Ducky Havok

I think the key here is that the car has a constant acceleration. From the info given, you can find the average velocity (140/3.6=38.89) and from that and the final velocity, you can get the initial velocity (53-38.89=14.11). Also, you can divide the average velocity by the time to get the acceleration (since it's a constant). Hope that helped and good luck on your test.

But that's moving a number and a sign, not just a number, and would be quite similar to John's post (#5).

The range of a natural log is all real numbers, (as x-2 approaches 0, ln(x-2) approaches negative infinity, as x-2 approaches infinity, ln(x-2) approaches infinity) so you should be able to get your domain from that.

I would love to post these here, but they are flash animations so I can only provide the link. It's called Jokes with Einstein... that should enough to get your interest . They're hilarious. http://www.spinnerdisc.com/jokeswitheinstein.html

And, something I only discovered recently, if you turn over those papers they put on your trays, they have all the nutritional information on them. I think they're obligated to give them to you, they just don't want you to read them or else you'll realize you might just die. Comparing those though would be a good way to start.

No, it was just numbers, and brabos got it right. It's supposed to be 101-10^2=1. I do like Dak's idea, though.

I completely stole this from iGoogle brainteasers. Move one number to make this equation correct: 101-102=1

Tli tate hlluk ht ealaee e eme lopa vav lvlris eyeat lwnt rnh. This isn't nearly as hard as the some of the others, and can very easily be solved with a paper and pencil, but I thought it was fun. Plus, this section doesn't have too many recent posts, so I'm hoping to spark some interest. The sentence doesn't really make sense (It is all real words in a grammatically correct sentence, I just couldn't think of anything serious to write so I had fun with it). I'm just going to go ahead and take a guess, and say it'll last 4 hours

http://www.physorg.com/news98468776.html Here's an article I came across this article and thought you all might enjoy. I'm curious to here the opinions of people who are much more knowledgeable than in physics, since I'm just a student. How much would we have to relearn if this was true?

I might be wrong, but I believe pressure and surface area also effect solubility of a solid.

In the first example they fail to give you the original balanced equation, or the fact that you start with 1.5 mol of benzene and 7 mol of oxygen. It seems to me like they just left out the question itself. And as for the 157.9 g/mol, that's the molecular weight of Iron (3) oxide (found from a periodic table). As for the first part, do you not understand what a limiting reagant is, or just how to find one?

I thought all gases did, but it had to do with their pressure since most gases are very insoluble due to entropy? Thanks, btw. I typed it out, and somehow I still failed to realize that it was a solid. I guess I should slow down just a little bit in the future.

I took my final about a week ago, and I was curious about the answer to one of the questions. I haven't been able to ask my professor about it (since I'm no longer in that class), so I figured I'd try here. Consider the following balanced equation. The reaction is endothermic as written. BaSO4(s) + H(+)(aq) in equilibrium with Ba(2+)(aq) + HSO4(-)(aq) (I was going to use the LaTeX, but when I looked at the guide all the appeared were errors, sorry). Which of the following effects will favor a shift to the right side? (A) Decrease temperature (B) Decrease pH © Add a soluble barium salt (D) Increase the pressure in the container (E) Add more BaSO4 (s) I put E right away and moved on, but the correct answer is B. I understand why B is correct (lower pH, more H+), but why isn't E correct as well?

Okay, thanks Dave, your approach makes perfect sense, and I kind of forgot about that at the time (this is all self taught so I may have just overlooked it). But, that was for the first one which I think was a little easier because the 2nd part simplified. Using Tom's method, I get [math]\frac{x^6\ln(x+3)}{6}-\frac{1}{6}\int{\frac{x^6}{x+3}dx}[/math] And thus comes a new dilemma for me. Maybe I should just give up for the night and look at it again after sleep... that usually seems to help. Thanks for the help though. (I think I've become discouraged because I used an online integrator to solve it to check for an answer and I got something extremely long)

Thanks, but that stills leaves [math]\int{x^5\ln(x)dx}[/math] which is pretty much the same problem. I hadn't thought about that or else I probably could have gotten partial credit, though. My thoughts were to turn it into a Maclaurin series and then integrate it (because that was one of the last things we went over), but I had some trouble with that. And after all that I just realized something. Since I never wrote the question down, I typed it wrong. I'm sorry. Here's the corrected version. [math]\int{x^5\ln(x+3)}[/math] Thanks for the help on the first one though, hopefully you can help on the real version now.