorithm";  

const array<string, 3> MENU_OPTIONS = {  

"Simulate with Random data",  

"Enter data",  

"Exit program"  

};  

void printMenuOptions() {  

cout << endl << "---------------------------" << endl;  

cout << APP_NAME << endl;  

cout << "---------------------------" << endl;  

for (int i=0; i<MENU_OPTIONS.size(); i++) {  

cout << i+1 << ". " << MENU_OPTIONS[i] << endl;  

}  

cout << endl << "Select an option: ";  

}  

int getRandomInt(int min, int max) {  

return min + (static_cast<int>(rand() % (max - min + 1)));  

}  

bool inArray(int value, int* arr, int size) {  

bool found = false;  

for (int i=0; i<size; i++) {  

if (arr[i] == value) {  

found = true;  

}  

}  

return found;  

}  

void generateRandomArrays(int size, int* arr0, int* arr1, int* arr2, int* arr3) {  

int value;  

bool ok = false;  

for (int i=0; i<size; i++) {  

while (!ok) {  

value = getRandomInt(1, size*10);  

if (!inArray(value, arr0, size)) {  

arr0[i] = value;  

arr1[i] = value;  

arr2[i] = value;  

arr3[i] = value;  

ok = true;  

}  

}  

ok = false;  

}  

}  

void print(int* data, int size) {  

for (int i=0; i<size; i++) {  

cout << data[i] << " ";  

}  

}  

int getPivot(int first, int last, int approach) {  

int pivot;  

switch (approach) {  

case 2:  

pivot = first;  

break;  

case 3:  

pivot = last;  

break;  

case 1:  

default:  

pivot = (first + last) / 2;  

}  

return pivot;  

}  

void swap(int* data, int i, int j) {  

int temp = data[i];  

data[i] = data[j];  

data[j] = temp;  

}  

int quickSort(int* data, int first, int last, int approach) {  

int ops = 0;  

int i = first;  

int j = last;  

int pivot = getPivot(i, j, approach);  

while (i <= j) {  

while (data[i] < data[pivot]) {  

i++;  

}  

while (data[j] > data[pivot]) {  

j--;  

}  

if (i <= j) {  

ops++;  

swap(data, i, j);  

i++;  

j--;  

}  

}  

if (j > first) {  

ops += quickSort(data, first, j, approach);  

}  

if (i < last) {  

ops += quickSort(data, i, last, approach);  

}  

return ops;  

}  

void simulate(int size, bool display) {  

int* data0 = new int[size];  

int* data1 = new int[size];  

int* data2 = new int[size];  

int* data3 = new int[size];  

int ops1, ops2, ops3;  

generateRandomArrays(size, data0, data1, data2, data3);  

ops1 = quickSort(data1, 0, size-1, 1);  

ops2 = quickSort(data2, 0, size-1, 2);  

ops3 = quickSort(data3, 0, size-1, 3);  

if (display) {  

cout << "Unsorted Array: ";  

print(data0, size);  

}  

cout << endl << endl << "> QuickSort #1: pivot is at the median" << endl;  

cout << "Swaps done: " << ops1 << endl;  

if (display) {  

cout << "Sorted Array: ";  

print(data1, size);  

}  

cout << endl << endl << "> QuickSort #2: pivot is at the start" << endl;  

cout << "Swaps done: " << ops2 << endl;  

if (display) {  

cout << "Sorted Array: ";  

print(data2, size);  

}  

cout << endl << endl << "> QuickSort #3: pivot is at the end" << endl;  

cout << "Swaps done: " << ops3 << endl;  

if (display) {  

cout << "Sorted Array: ";  

print(data3, size);  

}  

}  

void enterArray(int size, bool display) {  

// declare some variables  

int* data0 = new int[size];  

int* data1 = new int[size];  

int* data2 = new int[size];  

int* data3 = new int[size];  

int ops1, ops2, ops3;  

int value;  

for (int i=0; i<size; i++) {  

cout << "Enter value " << i+1 << " of " << size << ": ";  

cin >> value;  

data0[i] = value;  

data1[i] = value;  

data2[i] = value;  

data3[i] = value;  

}  

ops1 = quickSort(data1, 0, size-1, 1);  

ops2 = quickSort(data2, 0, size-1, 2);  

ops3 = quickSort(data3, 0, size-1, 3);  

if (display) {  

cout << "Unsorted Array: ";  

print(data0, size);  

}  

cout << endl << endl << "> QuickSort #1: pivot is at the median" << endl;  

cout << "Swaps done: " << ops1 << endl;  

if (display) {  

cout << "Sorted Array: ";  

print(data1, size);  

}  

cout << endl << endl << "> QuickSort #2: pivot is at the start" << endl;  

cout << "Swaps done: " << ops2 << endl;  

if (display) {  

cout << "Sorted Array: ";  

print(data2, size);  

}  

cout << endl << endl << "> QuickSort #3: pivot is at the end" << endl;  

cout << "Swaps done: " << ops3 << endl;  

if (display) {  

cout << "Sorted Array: ";  

print(data3, size);  

}  

}  

int main(int argc, char** argv) {  

int choice;  

char option;  

int num;  

bool end = false;  

bool display = false;  

while (!end) {  

printMenuOptions();  

cin >> choice;  

switch (choice) {  

case 1:  

cout << endl << "Enter size of array (elements will be integers randomly generated): ";  

cin >> num;  

if (num > 0) {  

cout << "Values will be randomly generated from 1 to " << num*10 << endl;  

cout << "Do you want to display the sorted arrays? <y/N>: ";  

cin >> option;  

display = (option == 'y') ? true : false;  

simulate(num, display);  

} else {  

cout << endl << "Incorrect size." << endl;  

}  

break;  

case 2:  

cout << endl << "Enter size of array (you will enter the numbers): ";  

cin >> num;  

if (num > 0) {  

cout << "Do you want to display the sorted arrays? <y/N>: ";  

cin >> option;  

display = (option == 'y') ? true : false;  

enterArray(num, display);  

} else {  

cout << endl << "Incorrect size." << endl;  

}  

break;  

case 3:  

end = true;  

break;  

default:  

cout << endl << "Incorrect option. Try again." << endl;  

}  

}  

return 0;  

}

8 0
3 years ago
Which of the following answers regarding Mealy and Moore Machines are true?
Rom4ik [11]

Answer:

suck bro ☺️☺️ lol

5 0
3 years ago
Find the altitude of the right cylinder of maximum convex surface that can be inscribed in a given sphere.
strojnjashka [21]

Answer:

The radius 4 is maximum in convex surface

5 0
2 years ago
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