If you're reading this, then I assume you're stuck on something physics-y or maths-y. If so, then I can help! I'm a Physics student at Exeter University and I can be your guru. Physics reduces the complex world to something so simple, and so beautiful. To do this, it uses Maths as a tool. They're amazing subjects, or at least I think so. Hopefully you will agree after we've finished.
I'll be patient with you, everyone struggles with new ideas sometimes! I've also been helping people who aren't as fast at picking up ideas as me for as long as I can remember, so you can trust that I know what I'm doing.
Obviously you decide what you want to cover in our sessions. It's important to note that in science, you must really understand a concept before you can answer questions on it. If you can't explain it to your 10 year old brother, then you don't understand it quite yet. You will be amazed how much easier questions seem when you truly understand the topic!
Different people have different styles of learning, be it pure equations, or diagrams etc. Whatever it is that works for you, we'll find it.
If you still fancy it, and I hope you do, then I look forward to meeting you!
Just a quick note to say that I'm on a year abroad in New Zealand at the moment, so please bear in mind that I'm asleep for a large part of your day! I'm still happy to help whilst I'm awake though.
|Maths||A Level||£20 /hr|
|Physics||A Level||£20 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Nabil (Parent) May 22 2016
Haider (Student) March 13 2016
Haider (Student) March 7 2016
Nabil (Parent) June 8 2016
The first law is that an object will remain at a constant velocity or stationary (which is just a constant velocity of zero) unless acted upon by an external force.
Once acted upon by this force, then the second law comes into play. This says that F = ma, or Force equals mass times the acceleration of the object.
The third law is easily stated as every action has an equal and opposite reaction. This means the sizes of the forces on the objects will be the same, but the forces will be in opposite directions.see more
Rutherford had a sheet of gold foil just a few atoms thick, and surrounded it with detectors. He then fired alpha particles at the gold sheet, knowing that these were positively charged. Some of these were deflected from their paths, a very small number were reflected, but most passed straight through the foil. This told Rutherford that most of the atom must be empty space. He explained the deflections and reflections with a concentration of charge in the middle of the space of each atom. As so few were reflected and deflected compared to those that went straight through, the charge must be tiny compared to the atom. As it was repelling the positive alpha particles, it must also be positive. This was the nucleus.see more
This is one of the trickier methods of integration, and it requires some practise. The basic idea is to split a function which would be difficult to integrate into two parts. Differentiating one part and integrating the other will then lead to a function which is much easier to integrate.
The formula is that the integral of u dv = uv - the integral of v du. It is best demonstrated with an example:
Let's integrate f(x) = xcos(x)
We can see that x will disppear if we differentiate it, so let's set x = u and cos(x) = dv.
Differentiating u and integrating dv then gives du = 1 and v = sin(x)
Now we substitute these into the formula: xsin(x) - integral of sin(x)
Sin(x) is easy to integrate, it is just -cos(x). Now we have our answer! The integral of xcos(x) = xsin(x) + cos(x) + c, where c is our unknown (and always necessary!) constant of integration.see more