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kaheart [24]
2 years ago
14

What is analog computer?​

Computers and Technology
1 answer:
Evgen [1.6K]2 years ago
3 0

Answer:

Analog computers are special purpose computer which can mesuare continuously changing data such as temperature, pressure, voltage,etc.

You might be interested in
Is jesus dead or alive
Morgarella [4.7K]

Answer:

Explanation:

i think he alive but in heaven i dunno

8 0
3 years ago
Suppose that a computer can run an algorithm on a problem of size 1,024 in time t. We do not know the complexity of the algorith
Setler79 [48]

Answer:

Time Complexity of Problem - O(n)

Explanation:

When n= 1024 time taken is t. on a particular computer.

When computer is 8 times faster in same time t , n can be equal to 8192. It means on increasing processing speed input grows linearly.

When computer is 8 times slow then with same time t , n will be 128 which is (1/8)th time 1024.

It means with increase in processing speed by x factor time taken will decrease by (1/x) factor. Or input size can be increased by x times. This signifies that time taken by program grows linearly with input size n. Therefore time complexity of problem will be O(n).

If we double the speed of original machine then we can solve problems of size 2n in time t.

5 0
3 years ago
There will be 10 numbers stored contiguously in the computer at location x 7000 . Write a complete LC-3 program, starting at loc
Artist 52 [7]

Answer:

The LC-3 (Little Computer 3) is an ISA definition for a 16-bit computer. Its architecture includes physical memory mapped I/O via a keyboard and display; TRAPs to the operating system for handling service calls; conditional branches on N, Z, and P condition codes; a subroutine call/return mechanism; a minimal set of operation instructions (ADD, AND, and NOT); and various addressing modes for loads and stores (direct, indirect, Base+offset, PC-relative, and an immediate mode for loading effective addresses). Programs written in LC-3 assembler execute out of a 65536 word memory space. All references to memory, from loading instructions to loading and storing register values, pass through the get Mem Adr() function. The hardware/software function of Project 5 is to translate virtual addresses to physical addresses in a restricted memory space. The following is the default, pass-through, MMU code for all memory references by the LC-3 simulator.

unsigned short int get Mem Adr(int va, int rwFlg)

{

unsigned short int pa;

// Warning: Use of system calls that can cause context switches may result in address translation failure

// You should only need to use gittid() once which has already been called for you below. No other syscalls

// are necessary.

TCB* tcb = get TCB();

int task RPT = tcb [gettid()].RPT;

pa = va;

// turn off virtual addressing for system RAM

if (va < 0x3000) return &memory[va];

return &memory[pa];

} // end get MemAdr

Simple OS, Tasks, and the LC-3 Simulator

We introduce into our simple-os a new task that is an lc3 Task. An lc3 Task is a running LC-3 simulator that executes an LC-3 program loaded into the LC-3 memory. The memory for the LC-3 simulator, however, is a single global array. This single global array for memory means that alllc3 Tasks created by the shell use the same memory for their programs. As all LC-3 programs start at address 0x3000 in LC-3, each task overwrites another tasks LC-3 program when the scheduler swaps task. The LC-3 simulator (lc3 Task) invokes the SWAP command every several LC-3 instruction cycles. This swap invocation means the scheduler is going to be swapping LC-3 tasks before the tasks actually complete execution so over writing another LC-3 task's memory in the LC-3 simulator is not a good thing.

You are going to implement virtual memory for the LC-3 simulator so up to 32 LC-3 tasks can be active in the LC-3 simulator memory without corrupting each others data. To implement the virtual memory, we have routed all accesses to LC-3 memory through a get Mem Adr function that is the MMU for the LC-3 simulator. In essence, we now have a single LC-3 simulator with a single unified global memory array yet we provide multi-tasking in the simulator for up to 32 LC-3 programs running in their own private address space using virtual memory.

We are implementing a two level page table for the virtual memory in this programming task. A two level table relies on referring to two page tables both indexed by separate page numbers to complete an address translation from a virtual to a physical address. The first table is referred to as the root page table or RPT for short. The root page table is a fixed static table that always resides in memory. There is exactly one RPT per LC-3 task. Always.

The memory layout for the LC=3 simulator including the system (kernel) area that is always resident and non-paged (i.e., no virtual address translation).

The two figures try to illustrate the situation. The lower figure below demonstrates the use of the two level page table. The RPT resident in non-virtual memory is first referenced to get the address of the second level user page table or (UPT) for short. The right figure in purple and green illustrates the memory layout more precisely. Anything below the address 0x3000 is considered non-virtual. The address space is not paged. The memory in the region 0x2400 through 0x3000 is reserved for the RPTs for up to thirty-two LC-3 tasks. These tables are again always present in memory and are not paged. Accessing any RPT does not require any type of address translation.

The addresses that reside above 0x3000 require an address translation. The memory area is in the virtual address space of the program. This virtual address space means that a UPT belonging to any given task is accessed using a virtual address. You must use the RPT in the system memory to keep track of the correct physical address for the UPT location. Once you have the physical address of the UPT you can complete the address translation by finding the data frame and combining it with the page offset to arrive at your final absolute physical address.

A Two-level page table for virtual memory management.

x7000 123F x7000 0042

x7001 6534 x7001 6534

x7002 300F x7002 300F

x7003 4005 after the program is run, memory x7003 4005

x7004 3F19

7 0
3 years ago
Read 2 more answers
Write a script that calculates the common factors between 8 and 24. To find a common factor, you can use the modulo operator (%)
AnnyKZ [126]

Answer:

  1. common = []
  2. num1 = 8
  3. num2 = 24
  4. for i in range(1, num1 + 1):
  5.    if(num1 % i == 0 and num2 % i == 0):
  6.        common.append(i)
  7. print(common)

Explanation:

The solution is written in Python 3.

Firstly create a common list to hold a list of the common factor between 8 and 24 (Line 1).

Create two variables num1, and num2 and set 8 and 24 as their values, respectively (Line 3 - 4).

Create a for loop to traverse through the number from 1 to 8 and use modulus operator to check if num1 and num2 are divisible by current i value. If so the remainder of both num1%i and num2%i  will be zero and the if block will run to append the current i value to common list (Line 6-8).

After the loop, print the common list and we shall get [1, 2, 4, 8]

8 0
3 years ago
You created an interactive application named GreenvilleRevenue. The program prompts a user for the number of contestants entered
Korolek [52]

Answer:

Explanation:

The program in this code is written correctly and has the messages applied in the code. Therefore, the only thing that would need to be done is pass the correct integers in the code. If you pass the integer 100 contestants for last year and 300 for current year. Then these inputs will provide the following output as requested in the question.

The competition is more than twice as big this year!

This is because the current year would be greater than double last years (100 * 2 = 200)

4 0
3 years ago
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