Recent content by JimChampion

I don't really know what level of explanation you're after! When I did A-level physics (in the UK, 1994) knowing about potential and how it relates to force was part of the course. I now teach A-level physics, and this is no longer part of the course: anything with calculus is avoided and...

F=ma is a common way of expressing Newton's second law of motion for an object with constant mass m. The mv^2/ r = | dV/dr | is equivalent to writing ma=F because v^2/r is the centripetal acceleration |dV/dr| is the magnitude of the centripetal force (given by the gradient of the...

Writing mv^2/ r = | dV/dr | is like writing ma=F (which you might be more familiar with?). Does this help?

Are you working at a level where you can relate forces to gradients of potential?

What do the various symbols in this equality represent?

... and put a person on the horse's back, facing backwards. The person takes off their shoe and throws it. Newton's 3 law: person pushes shoe & shoe pushes person. The result - shoe moves off in one direction, the horse-person-rope-rock moves off more slowly in the other direction (assuming the...

No. If you are within a space-station orbiting the Earth then both you and the space-station experience a gravitational attraction towards the Earth. The result is that you and the space-station accelerate towards the Earth at the same rate. The combination of inertia (the tendency to keep...

Sorry, in the above post I should have said amplitude not intensity. In the above did you mean I = I_0 \sin 30^o or I = I_0 \sin^2 30^o ?

Malus' Law is spot on. The intensity is greatest when angle=0deg, zero when angle=90 deg. The cos(angle) function fits the bill, especially as the situation is a rotation.

Zero resistance in this case means that a finite current can flow with zero potential difference across it (from this definition of resistance: V=IR). Also a current can persist in a superconducting ring as the potential difference is (necessarily) zero if you go all the way around the ring.

Define the terms in this equation: what exactly are B, N1, N2, and how do they relate to the problem you've quoted?

Method looks fine to me, if r is the distance from midpoint to star.

You should be able to find something about charged particle trajectories in magnetic fields. For example circling or spiralling trajectories. A U-turn is part of a circle.

"Completely dark" means no electrons are hitting the screen there. Find out how a magnetic field could make an electron do a U-turn.

I don't know what your knowledge of calculus is like, but (a) can be solved (to a first level of approximation) using calculus... (1) If F is the gravitational attraction between you and Moon, and R is separation between you and Moon... \frac{dF}{F} = 2\:\frac{dR}{R} If you can work out why...