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skelet666 [1.2K]
3 years ago
13

A semiconductor kit is used on a

Computers and Technology
1 answer:
iragen [17]3 years ago
5 0

The correct answer is PSC motor.

If you need more help, you can check on quizlet, I learnt about ecd from there lol

You might be interested in
#include #include int main( ) { int value = 10; int pid; value += 5; pid = fork( ); if (pid > 0 ) { value += 20; } printf(val
ss7ja [257]

Explanation:

To understand how this program is working let us print the variable value at different stages of the program so that we can understand how it is working.

Intitally, value=10 when it was declared.

Then we added 5 and it become value=15

then we used fork() function which creates a parent(orignal) and child(duplicate)

When fork() succeeds it returns the child pid to parent and returns 0 to the child. As you can see (pid > 0) condition is always true therefore the parent pid value becomes 35 ( 15+20) and the child pid value becomes 0.

#include <stdio.h>

#include <unistd.h>

int main( ) {

   int value = 10;

     printf("%d\n",value);

   int pid;

   value += 5;

     printf("%d\n",value);

   pid = fork( );

     printf("%d\n",pid);

   if (pid > 0 )

   {

       value += 20;

   }

   printf("%d\n",value);

   return 0;

}

Output:

10 (initial value)

15 (modified value)

5343 (pid when fork is used)

35 (final modified value)

0 (child value)

15 (the parent value when fork was used)

8 0
4 years ago
Determine the number of cache sets (S), tag bits (t), set index bits (s), and block offset bits (b) for a 1024-byte 4-way set as
Cerrena [4.2K]

Answer:

The answer is The Cache Sets (S) = 32, Tag bits (t)=24, Set index bits(s) = 5 and Block offset bits (b) = 3

Explanation:

Solution

Given Data:

Physical address = 32 bit (memory address)

Cache size = 1024 bytes

Block size = 8 bytes

Now

It is a 4 way set associative mapping, so the set size becomes 4 blocks.

Thus

Number of blocks = cache size/block size

=1024/8

=128

The number of blocks = 128

=2^7

The number of sets = No of blocks/set size

=128/4

= 32

Hence the number of sets = 32

←Block ←number→

Tag → Set number→Block offset

←32 bit→

Now, =

The block offset = Log₂ (block size)

=Log₂⁸ = Log₂^2^3 =3

Then

Set number pc nothing but set index number

Set number = Log₂ (sets) = log₂³² =5

The remaining bits are tag bits.

Thus

Tag bits = Memory -Address Bits- (Block offset bits + set number bits)

= 32 - (3+5)

=32-8

=24

So,

Tag bits = 24

Therefore

The Cache Sets = 32

Tag bits =24

Set index bits = 5

Block offset bits = 3

Note: ←32 bits→

Tag 24 → Set index  5→Block offset 3

7 0
3 years ago
The process in which a server is partitioned into smaller components virtually.
ExtremeBDS [4]
The answer is Allocation
5 0
3 years ago
Leo noticed that attackers have breached his wireless network. They seem to have used a brute-force attack on the WiFi protected
kakasveta [241]

Answer:

It is called a WPS brutal force attack.

Explanation:

Wired and wireless networks are both susceptible to attacks. The wired network, the advantage as a cable connection, is more secure than wireless networks, but wireless network also have security measures like the wifi protected set up (WPS).

WPS is used to connect to a network without passphrase, but with a key combination or a PIN.

Brutal force attacks are used on WPS to forcefully generate the PIN, using a third party software.

3 0
3 years ago
Now the y0utube home screen is gone
erik [133]

Answer:

oh no thanks not good

Explanation:

did you try it on a different device mabye a phone

7 0
3 years ago
Read 2 more answers
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