The answer is A. The South Pole of another magnet. I hope this helps
#include<studio.h>
int main( )
{
int n;
int a,b,c,d,x,y;
int avarage;
printf("enter value of n:\n");
scanf("%d",&n);
printf("enter value of a:\n,b:\n,c:\n,d:\n,x:\n,y:\n);
scan f("%d\%d\n%d\n%d\n%d\n%d\n",&a,b,c,d,x,y);
sum=(a+b+c+d+x+y);
avarage=(sum/n);
print f("%d",avarage);
if
{
n=positive interger
}
else
{
printf ("n must be positive");
}
return 0;
}
Answer:
10000 Bq / 625 Pq = 16
Radioactivity has decreased by a factor of 16
2^4 = 16
So the sample has gone thru 4 half-lives
24 da / 4 = 6 da
6 da is the half-life
The object in the lower orbit will move faster in its orbit, and will have a shorter orbital period. None of that depends on their masses either.
