Sorry to keep Posting on peoples comments advocating my videos,
But I do some past paper tutorials for the multiple choice questions explaining why each option is chosen etc... and going through each option which does mean of course the videos are fairly long, but do have a look
bills7187 on youtube
Cheers,
Will
Will take a look after i get biol4 done with today
Rearrange charge over mass to find an equation for charge, equate the electrostatic force (F=EQ) to the weight of the ion (F=mg), the value for mass will cancel and you can rearrange for E.
I thought the whole thing about these questions was that you have to switch whatever direction of current you're given to the opposite to get conventional current?
Have they always used simply "current" in questions to mean conventional current by default?
I thought the whole thing about these questions was that you have to switch whatever direction of current you're given to the opposite to get conventional current?
Have they always used simply "current" in questions to mean conventional current by default?
Yes they have, I think the textbook shows this in the first few pages of this topic. If electrons are fired perpendicular then you take the conventional current direction.
Yes they have, I think the textbook shows this in the first few pages of this topic. If electrons are fired perpendicular then you take the conventional current direction.
That's my point, you surely switch the current direction to get conventional current, which means that the force is acting upwards. They've given the direction of current in the question, not conventional current. They surely can't mean conventional current by referring simply to current, otherwise that would mess up any question that refers to current in general
That's my point, you surely switch the current direction to get conventional current, which means that the force is acting upwards. They've given the direction of current in the question, not conventional current. They surely can't mean conventional current by referring simply to current, otherwise that would mess up any question that refers to current in general
I know what you mean, but I've followed what I've said above and that has always worked.
Surely at the minimum at 0.2 seconds (assuming they mean that this represents the equilibrium point), some of the kinetic energy will have already been transferred to gravitational potential just by virtue of the fact that the mass has moved vertically up from its starting point, i.e. why does the graph display no potential energy when it must clearly have some potential energy?
Surely at the minimum at 0.2 seconds (assuming they mean that this represents the equilibrium point), some of the kinetic energy will have already been transferred to gravitational potential just by virtue of the fact that the mass has moved vertically up from its starting point, i.e. why does the graph display no potential energy when it must clearly have some potential energy?
Doesn't it have a maximum gravitational potential when it's at the top and maximum elastic potential at the bottom? I may be entirely wrong here but I think in the middle those two cancel out?
Surely at the minimum at 0.2 seconds (assuming they mean that this represents the equilibrium point), some of the kinetic energy will have already been transferred to gravitational potential just by virtue of the fact that the mass has moved vertically up from its starting point, i.e. why does the graph display no potential energy when it must clearly have some potential energy?
I think in these questions it classes grav potential energy at zero when its at equlibrium, otherwise how high you are when you do the experiment would affect it.... i think this i just another case of having to simplify data for a level spec....unless i completely misunderstood you
It would make these questions much more difficult, as the only time kinetic energy is actually gained would be when it oscillates from highest point to equilibrium.
Technically i dont think any kinetic energy would actually be gained moving from lowest displacement to equilibrium, the elastic potential would be directly converted to gpe
I think in these questions it classes grav potential energy at zero when its at equlibrium, otherwise how high you are when you do the experiment would affect it....
yeah, I think that fixes my problem with it, thanks. Though the question does still seem pretty nebulous to me
And these questions are tough....seen the same style come up before..
I think its 1.95 over diff in times equals 39
Because 39 oscilations times 1.95 and 39 times 1.9 the diff between total times is 1.95....soooo they will be at the same point but the 1.9 oscilator will be 1 whole oscilation ahead...but still in phase....its harder to word than i though :-/
yeah, I think that fixes my problem with it, thanks. Though the question does still seem pretty nebulous to me
Yeah some questions are a bit odd....your lucky you dont do biology....its not that some of the answers are vague its that they are out right wrong ......its infuriating!