500 miles long
HERE IS THE CODE FOR C++ 2010 #include <stdio.h> int main () { int i,j; char prnt='*'; for(j=0;j<10;j++){ for(i=0;i<=j;i++){ printf("%2c",prnt); } printf("\n"); } return 0; }  
Nested loop, you mean; one loop in the other loop, eg: for (i=0; i<10; ++i) { for (j=0; j<i; ++j) { printf ("i=%d, j=%d\n", i, j); } }
/* Bubble sort: code snippet only nos to be sorted are in the array named 'n' of size 'N' for(int i=0;i<N-1;i++) for(int j=i+1;j<N-1-i;j++) if(n[j]>n[j+1]) swap(n[j],n[j+1]); */ /* insertion sort int v,j; for(int i=1;i<N;i++) { v=n[j]; for(int j=i-1;j>0&&n[j]>v;j--) n[j+1]=n[j]; n[j+1]=v; } */
well you could go with something like this (this code gives you all triples within a < 500 and b < 500): --------------------------------------------- for (int i = 0; i < 500; i++) for (int j = 0; j < 500; j++) { double squareroot = Math.Sqrt(i*i + j*j); if (squareroot == Math.Round(squareroot)) { // i, j and squareroot is your triple. Add them to list or what ever. } }
It is a locked rotor kva code. See related link for an excellent description.
500 miles long
If 92 J of electrical energy are converted into 92 J of mechanical energy in an electric motor, then all of the electrical energy is being converted into mechanical energy. No thermal energy is being produced in this scenario. The efficiency of the motor in this case is 100%, meaning all the input energy is being converted into useful work without any energy loss in the form of heat.
not sure
The J code for bupivacaine is J3490.
At least 320 J
J. Richard Johnson has written: 'Electric circuits' -- subject(s): Electric circuits
J. A. Harrison has written: 'The essence of electric power systems' -- subject(s): Electric power systems
John J. O'Neill has written: 'Prodigal genius' -- subject(s): Electric engineers, Biography, Electric currents, Alternating, Electric engineering, History, Alternating Electric currents
It's a code for oral sex, since b is letter 2 and j is letter 10.
Electric potential is the electric potential energy per unit coulomb. So unit for electric potential is J/C and that of electric potential energy is simply J
To find the time it takes for a 500 W electric motor to do (1.50 \times 10^5) J of work, we can use the formula: [ \text{Power} = \frac{\text{Work}}{\text{Time}}. ] Rearranging gives us: [ \text{Time} = \frac{\text{Work}}{\text{Power}} = \frac{1.50 \times 10^5 \text{ J}}{500 \text{ W}} = 300 \text{ seconds}. ] Thus, it would take 300 seconds, or 5 minutes, for the motor to perform that amount of work.