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Elodia [21]
3 years ago
5

Radio Frequency IDentification (RFID) tags and readers are a category of low-end wireless devices that people may not recognize

as forming a computer network. Which of the statements below are true about RFID?
i. RFID technology takes many forms, used in smartcards, implants for pets, passports, library books and more
ii. with the EPC (Electronic Product Code) RFID tags are small, inexpensive devices that have a unique identifier and a small amount of memory that can be read and written by an RFID reader
iii. the tags are the intelligence in the system, analogous to base stations and access points in WiFi networks
iv. in the EPC Gen 2 Physical Layer the data is transmitted in the same way as in all other wireless situations we have seen so far
Engineering
1 answer:
Fittoniya [83]3 years ago
8 0

Answer:

See explaination.

Explanation:

Radio Frequency Identification (RFID) tags and readers uses the electromagnetic waves to identify and track the attached objects.

A tag is attached to the object which is to be identified or tracked, and reader is used to read the response and send the acknowledgement. Therefore, RFID tags and readers are used in many industries, passports, transportations and pet identification etc.

i.

RFID technology is used in smartcards, implants for pets, passports and library books to identify and track the persons, objects and pets etc.

Hence, we can say that option (i) is true.

ii.

Electronic Product Code (EPC) is a small code stored in the RFID tag. The code stored in the memory is 96 bits which are used to identify the organization which manages the data, unique number to identify the product and a number to identify the particular tag and etc.

EPC is a unique identification number, it can read and be written by the RFID reader. It is used in supply chains instead of a bar code even though expansive.

Hence, option (ii) is true.

iii.

Tags are used to identify and track the objects. It doesn’t belong to base stations and access points as a Wi-Fi networks.

Therefore, option (iii) is false.

iv.

The EPC Generation 2 RFID tag is used to improve the security by enabling the authentication features. It is not the similar way of data transmission in the other wireless situations.

Therefore, option (iv) is false.

Finally, the options (i) and (ii) are TRUE, while (iii) and (iv) are FALSE.

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Find the remaining trigonometric function of 0 if
amm1812

Answer:

Hope this will help you

have a great day

8 0
3 years ago
A 10-mm-diameter Brinell hardness indenter produced an indentation 2.50 mm in diameter in a steel alloy when a load of 1000 kg w
zysi [14]

Answer:

HB \approx 200.484

Explanation:

The Brinell Hardness is obtained from the following formula:

HB = \frac{2\cdot P}{\pi \cdot D^{2}}\cdot \left(\frac{1}{1 - \sqrt{1-\frac{d^{2}}{D^{2}} } } \right )

Where P is the load measured in kilograms-force, D is the diameter of indenter measured in millimeters and d is the diameter of indentation measured in millimeters.

HB=\frac{2\cdot (1000\,kgf)}{\pi\cdot (10\,mm)^{2}}\cdot \left[ \frac{1}{1-\sqrt{1-\frac{(2.50\,mm)^{2}}{(10\,mm)^{2}} } }  \right ]

HB \approx 200.484

4 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
(100 POINTS) {BRIANLIEST} PLEASE HELP ME
zhuklara [117]

Answer:

I think option d is the answer

6 0
3 years ago
NO reacts with Br2 in the gas phase according to the following chemical equation: 2NO(g) +Br2(g)2NOBr(g) It is observed that, wh
klemol [59]

Answer:

a) rate=r=k[NO]^{2} [Br_{2}]^{1}

b) k=\frac{1}{s*M^{2}}

Explanation:

First of all you need to indicate the reaction order of each reactant ( NO and Br_{2}):

1.  Br_{2}

Note that if Br_{2} concentration ([Br_{2} ]) is reduced to 1/3 of its initial value, the rate of the reaction is also reduced to 1/3 of its initial value, it means:

[Br_{2} ]=1/3 then r=1/3

As the change in the rate of the reaction is equal to the change of the initial concentration of  Br_{2}, you could concluded that the reaction is first order with respect to  Br_{2}

2. NO

Now, note that if NO concentration ([NO]) is multiplied by 3.69, the rate of the reaction increases by a factor of 13.6. In this case, to know the ratio could be advisable divide the rate of the reaction (13.6) over the factor whereby was multiplied the concentration (3.69), as follows:

\frac{13.6}{3.69}=3.69

As the result is the same factor 3.69 you could concluded that the change of the rate of reaction is proportional to the square of the concentration of A:

r=[NO]^{2} =3.68^{2} =13.6

It means that the reaction is second order with respect to NO

3. Rate Expression

Remember that the rate expression of the reactions depend on the concentration of each reactant and its order. In this case we have 2 reactants: NO and Br_{2}, then we have a rate law depending of  2 concentrations, as follows:

<h2>rate=r=k[NO]^{2} [Br_{2}]^{1}</h2>

Note that the expression is the result of the concentration of each reactant raised to its reaction order (previously determined)

<em>Note: I hope that you do not mix up the use of the rates of reaction of each reactant, that is experimentally determined, with the stoichiometric coefficient, are different.</em>

4. Rate constant units (k)

Assuming concentration is expressed as \frac{mol}{L}=M and time is in second, to find the units of k we need to solve an equation with units and with supporting of the rate equation previously obtained, as follows:

r=k[NO]^{2} [Br_{2}]^{1}

Where:

[r]=[\frac{M}{s}]

[[NO]]=[M]

[ [Br_{2}]]=[M]

Then:

\frac{M}{s}=kM^{2} M^{1}

\frac{M}{s}=kM^{3}

\frac{M}{s*M^{3}}=k

The units of the rate constant k are:

k=\frac{1}{s*M^{2}}

8 0
3 years ago
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