Kirchhoff's Current Law (KCL) states that for a parallel path circuit with current flow the sum of the current entering the circuit's junction is equal to the sum of the current moving away from it.
That is 10A enters the junction
and 4A is observed in one of the two branches, that means
Given that the current that enters the junction is 10A, According to Kirchoff's current law which states that the total amount of current entering a junction must be equal to the total amount of current leaving the junction, then the total current that also leaves the junction must be equal to 10A, and since one of the two branches by which current leaves the junction has a current of 4A, and the total current leaving the junction must be equal to 10A. So, current passing through the second wire can be given as;
I(total) = I1 + I2
I1 = 4A
I(total) = 10A
I2 = I(total) - I1 = 10A - 4A = 6A
Therefore, the amount of current leaving through the other branch is 6A.
The answer is 9.8 ms^-2, because there is only one force acting on the object so the acceleration will be numerically equal to the gravitational field strength.
