I believe the answer should be the last option. upon interaction, both objects should have the same charge after the electrons are transferred.
The magnetic field or force seems to be associated with the lineup of electrons withim the magnet
Answer:
v = 6t² + t + 2, s = 2t³ + ½ t² + 2t
59 m/s, 64.5 m
Explanation:
a = 12t + 1
v = ∫ a dt
v = 6t² + t + C
At t = 0, v = 2.
2 = 6(0)² + (0) + C
2 = C
Therefore, v = 6t² + t + 2.
s = ∫ v dt
s = 2t³ + ½ t² + 2t + C
At t = 0, s = 0.
0 = 2(0)³ + ½ (0)² + 2(0) + C
0 = C
Therefore, s = 2t³ + ½ t² + 2t.
At t = 3:
v = 6(3)² + (3) + 2 = 59
s = 2(3)³ + ½ (3)² + 2(3) = 64.5
Answer: a) 127 eV; b) there is no change of kinetic energy.
Explanation: In order to explain this problem we have to use the change of potentail energy ( conservative field) is equal to changes in kinetic energy. So for the proton ther move to lower potential then they gain kinetic energy from the electric field. This means the electric force do work in this trayectory and then the protons increased changes its speed.
If we replace the proton by a electron we have a very different situaction, the electrons are located in a lower potental then they can not move to higher potential if any external force does work on the system.
In resumem, the electrons do not move from a point with V=87 to other point with V=-40 V. The electric force point to high potential so the electrons can not move to lower potential region (V=-40V).
Answer
given,
mass of copper rod = 1 kg
horizontal rails = 1 m
Current (I) = 50 A
coefficient of static friction = 0.6
magnetic force acting on a current carrying wire is
F = B i L
Rod is not necessarily vertical
the normal reaction N = mg-F y
static friction f = μ_s (mg-F y )
horizontal acceleration is zero
B_w = B sinθ
B_d = B cosθ
iLB cosθ= μ_s (mg- iLB sinθ)
B = 0.1 T
