Answer:
D. 4x as much
Explanation:
The answer is pretty simple. The question is asking about 10m/s and 20m/s right? Take a look at the energy at each point on the graph. At 10m/s the kinetic energy is 100 on the graph. At 20m/s the kinetic energy is 400. You'll notice that 100 x 4 is 400 - That's the answer.
You can also prove this mathematically. The formula for kinetic energy is
The question says: constant mass so we can ignore that for now.
(1/2) * 10^2 is (1/2) * 100 = 50
for the 20m/s ball it would be
(1/2) * 20^2 = 400 * 1/2 = 200
Once again, you see that it is 4x
Answer:
Check the explanation
Explanation:
#define _MULTI_THREADED
#include <pthread.h>
#include <stdio.h>
#include <errno.h>
#define THREADS 2
int i=1,j,k,l;
int argcG;
char *argvG[1000];
void *threadfunc(void *parm)
{
int *num;
num=(int*)parm;
while(1)
{
if(i>=argcG)
break;
if(*num ==1)
if(argvG[i][0]=='a' ||argvG[i][0]=='2'||argvG[i][0]=='i' ||argvG[i][0]=='o' ||argvG[i][0]=='u')
{
printf("%s\n",argvG[i]);
i++;
continue;
}
if(*num ==2)
if(!(argvG[i][0]=='a' ||argvG[i][0]=='2'||argvG[i][0]=='i' ||argvG[i][0]=='o' ||argvG[i][0]=='u'))
{
printf("%s\n",argvG[i]);
i++;
continue;
}
sched_yield();
}
return NULL;
}
int main(int argc, char *argv[])
{
pthread_t threadid[THREADS];
int rc=0;
int loop=0;
int arr[2]={1,2};
argcG=argc;
for(rc=0;rc<argc;rc++)
argvG[rc]=argv[rc];
printf("Creating %d threads\n", THREADS);
for (loop=0; loop<THREADS; ++loop) {
rc =pthread_create(&threadid[loop], NULL, threadfunc,&arr[loop]);
}
for (loop=0; loop<THREADS; ++loop) {
rc = pthread_join(threadid[loop], NULL);
}
printf("Main completed\n");
return 0;
}
The below attached image is a sample output
Linear search
You implement this algorithm by iterating over each item, and checking if the item matches what you are searching for.
It is linear because it takes a linear amount of time to search for an item.
Answer:
Future Business Leaders of America (FBLA) for high school students; FBLA-Middle Level for junior high, middle, and intermediate school students; Phi Beta Lambda (PBL) for post secondary students; and Professional Division for those not enrolled in school or post secondary school
Explanation:
