What can you infer about fossil fuels from their name
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In step 1, to increase the potential energy, the iron will move towards the electromagnet.
In step 2, to increase the potential energy, the iron will move towards the electromagnet.<h3>Potential energy of a system of magnetic dipole</h3>The potential energy of a system of dipole depends on the orientation of the dipole in the magnetic field.
where;
is the dipole moment
- B is the magnetic field
Increase in the distance (r) reduces the potential energy. Thus, we can conclude the following;
- In step 1, to increase the potential energy, the iron will move towards the electromagnet.
- In step 2, when the iron is rotated 180, it will still maintain the original position, to increase the potential energy, the iron will move towards the electromagnet.
Learn more about potential energy in magnetic field here: brainly.com/question/14383738
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
- Replacing by the givens in (5), we can solve for μsmín, as follows: