Recent content by SirHall

Thanks, and if it's not too much to ask, would there be a way to approximate or find that?

What I've not been able to find after literally spending several hours researching is if you can find the formula to find the magnetic field strength from a temporary magnet. Or, to be put in better terms, what would be the force between two objects if one is a magnet and the other is something...

Well, I certainly could be incorrect, but firstly take a look at this. The coefficient of friction between rubber and concrete is 0.65 while the coefficient for plastic hovers somewhere between 0.2-0.55 depending on the type of plastic. So, I must say that the tires of a bicycle are great for...

Realistically, this pretty much comes down to the friction between the wheels and the ground. Firstly, assuming friction between the ground and the wheels were identical for both the bike and the skateboard, they 'should' both reach the end at the same time(remember, heavy objects fall at the...

You are exactly right. However after looking a little further with our great friend Google Scholar and taking advice given here, I've found my answer. So no further replies are needed. Thanks again for the assistance!

Well, after a little experimentation I think I may have found v(x)! ##v\left(x\right)\ =\ \sqrt{2\cdot \frac{-\frac{1}{2}pACx^2}{m}\left|x\right|+u^2}##. So, back to the original reason for this thread, what do I do with v(x)? Well, if I want to find the distance it took for the object to stop...

I see your point... While I wasn't too happy to do it this way, here we go. I'm attempting to find the total distance a projectile would travel in a fluid given: m = Mass u = Initial Velocity p = Fluid Density A = Projectile Reference Area C = Drag Coefficient b = Total Distance Travelled x =...

I have recently been attempting to solve a problem that has been bugging me for quite some time. I've gotten back into calculus and integrals to attempt to solve a little formula I'm trying to build for a simulation test. Over-all, if I have ##\int _0^bv^2x\ dx## I'd expect the outcome to be...

Okay, (yes, still trying to figure this out) I've re-learnt derivatives and integrals. I think I've figured out the answer to : But, before I come to any conclusions I was wondering, what is V? You have u for initial velocity, and there isn't any t if I was supposed to find the instantaneous...

... Taunt appreciated :) Great question! Well aren't I damned, you pointed out three to four things I forgot which were trivial, made three to four steps which I needed some researching to get a basic understanding of, and to top it all off, I'm not entirely even sure exactly how to script...

I don't recall stating that it worked, more just asking if or whether it did indeed work. I'd assume you were talking about: Yes, I understand what you said, which is why I used the original drag force equation which does indeed use velocity squared. d = Distance it takes for the moving...

Yes, I'm still trying to find the stopping distance. But now I'm sort of at a loss. Would I simply throw that equation into the original to replace the dx/dt?

In stopping distance I'd assume you mean ##dx\ =\ x_1-x_2## where ##x_2## is the stopping distance. Sorry if I'm coming a little slow to this, but I do appreciate actually being led to the solution.

*Shot down*

Hmm, I remember something along those lines from back in high school. While sort of a cheap solution, since dt = t2 - t1, I'd assume it's safe to say: ##dx\ =\ \sqrt{\frac{\left(2fdt^2v\right)}{acm}}##