Recent content by AceInfinity

Alright, so I added them as If they weren't absolute values and ended up with the smaller number (exponent -9) I should have assumed that was the case since there's no other way that I would end up with a smaller number than to end up with a smaller number through the adding process, and the...

I know, There's just that number of charges MORE as those numbers are only net charges, but how would you solve this? Still doesn't answer that

Not sure. I'm still convinced that the answer they gave in the answer key is correct because this is from a practice Diploma. @jtbell I got 5.00x108 Dividing that by 2 would equal 2.50x108 I'm still confused though, as the way I was doing it must be wrong, so doing that is irrelevant. I was...

Alright I got it: B = [mv/r] ÷ q B = [(1.67x10-27)(1.38x106)/(3.00m)] ÷ 1.60x10-19 B = 4.8012...x10-3 However I just took the answer of the velocity from the answer key. I'm not quite sure how to solve for velocity in this case. If that could be the second (should have been the first) part of...

Still requesting some help on this one. At this point I'm at a loss for how to solve this kind of problem at all. Sorry for being so impatient, but I'm trying to learn as much as I can for today about classical physics involving charge, momentum, mainly. I know that the curvature of the...

Example: A conducting sphere X that has an initial charge of +2.0 × 10–8 C and an identical conducting sphere Y that has an initial charge of –3.0 × 10–8 C are touched together. After they are separated, the charge on sphere X is? Answer: (–5.0 × 10–9 C) And what I would have done is tried to...

ohhh... I think you put me on the right track for beginning to solve this. The initial momentum was [right] in this case. Therefore the conservation of momentum is seen for that X component, I'll have to split them off into the X and Y components of momentum and look specifically at the...

That's obvious though, but the code does do the job with the right permissions, saying that the browser decides with your example would be like asking how to open a webpage in a new tab with IE6. IE6 doesn't support tabbed webpages, and therefore it can't be done. But the html code...

E = V/m E = (6.75x102V)/(0.01m) E = 67500 N/C Fg = Fe m x ag = Eq (3.30x10-15kg)(9.81m/s2) = (67500 N/C)(q) q = [(3.30x10-15kg)(9.81m/s2)] / (67500 N/C) q = 4.796x10-19 *Doing this as I type it out and that -19 looks promising... 4.796x10-19 / (1.60x10-19) = 2.9975 and we would round up to...

First off, i'd like to note that this isn't homework, and I've seen other threads in here that deal with question/equation/problems, so I hope this isn't against the rules. I found this on a practice physics test online. I'm just using it for the benefit of my knowledge, nothing more. I can...

I was wondering, what about the problems that involve questions where 2 "objects" rebound off each other, or one stops and the other continues moving after collision. Would it be the same kind of "solution" to finding the momentum of one or the other?

Ahh, thanks, I'm going to try that... ... > v = qBr/m v = (1.60x10-19)(B)(3.00m)/(1.67x10-27kg) What exactly is B though in this equation? How do I get it

This was the original way that we learned how to do it. Thanks for the second method :) I'll be able to do this kind of problem easily now, the key here is to know that it's the conservation of momentum.

E = V/m = E = (6.75x102V)/(0.10m) Therefore E = 6750N/C E = N/C Therefore N = E/C N = (6750N/C)/(1.60x10-19C) = 4.21875x1022 = ANS / 1.60x10-19 <--- to find out how many electrons there are by finding how many charges of one electron can fit into the oil drop's total charge. = 2.63671...

alright, so I have this: mv2/r = (1.67x10-27kg)(v2)/(3.00m) How do I solve for v in either equation when I don't have the Fm?