1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Tema [17]
3 years ago
5

2. How many amperes of current will flow through a circuit that has 3 ohms of resistance if

Engineering
1 answer:
Dmitriy789 [7]3 years ago
4 0

Answer:

4A

Explanation:

From Ohms law ;

R= V/I where R is resistance, V is voltage and I is current

Given that;

R= 3Ω

V= 12

Using the values in the formula as;

R= V/I

3=12/I

I= 12/3

I= 4 A

You might be interested in
A vertical plate has a sharp-edged orifice at its center. A water jet of speed V strikes the plate concentrically. Obtain an exp
Vesnalui [34]

Answer:

Force = 455672061 N

Explanation:

given data

V = 515 ft/s

D = 54 in = 4.5 ft

d = 51 in = 4.25 ft

to find out

Evaluate the force

solution

we know that force is express as

force F = ρ ΔA v²     ..................1

put here value we get

force F = 1000 × \frac{\pi}{4} [ D² - d² ] × v²

F = 1000 × \frac{\pi}{4} [ 4.5² - 4.25² ] × 515²

Force = 455672061 N

8 0
4 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
The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the allowable shearing stress is 25 MPa. Rod AB is
nexus9112 [7]

Answer:

The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the allowable shearing stress is 25 MPa. Rod AB is hollow and has an outer diameter of 25 mm; it is made of a brass for which the allowable shearing stress is 50 MPa.

4 0
3 years ago
Germanium forms a substitutional solid solution with silicon. Compute the number of germanium atoms per cubic centimeter for a g
brilliants [131]

Answer:

The number of germanium atoms per cubic centimeter for this germanium-silicon alloy is 3.16 x 10²¹ atoms/cm³.

Explanation:

Concentration of Ge (C_{Ge}) = 15%

Concentration of Si (C_{Si}) = 85%

Density of Germanium (ρ_{Ge}) = 5.32 g/cm³

Density of Silicon (ρ_{Si}) = 2.33 g/cm³

Atomic mass of Ge (A_{Ge})= 72.64 g/mol

To calculate the number of Ge atoms per cubic centimeter for the alloy, we will use the formula:

No of Ge atoms/cm³=[Avogadro's Number*C_{Ge}]/([C_{Ge}*A_{Ge}/ρ_{Ge})+(C_{Si}*A_{Ge}/ρ_{Si})]

                              = (6.02x10²³ * 15%) / [(15% * 72.64/5.32)+(72.64*85%/2.33)]

                              = (9.03x10²²)/(2.048+26.499)

                              = (9.03x10²²)/(28.547)

No of Ge atoms/cm³ = 3.16 x 10²¹ atoms/cm³

3 0
3 years ago
Explain biometric senser.​
Contact [7]

Biometric sensors are used to collect measurable biological characteristics from a human being, which can then be used in conjunction with biometric recognition algorithms to perform automated person identification.

8 0
3 years ago
Read 2 more answers
Other questions:
  • A heat pump with an ideal compressor operates between 0.2 MPa and 1 MPa. Refrigerant R134a flows through the system at a rate of
    15·1 answer
  • 10 kg/s Propane at 10 bar and 20 C is directed to an adiabatic rigid mixer and is mixed with 20 kg/s Propane at 10 bar and 40 C.
    11·1 answer
  • If the value of the feedback resistor in the filter is changed but the value of the resistor in the forward path is unchanged, w
    14·1 answer
  • Analysis of the relationship between the fuel economy​ (mpg) and engine size​ (in liters) for 35 models of cars produces the reg
    15·1 answer
  • In this assignment, you will demonstrate your ability to write simple shell scripts. This is a cumulative assignment that will c
    8·1 answer
  • Giving 100 points for answer plzz
    10·2 answers
  • Pleaseeeeeeeeeeeeee answer
    15·2 answers
  • Whats the thing in the picture? Its a part of a computer.
    14·2 answers
  • How can any student outside apply for studying engineering at Cambridge University​
    7·1 answer
  • Engineering Specificaitons is ...
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!