Answer:
<h2>Three laws of motion haha.....</h2>
Speed= (4.92-2.45) 2.47 m/s
Distance= ?
You didn't mention about the distance?
The chemical change is are to water breaking down into hydrogen and oxygen.
Answer is D.
Answer:
a) attractiva, b) dF =
, c) F =
, d) F = -1.09 N
Explanation:
a) q1 is negative and the charge of the bar is positive therefore the force is attractive
b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x
dF =
where k is a constant, Q₁ the charge at the origin, x the distance
c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L
∫ dF =
as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density
λ = dQ₂ / dx
DQ₂ = λ dx
we substitute
F = 
F = k Q1 λ (
)
we evaluate the integral
F = k Q₁ λ
F = k Q₁ λ 
we change the linear density by its value
λ = Q2 / L
F =
d) we calculate the magnitude of F
F =9 10⁹ (-4.2 10⁻⁶)
F = -1.09 N
the sign indicates that the force is attractive
The correct answer is "an attractive force" between the wires.
Let's see why. Assume we have wire A on the left and wire B on the right, and that the current in both wires go upward. First, let's find the direction of the magnetic field produced by wire A at wire B: by using the right-hand rule, we see that since the current (the thumb) goes upward, the magnetic field (given by the other fingers) at wire B is directed inside the paper.
Then we can apply again the right-hand rule to see what is the force on wire B. The index gives the direction of the current (upward), the middle finger the direction of the magnetic field (inside the paper), and the thumb gives the direction of the force: to the left, so toward wire A. This means the force is attractive. (you can re-do the procedure on wire A, and you will find the force on wire A is directed toward wire B)