Recent content by Thijske

I genuinely apologize and I understand it's a bit irritating that my numbers are a bit off, but I want you to know that the numbers were educated guesses (400Pa should be 400kPa). Also, I am no absolute rookie in science (done 6 years of pre-university), so I think I do know about the things you...

Thanks to everyone for taking time to write a response! This was my idea as well. The problem is; I assume a part of the energy will be lost to the cylinder walls and other stuff. Therefore, not all generated energy will go into heating the air-fuel mixture. I don't know how to calculate the...

Say there's 2 grams of gasoline (vapor) (E85) and sufficient air, how can you calculate the force/pressure of the explosion when this mixture is ignited? The pressure of the mixture before ignition is 400 pascal. The calorific value of the fuel is 45 MJ/kg. The volume in which the explosion...

Hey all, I was recently searching for kinetic friction coefficients, and I noticed that for a reason which is unknown to me, some materials had not been appointed such coefficient. The specific kinetic friction coefficient that I'm looking for is aluminum on greasy/lubricated cast iron. Any type...

Yeah I've experienced that a lot lately, so it kind of made me laugh. Anyway, I want to thank you for your time again! However I'd like you to consider the following: https://cdn1.imggmi.com/uploads/2019/12/20/5f2ca236e50492ac9894f0f286e6276c-full.png In here, Fpressure is what I meant...

Hi, thanks for your response! In answer to your concerns: - This drawing is a top view, so gravity doesn't play a role. - The two components are both made of thick aluminum (flexibility is therefore probably insignificant). - The turning rod and the circular thing are airtight. Therefore, a...

This is quite a specific case: I would like to calculate how much friction force is exerted between the rotating rod and the stationary circular thing. I guess we can't just use the formula of static/kinetic friction, because I don't think a normal force is of application... In the image, Fres>...