The magnitude of the magnetic dipole moment of the bar magnet is 1.2 Am²
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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.
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Answer:
Mass and velocity.
Explanation:
Kinetic energy <u>is the energy that an object has due to its movement</u>, mathematically it is represented as follows:
where 
is the mass of the object, and 
is its velocity at a given point in time.
So we can see that to find the kinetic energy just before the ball hits the gound, we need the quantities:
- mass of the ball
- velocity of the ball before it hits the ground
With the knowledge of these two quantities the kinetic energy of the ball before touching the gound can be determined.
Answer:
F = m a = m v / t where v is the change in velocity in time t
F = p / t since m v is equal to p
F = 2.2 (kg m / s) / 1.1 s = 2 kg-m / s^2 = 2 N
Or you can use the impulse equation
Answer
a) For the rock
b) 
for maximum range
c) The value of θ is the same on every planet as g divides out.
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