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.
Learn more about dipole moment here: brainly.com/question/27590192
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Because radio waves can travel in space but sound waves cannot.
Answer:
52 mm/s (approximately)
Explanation:
Given:
Initial speed of the projectile is, 
Angle of projection is, 
Time taken to land on the hill is, 
In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.
So, the velocity in the horizontal direction always remains the same.
The horizontal component of initial velocity is given as:
Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.
Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.
