What accelerating potential difference would be required to give an alpha particle a de Broglie wavelength of 2.0nm?
mass of alpha is 6.67x10^-27kg
I attached the solution and am just unsure why they doubled the coulomb of charge. The way i tried to work it out is find the velocity using de broglie's wavelength formula which gave me 49.7m/s. Then use this to find the kinetic energy which was 8.24x10^-24J. Once i did this i used the formula W=qV and subbed in 8.24x10^-24=(1.6x10^-19)V which gave me the answer of 5.1x10^-5V required.
Think about the charge on an alpha particle
It's a bit of a trick question because you have to had remembered what the charge is from year 11, or do it the way you did it with year 12 methods
what is the importance of frequency in determining the non-ideal behaviour of transformers. are there any other parameters?
Hmm interesting question but usually not examinable. A quick run down of things that can make transformers less ideal:
Eddy currents
Non ideal joins (side note is that Wilson Transformers actually makes all their transforms by hand and its an amazing process to get maximum efficiency)
Heat in the coil (usually transformers are oil-convection cooled)
Whether or not the core is laminated (this goes back to the eddy currents issue)
And the last one I can think of is flux leakage, but tbh I don't really get this either
But basically in essence the most important one as you've pointed how is the frequency as this will have an effect on eddy currents (imagine it as a sort of like a residue current in the coil)
question: in a DC motor why does the armature continue to spin even when at one point on force (the forces cancel out) are acting on it?
I know it has something to do with momentum but I don't quite get it....
I assume you mean something like this
I'll start with B first, because it's the most obvious: Force on side is going up and the force on the other is going down, so the coil spins, seems simple enough
But then when it's in points A and C, the force on the top is going upwards and the force on the bottom is acting downwards so I get there can be some confusion as how the coil keeps on turning. In fact, it's because of the commutator. At that instant the coil briefly loses connection with the batter (the 'split' part of the split ring commutator) so there is no current flowing and thus no force acting on the coil. Now, the coil was originally spinning (see point B) so it will continue to spin as there is no braking or frictional force acting on the coil. The terminals of the coil then touch the other side of the split ring commutator (reversing the current) and now the forces are reversed as well meaning the coil keeps on spinning.
So in essence, the coil's spinning momentum just before the perpendicular position is what carries to through the vertical position and allows the direction of the current to be changed via the commutator