Yes if you want to do that it would make a good experiment. LC circuits display a phenomenon called resonance that would be interesting to study. It depends on how much work you want to do.
Inductors have an inductance usually written as L, and if you put an inductor and capacitor together they have a resonant frequency given by: f = 1/(2𝜋√(LC))
You can see this by varying the frequency and measuring the voltages as you did for the RC circuit. The maths is a bit more complicated but not terrible.
You'd need to ask the technicians if they have inductors, and if so what values of L they have.
@Wolgwang you need to be clearer, which particular two versions are you talking about? Are you wanting to ask if you should be using $\frac{D}{D^2+y^2}$ or to use $\frac{1}{1+\left(\frac yD\right)^2}$?
If you are asking about $d\left(1-\frac{y^2}{2D^2}\right)$, then obviously you will get something different because this is an approximation, as opposed to the exactly ones above
You are correct. If you want to compute with greater precision, you could set up that, but then you will lose the ability to easily predict where they are, because the calculations become too difficult.
And all that work, will mostly only get you into numerical difficulties.
Because the physical reason why we considered the parallel rays approximation, is particularly good, and so it is really difficult to observe any deviation from parallel rays approximation.
In both cases they are making approximations. In the image, they should have raised the temp to BP, but it is a tolerable approximation not to, because the latent heat of vaporisation is a lot bigger than the heat needed to raise the temp to BP
In the link, unless the evaporated part is allowed to escape, the latent heat should be added back. So, it is not particularly clear how this question is meant to be setup/interpreted.
But this is the energy in the frame in which the total momentum is zero. That is we have the alpha particle coming in with some speed v₁ and the nitrogen atom coming in with speed v₂ and the total momentum m₁v₁ - m₂v₂ = 0.
So what we need to do is start in the centre of mass frame and do the calculation in that frame. Then we need to transform back into the lab frame so get the KE as measured in the lab frame.
p = √0.0128u and p = mv, so the velocities of the two nuclei are α = +(√0.0128u)/4 N = -(√0.0128u)/16 where the different signs are because the nuclei are moving in opposite directions. Yes?
Ah, OK. The problem is that below the BP we only get evaporation until the partial pressure of the evaporated water is equal to the vapour pressure of water at whatever temperature we are using.
So for example at 0°C the vapour pressure is very low (I forget the exact figure) so unless you had a huge volume for the water to evaporate into it would simply not evaporate.
If you allow evaporation below 100°C the question will get very complicated as you need to know the volumes involved, and the vapour pressure of water (which is a function of temperature) and indeed the latent heat of vaporisation is also a function of temperature.
I see the point you are making, and you are quite correct, but I'm sure the question means you to boil the water not just let it evaporate at a lower temperature.
The questions aren't that different from Mains. The skills you learn for Mains are what you need for Advanced. It's just that Advanced requires a bit of extra thought about how to set up the equations.
That question we just did looks to me like an Advanced question. Was it?
It was a good example of what makes Advanced harder. Nothing in the calculation was hard, but you needed to understand that you had to work in the COM frame.
That's because the minimum energy is when the reaction products have zero KE so all the reaction has gone into the mass deficit. Yes?
@JohnRennie That's the thing really,,, When I solve the questions as a test, I'm not able to do them... When I try them later, they just seem a bit easier
Remember that the advanced exam always includes some really, really hard questions - questions even I can't do. They deliberately put them in to let the few really good students shine.
So if you find an advanced question looks impossible that's probably because it is!
You don't need to answer the impossible questions, you only need to answer enough questions.
Problem Solving Strategies
Problem Solving Strategies