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.
1 m = 1 000 000 ym
converted other way we can say that:
1 ym =

m
Now, since we have ym^2 which is ym*ym which means:
1 ym^2 =

m
we have 1,5 ym^2 which means that answer is:
Instantaneous velocity is the velocity at a specific instant in time. I bet you are taking Honors Physics.
Answer:
Explanation:
When the central shaft rotates , the seat along with passenger also rotates . Their rotation requires a centripetal force of mw²R where m is mass of the passenger and w is the angular velocity and R is radius of the circle in which the passenger rotates.
This force is provided by a component of T , the tension in the rope from which the passenger hangs . If θ be the angle the rope makes with horizontal ,
T cos θ will provide the centripetal force . So
Tcosθ = mw²R
Tsinθ component will balance the weight .
Tsinθ = mg
Dividing the two equation
Tanθ = 
Hence for a given w , θ depends upon g or weight .
The x-acis of a trajectory represents its C