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:
58.8 N
Explanation:
The normal force is calculated as equal to the perpendicular component of the gravitational force.
Thus; N = mg
We are given m = 6 kg
Thus;
N = 6 × 9.8
N = 58.8 N
Thus, magnitude of normal force on the rock = 58.8 N
The magnitude of the magnetic dipole moment of the bar magnet is 1.2 Am²
<h3>
Magnetic dipole moment of the bar magnet</h3>
The magnitude of the magnetic dipole moment of the bar magnet at distance from its axis is calculated as follows;
where;
- B is magnetic field
- m is dipole moment
- μ is permeability of free space
m = (4π x 0.1³ x 2.4 x 10⁻⁴)/(2 x 4π x 10⁻⁷)
m = 1.2 Am²
The complete question is below:
What is the magnitude of the magnetic dipole moment of the bar magnet from 0.1 m of its axis and magnetic field strength of 2.4 x 10⁻⁴ T.
Learn more about dipole moment here: brainly.com/question/27590192
#SPJ11
We Know, P = m*v
Here, m = 30 Kg
v = 5 m/s
Substitute it into the expression,
P = 30*5 Kgm/s
P = 150 Kgm/s
So, your final answer is 150 Kg.m/s
Hope this helps!
