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Bad White [126]
3 years ago
12

James has a USB flash drive that he has used at work. The drive needs to be thrown away, but James wants to make sure that the d

ata is no longer on the drive before he throws it away. What can James use to wipe the data clean?
a. Zero-fill utility
b. Format the drive
c. ATA Secure Erase
d. Smash the USB drive
Computers and Technology
1 answer:
loris [4]3 years ago
7 0

Answer:

C. ATA Secure Erase

D. Smash the USB drive

Explanation:

Zero fill utility is a specialized way of formatting a storage device particularly secondary storage such as hard disk, flash drive e.t.c. In this process, the disk contents are overwritten with zeros. Once this has been done, undoing is essentially hard but possible. In most cases, this might just mean that the data content is corrupt and as such might still be recovered, maybe not in its pure entirety.

Formatting the drive on another hand does not essentially mean cleaning the drive and wiping off data. It just means that operating systems cannot see those data contents anymore. They are still present in the drive and can be recovered.

ATA Secure Erase is actually a way of completely and permanently erasing the content in a drive. Once the function has been done, undoing is not possible. Both the data content and even the data management table will be completely gone.

Smashing the USB drive is the surest way of cleaning data as that will permanently destroy the working components of the drive such as the memory chip. And once that happens then there's no drive let alone its contents.

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In this program you will read in the number of seconds and convert it to days, hours, minutes and remaining seconds. Your progra
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// here is code in c++.

// include headers

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

// variables

long long int inp_sec,days,hours,minutes,sec;

long long int s;

cout<<"please enter the seconds:";

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s=inp_sec;

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 // compute hours, 3600 seconds in an hour

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inp_sec=inp_sec%3600;

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minutes=inp_sec/60;

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sec=inp_sec%60;

 // print the output

cout<<s<<" seconds is equal to "<<days<<" days "<<hours<<" hours "<<minutes<<" minutes "<<sec<<" seconds "<<endl;

return 0;

}

Explanation:

Declare variables "inp_sec","days","hours","minutes" and "sec" of long long int type. read the seconds from user and assign it to variable "inp_sec".Make a copy of input  seconds.Then compute the days by dividing inp_sec with 86400 and update the inp_sec. then find hours by dividing inp_sec with 3600 and update the inp_sec.Similarly  find the minutes and remaining seconds.After this print the output.

Output:

please enter the seconds:83647362                                                                                                                            

83647362 seconds is equal to 968 days 3 hours 22 minutes 42 seconds

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AlekseyPX

Answer:

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Explanation:

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