Recent content by Falcon

'solid edge' is also a great intuitive solid modelling program whatever you do, try to stay away from 'ideas'.. and if you have to use it.. use the unix based version (it'll still make your brain bleed) pro eng W is prolly one of the best.. especially since you have all the analysis tools...

sum the forces in y, and the moment about point 1 define the point at which w1 acts as point 1, x=0 define the point at which w2 acts as point 2, x=35 Sum Fy=0 integral from 0 to 35 of [(w2-w1)/35in)x+w1]dx = 12000lbs Sum M1=0 integral from 0 to 35 of x[(w2-w1)/35in)x+w1]dx...

If you want to simplify it, think about this: You have two kinds of filters: A - light screen mesh, allows a high flow (10L/s) B - thicker carbon filter, allows a low flow (1L/s) You put A into a circuit, its allowing 10L/s to move through it. Now if you connect filter B parallel to A...

King of Spades, Ace of Diamonds, Jack of Clubs, Seven of Spades, Ten of Spades, Ace of Hearts Snake, Beet, Cricket Gray, Thin, move to the East "SIRHC" Thanks!

ln x^1/2 because of the rules of ln, you can 'pull' the 1/2 down and out in front 1/2 ln x and as you know, the deriv. of ln x, is 1/x 1/2 1/x result 1/2x ;)

i used a Ti-89 in high school.. and now I'm using an HP49G+ the hp is quicker (it seems), has a much nicer screen and personally, i just like it better i cannot comment on the functionality, since i don't have the TI to compare with now and i suppose it should be obvious, but i am now...

johnny, if you actually have the part, and you don't have to be extremely accurate, the easiest way to measure the amount cut off, may be to get another uncut part then see the difference in the amount of water they displace

you can't simply revolve what you see in the red, since it won't fill the entire cylinder rotate the section you see in the yellow, then subtract B (which will be a triangular prism, plus an erregular shape that is somewhat like a rectangular prism.. which i can't remember the name of.. lol)...

OK.. i got it now.. let me draw it out so it makes some sense

Try this.. rotate sections A about the GREY axis

that won't work Chen.. if you do that, you'll be revolving a different shape than what was removed its hard to explain, but think about it, the shape will be different... you have to revolve it around an axis which is parallel to the long side of the cutout inorder for the shape to be the...

Yeah, in fact looking at it now, I think I would have just added them all to |v|2 = 25w2sin2(wt) + 25w2cos2(wt) then factored 25w2 all at once

ahh... OK.. I just got an idea.. good old trig identities.. let me see if I can work it out EDIT: ahhh... makes perfect sense now :D 9sin^2(wt) + 9cos^2(wt) + 16sin^2(wt) + 16cos^2(wt) =25sin^2(wt) + 25cos^2(wt) =25(sin^2(wt) + cos^2(wt)) =25(1) =25

Thanks Chen, I probably should have shown the steps that I had come up with. I can reach this: |v|^2 = w^2[9sin^2(wt) + 9cos^2(wt) + 16sin^2(wt) + 16cos^2(wt)] and i can see how we get the w^2 in the next line |v|2 = 25w^2 but how do we get the 25 from this mess: [9sin^2(wt) +...

This doesn't exactly have to do with the title, but generally comes up in this sort of thing. I'm feeling pretty dense right now to not be able to understand what's going on.. but my textbook isn't exactly helping. lets say there is a function which relates a particles position, r, with...