-- heat the magnet red-hot in a flame
-- drop it on the floor several times, or beat it with a hammer
-- place it in a strong, rapidly alternating external magnetic field
The whole question is talking about the amplitude of a wave
that's transverse and wiggling vertically.
Equilibrium to the crest . . . that's the amplitude.
Crest to trough . . . that's double the amplitude.
Trough to trough . . . How did that get in here ? Yes, that's
the wavelength, but it has nothing to do
with vertical displacement.
Frequency . . . that's how many complete waves pass a mark
on the ground every second. Doesn't belong here.
Notice that this has to be a transverse wave. If it's a longitudinal wave,
like sound or a slinky, then it may not have any displacement at all
across the direction it's moving.
It also has to be a vertically 'polarized' wave. If it's wiggling across
the direction it's traveling BUT it's wiggling side-to-side, then it has
no vertical displacement. It still has an amplitude, but the amplitude
is all horizontal.
Answer:
Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all - option A
Explanation:
Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all; the reason being that: no magnetic field is being produced by a charge that is static. Only a moving charge can produce a magnetic effect. And the magnet can not have any torque due to its own magnetic lines of force.
Distance = (1/2) (acceleration) (time)²
1.4m = (0.835 m/s²) (time)²
(time)² = (1.4/0.835) s²
<em>time = 1.295 s</em>
1.velocity and acceleration
2.
3.inertia
4.
5.speed
