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Does pV = \dfrac 1 3 N m \left(c_{rms}\right)^2 apply in containers that aren't cuboids? The derivation I have seen uses a cuboid container so I'm not sure if or how this can be generalised.

Don't pay attention to rankings; they're meaningless. Just have a look at the 2015 UK Mathematics league tables according to The Guardian. http://www.theguardian.com/education/ng-interactive/2014/jun/03/university-guide-2015-league-table-for-mathematics I suppose firming Cambridge and...

My money is on Rudin.

Yes, I was planning on doing that, I thought it might be quite difficult to learn it straight from Rudin though. I'll see how it goes.

I am currently reading baby Rudin, but I only know single-variable calculus at the moment, so I think it would be a good idea to learn the multi-variable stuff non-rigorously before I do the analysis in Rudin (chapters 9-11). So I was thinking of either getting one of the many 'Mathematical...

Yes, I would have picked that in the exam, but the question is still wrong. It can't be exactly horizontal if there is no vertical force to balance gravity.

Indeed. If it is a mistake, then it's shocking that it made its way to an A level exam. Source: http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-QP-JUN12.PDF (question 6)

What's interesting is the "correct" answer is \dfrac{mv^2}{rg}, according to the exam board AQA.

Yes, only in vertical equilibrium. Maybe M could be moving in a horizontal circle, though we aren't told that it is.

Homework Statement The diagram shows a smooth thin tube through which passes a string with masses m and M attached to its ends. The tube is moved so that the mass m travels in a horizontal circle of radius r at constant speed v http://quickpic.info/z/yb.jpg [Broken] Find an expression...

Cambridge University recommend Calculus by Spivak/Apostol, then Rudin, but they seem to prefer Burkill for some reason.

I was told that the field is from north to south, so is that not really true then? It seems the field is both perpendicular and parallel from that diagram. Also how can I use the left hand rule to determine the direction of current in this case if the field is going in infinitely many different...

Yes (I presume so).

The south is on the other end of the magnet, above the north.

This is what is confusing me then. Consider a set-up like this: The magnet field points up (north to south) and the movement is also vertical, so the field and movement are parallel.

Can I use Fleming's right hand rule to determine the direction of current when a magnet is moved into a coil(s) of wire? I am finding it difficult because the magnetic field and movement are parallel instead of at right angles. If not, does this mean I am only allowed to use it when a wire...

In the case of the hollow sphere, the ribbon would also be an infinitesimal slice, so I don't really understand why ##\delta x## wouldn't work in that case, when the approximation(which it is, at least before the limit is taken) works in the case of the solid sphere. I'm not understanding the...

How is it not an approximation? A slice of a sphere isn't cylindrical, it has some curvature.

I know, but I'm wondering why it's OK to approximate the width as ##\delta x## to find the volume, but not to find the surface area.

Ah, I understand. This method won't work then, the previous one involved ##r\delta\theta##. For the derivation of the moment of inertia of a sphere that isn't hollow, the disc is assumed to have volume ##\pi y^2 \delta x##. Why is this ok?

Sorry, I forgot to add it. It's just showing what my x,y, and r mean. Wouldn't the width of the hoop just be ##\delta x## ?

I need to find the moment of inertia of a sphere of radius ##r## and mass ##m## about an axis through it's centre. I've already done it and got the correct answer of ##\frac{2}{3}mr^2## however I have tried doing it using a different method to see if I get the same answer, but I don't, and I...

Ah, didn't know we had LaTeX.

I read that for a rotating body the kinetic energy ##E_k = \sum \frac{1}{2}mv^2 = \frac{1}{2}{\omega}^2∑mr^2 = \frac{1}{2}I{\omega}^2## where ##I## is the moment of inertia. If we did the same thing for momentum then ##P = ∑mv = \omega\sum mr## So why is angular momentum ##I\omega=\omega\sum...

My question is about the centripetal acceleration formula |a| = ω^2*r. If we keep the angular speed constant then why does increasing the radius increase the centripetal acceleration? I don't find this intuitive because the velocity vector is being turned by the same amount each second, if ω is...

Thanks a lot!

I have seen the derivation of the centripetal acceleration formula a=v^2/r by saying r= rcosθi+isinθj=rcosωti+rsinωtj and differentiating twice. Since ω is constant we get a=-ω^{2}r. I've started looking at non-uniform circular motion where there is also the tangential acceleration vector...

Thanks for that, I understand now. What is the correct word to describe λ then? My textbooks refer to λ as the modulus of elasticity of the spring, rather than the modulus of elasticity of the material.

I'm learning about Hooke's law and modulus of elasticity (also known as youngs modulus) but it seems I am being taught it differently in maths and physics. In maths I am taught that T=λx/l and λ is the modulus of elasticity, measured in Newtons. In physics I am taught that T=λAx/l where λ is...

Photons need specific energy levels, equal to the difference between two energy levels to excite an electron in an atom. Is this the same case with electrons that collide with atoms?

Am I right in saying that, in an exothermic reaction for example, the reactants have a higher enthalpy, and the products have a lower enthalpy, as seen from the energy profile diagram. But the reactants have a lower bond enthalpy, and the products have a higher bond enthalpy, because a lower...

Thanks, I understand now. I never considered that carbon would have to have 5 bonds!

I've learned that tertiary alcohols can't be oxidised because the carbon bearing the OH contains no hydrogen atoms. But why can't the oxygen just take the hydrogen from the OH and another hydrogen from another carbon atom on the molecule? I also read that it would involve breaking a C-C bond...

Was about time, looks very nice.

I was thinking about wave particle duality, and it's always made out to seem so mysterious and paradoxical. But then it occurred to me that it might just be the same thing as mass energy equivalence, as waves are just energy and particles are 'mass'. Is this right or am I missing something?

I'm doing A level Chemistry, and my teacher isn't good at explaining things. I know activation energy is the minimum energy required for a reaction to take place - so is that just the energy required to break the original bonds? And is the activation energy for a particular bond to form the same...

So when doing the double slit experiment, one will see two diffraction patterns on top of each other then?

I'm confused about the single slit diffraction pattern. Why are light and dark patterns? Where is the constructive and destructive interference occurring if there is just one wave?

Thanks, I understand now.

Why is the energy stored in a spring 1/2 * distance * force? Isn't work just force * distance?

What defines what element results when energy is converted into matter? i.e. the protons/electrons/neutrons

Thanks so much! I can't believe I never thought to do that.

Differentiate and simplify: y=(x+1)(2x-3)^{4} I got: 8(x+1)(2x-3)^{3} + (2x-3)^{4}. But the answers in the answer booklet say: 5(2x+1)(2x-3)^{3} I put both answers in Wolfram Alpha and found they were both equal. So this is just a matter of simplifying/rearranging. Could someone please...