Complete question:
Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current, if the distance between the two wires is 40cm.
Answer:
The bottom current is 12.8 A to the right.
Explanation:
Given;
length of the wires, L = 3.0 m
current in the top wire, I₁ = 12.5 A
repulsive force between the two wires, F = 2.4 x 10⁻⁴ N
distance between the two wires, r = 40 cm = 0.4 m
The repulsive force between the two wires is given by;

Where;
I₂ is the bottom current
The direction of the bottom current must be in the same direction as the top current since the force between the two wires is repulsive.

Therefore, the bottom current is 12.8 A to the right.
100N describes the weight of the sandbag, while 100kg is the mass of the sandbag.
To calculate acceleration, divide your weight by the mass, thus the accleration is:
Answer:
Explanation:
Given
mass of spring 
extension in spring 
downward velocity 
Position in undamped free vibration is given by

where 
also 



it is given


substituting values we get






