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3 years ago

How much time does a gold chest take to open

Computers and Technology

1 answer:

MaRussiya [10]3 years ago

7 0

Gold chest takes 8 hours to open or unlock.

Explanation:

Clash royale is a game which has been developed and published with the help of a super cell. It is a video game. Many players can play this game at the same time. In the sense, clash royale is a multi player game.

In this game, the players who are playing the game are awarded with the gold chest. They are also awarded with the gems. In order to unlock or open a gold chest eight hours are needed. Apart from this there is an other way also with which the gold chest can be unlocked. It can be unlocked with the help of forty eight gems.

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The collection of all component frequencies iscalled _____________

yKpoI14uk [10]

Answer:

The answer is Frequency Spectrum.

Explanation:

The frequency spectrum is range of all component frequencies.It contains all the waves which are as following:-

Gamma Rays

X-Rays

Ultraviolet

Visible light.

Infrared

Micro wave

Radio wave

These all waves have their range of frequencies.The waves that are visible to us is only the visible light.

4 0

3 years ago

Can someone please help me? my audio visual teacher wants me to watch two movies. i've never seen them and i have no way to watc

kati45 [8]

Answer:

You can try afdah or 123 movies, maybe try Netflix if you have it

Explanation:

The sites do have ads but that's where I usually watch my movies

4 0

4 years ago

Create a program which will input data into a pipe one character at a time. Count the number of characters as they are written i

LenKa [72]

Answer:

#include<stdio.h>

#include<stdlib.h>

#include<fcntl.h>

#include<unistd.h>

#define BUFSIZE 16

int main(){

//pipe descriptors

char msg[BUFSIZE];

char buf[BUFSIZE];

int pipefd[2];

if(pipe(pipefd) == -1){

//pipe creation error

perror("pipe");

exit(EXIT_FAILURE);

}

//pipe creation successfull

//write four messages

sprintf(msg , "apple");

write(pipefd[1], msg, BUFSIZE);

printf("Written %s\n", msg);

sprintf(msg , "boy");

write(pipefd[1], msg, BUFSIZE);

printf("Written %s\n", msg);

sprintf(msg , "cat");

write(pipefd[1], msg, BUFSIZE);

printf("Written %s\n", msg);

sprintf(msg , "dog");

write(pipefd[1], msg, BUFSIZE);

printf("Written %s\n", msg);

//read

read(pipefd[0], buf, BUFSIZE);

printf("Read: %s\n", buf);

read(pipefd[0], buf, BUFSIZE);

printf("Read: %s\n", buf);

read(pipefd[0], buf, BUFSIZE);

printf("Read: %s\n", buf);

read(pipefd[0], buf, BUFSIZE);

printf("Read: %s\n", buf);

close(pipefd[0]);

close(pipefd[1]);

return 0;

}

(b)

#include<stdio.h>

#include<stdlib.h>

#include<fcntl.h>

#include<unistd.h>

#include<wait.h>

#define BUFSIZE 16

int main(){

//pipe descriptors

char msg[BUFSIZE];

char buf[BUFSIZE];

int pipefd[2];

if(pipe(pipefd) == -1){

//pipe creation error

perror("pipe");

exit(EXIT_FAILURE);

}

//pipe creation successfull

//write four messages

if(fork() == 0){

//this is child

//close unused write end

close(pipefd[1]);

read(pipefd[0], buf, BUFSIZE);

printf("This is child process reading first message. Content is %s\n", buf);

read(pipefd[0], buf, BUFSIZE);

printf("This is child process reading second message. Content is %s\n", buf);

read(pipefd[0], buf, BUFSIZE);

printf("This is child process reading third message. Content is %s\n", buf);

read(pipefd[0], buf, BUFSIZE);

printf("This is child process reading fourth message. Content is %s\n", buf);

close(pipefd[0]);

//exit from child

exit(EXIT_SUCCESS);

}

//this is parent process

//close unused read end

close(pipefd[0]);

sprintf(msg , "apple");

printf("This is parent process. Writing first message into pipe\n");

write(pipefd[1], msg, BUFSIZE);

sprintf(msg , "boy");

printf("This is parent process. Writing second message into pipe\n");

write(pipefd[1], msg, BUFSIZE);

sprintf(msg , "cat");

printf("This is parent process. Writing third message into pipe\n");

write(pipefd[1], msg, BUFSIZE);

sprintf(msg , "dog");

printf("This is parent process. Writing fourth message into pipe\n");

write(pipefd[1], msg, BUFSIZE);

close(pipefd[1]);

//wait for child

wait(NULL);

return 0;

}

(c)

#include<stdio.h>

#include<stdlib.h>

#include<fcntl.h>

#include<unistd.h>

#include<sys/wait.h>

#include<sys/types.h>

#include<signal.h>

typedef struct sigaction Sigaction;

unsigned long long size = 0;

void alarmhandler(int sig){

//alarm fired writing blocked

printf("Write blocked after %llu characters\n", size);

exit(EXIT_SUCCESS);

}

int main(){

//pipe descriptors

int pipefd[2];

if(pipe(pipefd) == -1){

//pipe creation error

perror("pipe");

exit(EXIT_FAILURE);

}

//install handler

sigset_t mask , prev;

sigemptyset(&mask);

sigaddset(&mask , SIGALRM);

sigprocmask(SIG_BLOCK , &mask , &prev);

Sigaction new_action;

sigemptyset(&new_action.sa_mask);

new_action.sa_flags = SA_RESTART;

new_action.sa_handler = alarmhandler;

sigaction(SIGALRM , &new_action , NULL);

sigprocmask(SIG_SETMASK , &prev, NULL);

while(1){

//print size on multiple of 4

if(size != 0 && size % 1024 == 0){

printf("%llu characters in pipe\n", size);

}

//reset previous alarm

alarm(0);

//set new alarm for 5 seconds

alarm(5);

//write to pipe one character

write(pipefd[1], "A", sizeof(char));

size++;

}

return 0;

}

Explanation:

Output for a, b, c are pasted accordingly

4 0

3 years ago

What is a system of computers that are joined together and connect to peripheral devices and can’t be accessed from home

harina [27]

What is a system of computers that are joined together and connect to peripheral devices and can’t be accessed from home = LAN for Local Area Network, unless it's connected to the internet and you allow VPN/RAS system access you will not be able to connect to a LAN remotely.

7 0

3 years ago

Assume that the population of Mexico is 128 million and that the population increases 1.01 percent annually. Assume that the pop

horrorfan [7]

Answer:

//Define class

public class Main {

//define main method

public static void main(String[] args)

{

//declare and initialize the double type variable to 128

double mexico = 128;

//declare and initialize the double type variable to 323

double us = 323;

//declare and initialize the integer type variable to 0

int yr = 0;

//set the while loop to check which is greater

while (mexico < us)

{

//increment in the variable by 1

yr++;

//initialize in the variables acc. to the percentage

mexico *= 1.0101;

us *= 0.9985;

}

//print the following results

System.out.println("Population of the Mexico will be exceed the population U.S. in " + yr + " years");

System.out.println("Population of the Mexico will be " + mexico + " million");

System.out.println("and population of the U.S. will be " + us + " million");

}

Output:

Population of the Mexico will be exceed the population U.S. in81 years

Population of the Mexico will be 288.88435953355025 million

and population of the U.S. will be 286.0198193927948 million

Explanation:

Following are the description of the program.

Firstly, we define the class 'Main' and inside it, we define the main method.
Then, declare two double data type variables which are 'mexico' and 'us' and initialize in it to 128 and 323.
Declare integer data type variable 'yr' and initialize in it to 0.
Set the while loop and pass the condition to check that the variable 'mexico' is less than the variable 'us' then, increment in the variable 'yr' by 1 and multiply the variables 'us' and 'mexico' by the following percentage.
Finally, print the following results with the message.

6 0

3 years ago