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tankabanditka [31]
3 years ago
10

An operating gear box (transmission) has 350 hp at its input shaft while 250. hp are delivered to the output shaft. The gear box

has a steady state surface temperature of 180. °F. Determine the rate of entropy production by the gear box.
Engineering
1 answer:
True [87]3 years ago
8 0

Answer:

Rate of Entropy =210.14 J/K-s

Explanation:

given data:

power delivered to input = 350 hp

power delivered to output = 250 hp

temperature of surface = 180°F

rate of entropy is given as

Rate\  of\ entropy  = \frac{Rate\ of \ heat\  released}{Temperature}

T = 180°F = 82°C = 355 K

Rate of heat = (350 - 250) hp = 100 hp = 74600 W

Rate of Entropy= \frac{74600}{355} = 210.14 J/K-s

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What are the causes and solutions of social problems ?what are the causes and solutions of social problems ​
aksik [14]

Answer:

causes: unemployment, poverty,Lack of education,urbanization e.t.c

solutions:making people educated, giving people jobs

7 0
3 years ago
In a mechanical assembly operation, the first work unit required 7.83 min to complete and the learning rate for mechanical assem
Marianna [84]

Answer:

(a) 6.5774 minutes

(b) 4.3879 minutes

(c) 2.459 minutes

(d) 245.902 minutes

Explanation:

The learning curve model follows the following equation:

Y=aX^b

where Y are the units time to produce X units, and a is the time for assembly the first unit. Additionally, b is calculated as:

b=\frac{ln(r)}{ln(2)}

Where r is the learning rate for mechanical assembly. So, b is equal to:

b=\frac{ln(0.84)}{ln(2)}=-0.2515

Then, the equation is:

Y=7.83X^{-0.2515}

Finally, the unit times to produce the second unit are:

Y=7.83(2)^{-0.2515}=6.5774

The 10th unit:

Y=7.83(10)^{-0.2515}=4.3879

The 100th unit:

Y=7.83(100)^{-0.2515}=2.459

Then, the total cumulative time T to produce 100 units in the Crawford model is calculated as:

T=7.83\frac{(100.5)^{1-0.2515}-(0.5)^{1-0.2515})}{1-0.2515} \\T=323.5383

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
4 years ago
How much will it cost to train the entire company to use a recycling program if the training includes paper handouts? (Remember,
bezimeni [28]

Answer:

With the current budget of $1,000, it is possible to train the company and buy 3 recycling carts given that recycling pickup within the area is free

Explanation:

The office waste management budget that is not being spent = $1,000

The given expense parameters are;

The training cost per hour per employee = $12.00

The number of people in the entire company = 45

The cost of preparing the training material per hour = $20.00

The number of people to create the training material = 3 people

The time it will take each person in creating the training material = 3 hours

The cost of recycling cart = $80.00 per cart

The cost of paper handouts per employee = $0.05

The cost of materials include;

The total cost for the training = $12.00 × 45 = $540

The cost for the handout = $0.05 × 45 = $2.25

The cost of preparing the materials = $20.00 × 3 × 3 = $180.00

The total costs of the training = $540 + $4 + $180.00 = $724

The amount available to buy cart = $1,000 - $724 = $276

Therefore;

The amount available to buy cart = $276

The number of carts that can be bought = 276/80 = 3.45 carts

Therefore, we round down to get;

The number of carts that can be bought = 3 carts

8 0
3 years ago
Two plates are separated by a 1/4 in space. The lower plate is stationary; the upper plate moves at 10 ft/s. Oil (viscosity of 2
Montano1993 [528]

Answer:

τ = 0.25 lbf/in²

Explanation:

given that the oil viscosity, μ =  2.415 lb/ft-s

gap between plates = 1/4 inches = 1/4*12 = 1/48 ft

recall from newtons law of viscosity;

shear stress τ = μ du/dy =

  τ = (2.415 lb/ft-s) (10 ft/s)/(1/48) ft

τ = 1159.2 lb/ft-s²

we know that, 1 slug = 32.174 lb

lb = 1/32.174 slug

∴  τ = 1159.2/32.174 slug/ft-s² = 36 slug/ft-s²

τ = 36 slug/ft-s²

multiply both the numerator and denominator by ft, this gives

τ = 36 slug-ft/ft²-s²

τ = 36 lbf/ft² where 1 slug-ft/s² = 1lbf

since 1 ft = 12 inch = 1 ft² = 12² in² = 144 in²

∴  τ = 36/144 lbf /in² = 0.25 lbf/in²

τ = 0.25 lbf/in²

6 0
3 years ago
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