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: x ≈ 36.3 cm
Explanation:
Conservation of momentum during the collision
0.0340(120) + 1.24(0) = (0.0340 + 1.24) v
v = 3.2025 m/s
The kinetic energy of the block/bullet mass will convert to spring potential
½kx² = ½mv²
x = √(mv²/k)
x = √(1.274(3.2025²) / 99.0)
x = 0.363293... ≈ 36.3 cm
Answer:
w = 1.976 rpm
Explanation:
For simulate the gravity we will use the centripetal aceleration , so:
where w is the angular aceleration and r the radius.
We know by the question that:
r = 60.5m
= 2.6m/s2
So, Replacing the data, and solving for w, we get:
W = 0.207 rad/s
Finally we change the angular velocity from rad/s to rpm as:
W = 0.207 rad/s = 0.207*60/(2)= 1.976 rpm