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.
they act as sunshades from the sun intensity, and filters
Answer:
The cable run exceeds the specifications for Ethernet over twisted pair
Explanation:
The ethernet network's router also serves as a bridge to the Internet. The router connects to the modem, which carries the Internet signal, sending and receiving data packet requests and routing them to the proper computers on the network.
Ethernet is a way of connecting computers together in a local area network or LAN. It has been the most widely used method of linking computers together in LAN s since the 1990 s.
The basic idea of its design is that multiple computers have access to it and can send data at any time.
I got -3.6 m/s but I had to do conservation of momentum for this question. Which involves Newtons third law but with simply that law I do not know how to complete this question. If you would like me to post my work I will though! Sorry