Pacific Plate<span> and </span>Juan de Fuca<span> 6.</span>
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.
<span> The indivisibility of an atom was proved wrong an atom can be further subdivided into protons, neutrons and electrons. However an atom is the smallest particle that takes </span>part<span> in chemical reactions</span>
<span>x = 129.9 m
y = 30.9 m
First, let's calculate the horizontal and vertical velocities involved
h = 50.0cos(30) = 43.30127 m/s
v = 50.0sin(30) = 25 m/s
The horizontal distance is simply the horizontal velocity multiplied by the time, so
43.30127 m/s * 3 s = 129.9 m
So the horizontal distance traveled is 129.9 m, so x = 129.9 m
The vertical distance needs to take into account gravity which provides an acceleration of -9.8 m/s^2, so we get
d = 25 m/s * 3s - 0.5*9.8 m/s^2 * (3 s)^2
d = 75 m - 4.9 m/s^2 * 9 s^2
d = 75 m - 44.1 m
d = 30.9 m
So the vertical distance traveled is 30.9 m, so y = 30.9 m</span>
A curved line on a position/time graph shows that the speed is changing.
So right there, we know there is acceleration.