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2 years ago

(C) Describe about the different types of computer peripherals and memory devices.

Computers and Technology

1 answer:

Alecsey [184]2 years ago

3 0

Computer peripheral device

Monitor.

Monitor. Keyboard.

Monitor. Keyboard. Mouse.

Monitor. Keyboard. Mouse. Trackball.

Monitor. Keyboard. Mouse. Trackball. Touchpad.

Monitor. Keyboard. Mouse. Trackball. Touchpad. Pointing stick.

Monitor. Keyboard. Mouse. Trackball. Touchpad. Pointing stick. Joystick.

Monitor. Keyboard. Mouse. Trackball. Touchpad. Pointing stick. Joystick. Light pen.

 MEMORY DEVICES : 

 The main storage is the primary memory, and data and programs are stored in secondary memory. However, memory is not stored in the CPU, but the CPU would only be a mess of wires without it!

RAM (Random Access Memory) and ROM (Read Only Memory) are examples of primary storage. Secondary Storage Devices: Secondary storage is a memory that is stored external to the computer. ... Hard Disk, CD, DVD, Pen/Flash drive, SSD, etc, are examples of secondary storage.

Thus, Computer storage is of two types: ... RAM (Random Access Memory) and ROM (Read Only Memory) are examples of primary storage. Secondary Storage Devices: Secondary storage is a memory that is stored external to the computer. It is mainly used for the permanent and long-term storage of programs and data.

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Given six memory partitions of 100 MB, 170 MB, 40 MB, 205 MB, 300 MB, and 185 MB (in order), how would the first-fit, best-fit,

nlexa [21]

Answer:

We have six memory partitions, let label them:

100MB (F1), 170MB (F2), 40MB (F3), 205MB (F4), 300MB (F5) and 185MB (F6).

We also have six processes, let label them:

200MB (P1), 15MB (P2), 185MB (P3), 75MB (P4), 175MB (P5) and 80MB (P6).

Using First-fit

P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
P2 will be allocated to F1. Therefore, F1 will have a remaining space of 85MB from (100 - 15).
P3 will be allocated F5. Therefore, F5 will have a remaining space of 115MB from (300 - 185).
P4 will be allocated to the remaining space of F1. Since F1 has a remaining space of 85MB, if P4 is assigned there, the remaining space of F1 will be 10MB from (85 - 75).
P5 will be allocated to F6. Therefore, F6 will have a remaining space of 10MB from (185 - 175).
P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using First-fit include: F1 having 10MB, F2 having 90MB, F3 having 40MB as it was not use at all, F4 having 5MB, F5 having 115MB and F6 having 10MB.

Using Best-fit

P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
P2 will be allocated to F3. Therefore, F3 will have a remaining space of 25MB from (40 - 15).
P3 will be allocated to F6. Therefore, F6 will have no remaining space as it is entirely occupied by P3.
P4 will be allocated to F1. Therefore, F1 will have a remaining space of of 25MB from (100 - 75).
P5 will be allocated to F5. Therefore, F5 will have a remaining space of 125MB from (300 - 175).
P6 will be allocated to the part of the remaining space of F5. Therefore, F5 will have a remaining space of 45MB from (125 - 80).

The remaining free space while using Best-fit include: F1 having 25MB, F2 having 170MB as it was not use at all, F3 having 25MB, F4 having 5MB, F5 having 45MB and F6 having no space remaining.

Using Worst-fit

P1 will be allocated to F5. Therefore, F5 will have a remaining space of 100MB from (300 - 200).
P2 will be allocated to F4. Therefore, F4 will have a remaining space of 190MB from (205 - 15).
P3 will be allocated to part of F4 remaining space. Therefore, F4 will have a remaining space of 5MB from (190 - 185).
P4 will be allocated to F6. Therefore, the remaining space of F6 will be 110MB from (185 - 75).
P5 will not be allocated to any of the available space because none can contain it.
P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using Worst-fit include: F1 having 100MB, F2 having 90MB, F3 having 40MB, F4 having 5MB, F5 having 100MB and F6 having 110MB.

Explanation:

First-fit allocate process to the very first available memory that can contain the process.

Best-fit allocate process to the memory that exactly contain the process while trying to minimize creation of smaller partition that might lead to wastage.

Worst-fit allocate process to the largest available memory.

From the answer given; best-fit perform well as all process are allocated to memory and it reduces wastage in the form of smaller partition. Worst-fit is indeed the worst as some process could not be assigned to any memory partition.

8 0

3 years ago

Mention two hardware groups

mars1129 [50]

Answer:

input Devices, Processing Devices, Output Devices, Memory/Storage Devices

Explanation:

4 0

3 years ago

17.

Radda [10]

Sorry to answer you so late but that would be truth because that is the first rule of science which is what thermodynamics depends in.

The first law of thermodynamics is a type of law of conservation but with thermodynamics systems.

So, in gogle I found this: "The law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but cannot be created or destroyed."

Like I said, sorry to reply so late but I hope this helped! =)

3 0

3 years ago

In this project, you will add a (self-proclaimed) priority attribute to xv6 processes. The priority does not actually do anythin

kondor19780726 [428]

Answer:

Priority programming is a process programming method based on priority. In this technique, the developer chooses the tasks to work according to priority, which is different from other types of programming, for example, a simple round-robin.

On UNIX and many other systems, higher priority values represent lower priority processes. Some of the systems, such as Windows, use the opposite convention: a higher number means a higher priority

<h3>Explanation: </h3>

Priorities can be dynamic or static. Static priorities are assigned during creation, while dynamic priorities are assigned according to the behavior of the processes while they are in the system. To illustrate, the planner could favor intensive input / output (I / O) tasks, allowing expensive requests to be issued as soon as possible.

Priorities can be defined internally or externally. Internally defined priorities make use of a measurable amount to calculate the priority of a given process. On the contrary, external priorities are defined using criteria beyond the operating system (OS), which may include the importance of the process, the type and sum of the resources used for the use of the computer, user preferences , trade and other factors such as politics etc.

i hope this is right lol

6 0

3 years ago

A mobile device user has entered her user ID and password to access an online account. The user immediately receives a text mess

tensa zangetsu [6.8K]

Answer:

A. Multifactor authentication

Explanation:

Multifactor authentication is a security system process that requires a user to verify his or her identity by providing two categories of credentials.

A mobile device user has entered her user ID and password to access an online account, she immediately receives a text message with a unique PIN or One Time Password (OTP) that must be entered before she is granted access to the account. This is an example of a multifactor authentication security method.

8 0

3 years ago