Recent content by Samky

I'm student-level at vectors and derivatives. I've done basic vector calculus and all that. I'm confident I can solve simple problems involving these things.

Sorry, I mean speed. The magnitude doesn't change but the direction does. It's basically a one dimensional oscillation that creates a two dimensional oscillation.

Thanks a lot, I'll look into that! I wrote some MATLAB code to get a good approximation, so I have some graphs and numbers, but that's sort of cheating. I'm sure a mathematician could solve the original problem in closed form but I have no clue how to begin... but this should help me out...

An old puzzle that was given to me long ago. I don't know how to approach this problem, any help is appreciated. I'm removing some specifics because I don't want it solved for me, I want to know how to approach it. Feel free to solve a simplified version if that helps you explain it to me...

You can sub in q = (E*r^2)/k into v = kq/r to get V=Er but even simpler is to just look at the units. You're asked for V which is joules per coulomb. You're given N/C and a distance. How can you make N/C into J/C ? A joule is simply a Newton x distance (recall work = force * distance for...

It's fine to find your charge first, but it's not strictly necessary, you can continue to manipulate equations until you're ready to solve for the final answer (which in almost every case saves you time). That said, you're on the right track, didn't check everything, but I noticed it seems...

I guess I'm trying to see the process you used. Sometimes it's possible to use an intuitive method and happen upon the right answer. That's great, but if you're not sure why it worked, or not sure how to use it if given a different problem then the exercise loses its purpose even if you did...

A quick tip that is a useful habit for nearly all problems, is to keep your work in variable form and only plug in the numbers at the very end when you're ready to solve. Another great tip I like to use when I feel stuck is to see if I can write my unknown(s) in terms of knowns. Either you'll...

Interesting, so the TH or hot reservoir would be the heat we add to the engine though combustion. That means the TL or cold reservoir is the surrounding air. And we see by the equation that as TL -> TH the efficiency drops to zero. This makes sense because if the combustion cannot add heat...

Yes, if you remove that much density then the ball would not sink. As he said your only control is to hollow it out, so the question is how much would you have to hollow out to reduce the density that much? It's quite likely that they will treat the mass of the air as negligible. So to solve...

Hmm, are you sure you converted it into slugs? You're on the right track, but in your answer you give the units ft/s^2 but I don't see slugs? Q1: What are the desired units for your final answer? You're going from kg*m / s^2 to _____ ? Q2: Can you write a fraction in the form...

You're close to the answer. Antiphon's point was key, how can you make them the same? Here's an in-between step. Can you write an equation for the mass of your lead ball using your new formula for volume? Then, as he asked, how can you make sure the mass of your lead ball will not be greater...

It's impossible to resolve a difficulty if you don't mention what your difficulty is. Two things to know for unit conversion though: 1). Units cancel just like variables of an algebraic expression. For example if I multiplied x/y and y together the lone y and the y on bottom of the fraction...

Hmm, not sure about the equation you've given with the many "x" variables, although if you mean V2 = V1 + at then that's fine. Where the last term is acceleration multiplied by time. What is the answer you got? Do a sanity check i.e. do your units match, does your answer even make sense in...

Is it strictly true that an engine will be more efficient if operating in a hotter environment... e.g. your car is more efficient in the summer than the winter? Just going through thermo 1, not a HW question, but it just seems to be true to me because the simple Newton's law of cooling says...