Recent content by Squires

Good morning SteamKing my friend, and thank you for the reply! Yeah I did some research, and I understood that part pretty well. But that just raises my question, instead of solve it, because as you say: The more turbulent a flow, the higher the turbulent intensity, I And we also...

Hello, I have a question regarding Reynolds numbers. As I understand, the higher a Reynolds number, the more turbulent a flow becomes, with Re>4000 considered turbulent. I have a model where I've calculated a Reynolds number of 10^7, and when going to set my model boundary conditions I'm...

Hey guys, back again for some help if that's okay! I've derived a formula for an electromagnetic wave entering a material, as \underline{E}e^{-\frac{x}{\delta}}e^{i(\frac{x}{\delta}- \omega t)} x=direction of propagation t=time \delta=skin depth As you can see this describes the wave...

Ahh yeah totally, T1 and T2 any day, but I had a question to derive it from Q1 and Q2, and made a schoolboy error of using wiki, to only find out Q1 and Q2 had opposite direction, physicsrage! Haaaahaha, I'm not loosing sleep over it just yet, don't worry ;) And as for London, believe me it...

Yeah I went back and looked over the work thing aswell, just to be sure. It appears that work done is negative when the system does work on the surroundings, and the dU=dQ+dW is still correct, it's just when we calculate our work, it's calculated as negative, so dU=dQ+-dW -> dU=dQ-dW for my...

Yesss this makes so much sense now, I had this theory earlier, but had convinced myself that the internal energy would have to change, and because of the constant temperature there was equal heat flow, but this makes so much more sense, thankyou so much! The question is actually for for an...

Thanks for the reply, appreciated! Okay, so that means that there is no change in internal energy? As, theoretically all the energy that would have been gained is lost due to the free flowing heat leaving the piston? giving dQ=dW, but I thought the 1st law was expressed as dU=dQ+dW, not...

Or do I use the enthalpy equation dH=dQ + VdP, calculate the dH by calculating H for each ViPi and VfPf, and then dQ is the heat flow?

Okay, so firstly sorry if this is a poor post/wrong topic, I'm kind of new here, and it's been a while! I'm given an example of a quasistatic process, in a frictionless piston. The piston compresses an ideal gas from Vi to Vf, and pressure increases from Pi to Pf, all at a constant temperature...

True that, thanks man, I googled an acot scientific calculator, just checked your method with it and got the same results, boom!

Nevermind, ultra blonde moment fml. cotcotcotcotcot!

if cos(theta)/sin(theta)=10.11, what is theta considering theta is a constant, just spent 2 hours on a collisions question and am stuck on the last hurdle, thanks for looking

Ahaaa, maybe not quite so much detail was needed, sorry for your pain! Thanks loads, you've made it make far more sense than any website I've looked at, and helped my coursework mark for sure, Thanks again man :)

Thankyou for responding. Yeah I understand how integrating will give me constants that show how it proves the equation, but am really unfamilliar with the differential techniques and how to get that far? Would it be complicated to explain how to separate the variables in this instance? No...

The rate of capital growth in a bank account is described by the differential equation dM/dt = aM Where dM/dt is the rate of change of the capital M and a is the annual interest rate. Show that the general solution for the time dependence of capital M(t) is given by: M(t) = M0 * e^at...