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slamgirl [31]
3 years ago
12

Digital data can also take the form of light and dark spots etched onto the surface of a CD or the positive and negative orienta

tion of_________
Computers and Technology
2 answers:
Rus_ich [418]3 years ago
7 0

Answer:

Magnetic particles on a hard disk.

Explanation:

Computer storage devices are permanent storage devices used to store large data for a long time. Unlike the memory in Computers, it is not volatile ( lost when power goes off). There are different types of storage devices and different forms for which they store data.

The types of storage are Optical storage like CD and DVD, magnetic storage as in hard disk drive ( HDD) and solid state drive (SSD), and holographic storage devices.

The optical disk uses dark and white light to represent data in its disk, while the magnetic disk represent data with the positive and negative values. But the holographic storage uses light refractions at different angles.

statuscvo [17]3 years ago
7 0

Answer: Magnetic particles on a hard disk

Explanation:

magnetic disk is a storage device that uses a magnetization process to write, rewrite and access data. Data is stored on the platters by magnetizing the tiny particles on the surface.

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Will technology be the destruction or salvation of human-kind
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Write an algorithm that prints out all the subsets of 3 elements of a set of n elements. The elements of the set are stored in a
Ksenya-84 [330]

void Print_3_Element_Subsets(int n,const element_type S[])

{

if(n<3) // condition to check if list have less than 3 elements

{

print("No subset found");

}

for(int i =0; i < n ; i++)

{

for(int j =i+1; j < n ; j++)

{

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{

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}

C++ Implementation;

#include<cstdlib>

#include<time.h>

using namespace std;

int S1[5]= {1,2,3,4,5};

int n1 = 5;

int S2[2]={1,2};

int n2 = 2;

void Print_3_Element_Subsets(int n,const int S[])

{

if(n<3) // condition to check if list have less than 3 elements

{

printf("No subset found\n");

}

printf("3 Subsets are\n");

for(int i =0; i < n ; i++)

{

for(int j =i+1; j < n ; j++)

{

for(int k =j+1; k < n ; k++)

{

cout<<S[i]<<" "<<S[j]<<" "<<S[k]<<endl;

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}

}

int main()

{

cout<<"Case 1"<<endl;

Print_3_Element_Subsets(n1,S1);

cout<<endl<<"Case 2"<<endl;

Print_3_Element_Subsets(n2,S2);

}

OUTPUT

Case 1

3 Subsets are

1 2 3

1 2 4

1 2 5

1 3 4

1 3 5

1 4 5

2 3 4

2 3 5

2 4 5

3 4 5

Case 2

No subset found

3 Subsets are

--------------------------------

Process exited after 0.0137 seconds with return value 0

Press any key to continue . . .

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Answer:

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