Recent content by mybsaccownt

i hope it's not bad form to re-use my post from a similar type of question:from this point: yes, you can convert it to exact by trying:

ok, t is not just 12, there are some extra digits as well (i'm horrible with sig digs as well :-P) also, the horizontal velocity is constant and velocity = distance/time so, distance = velocity * time you know the time, and you know the velocity

hmm...the time needed for the object to fallwell, vf = vo + at vf = 0 so...

Hmm...maybe this decision is not as difficult as I thought. I suppose I have a pretty big list of complaints! haha I guess it's just the fear of change, which should be a good experience anyway. Thanks for putting this into perspective, everyone. I appreciate it.

Didn't realize this was such a huge post...so I provide you with...a table of contents! (and the option of skipping part 1 without missing any really important information)1. Background Info - the epiphany! 2. Accepted! (to hell?) - the truth unfolds 3. The other - university, that is 4. The...

you want to know the force on the small piston that will cause a pressure large enough for the bigger piston to overcome the weight of the car be careful with your units, a kg/in^2 is not a pressure (you want a N/m^2 or a lb/in^2) you have the force for the big piston, (the force that you...

Yes...it does look like it would be an exact equation, but it's not quite there. Is there anything we can do to make it so? how about finding an integrating factor to multiply through in order to convert it to exact? if... \frac{My-Nx}{N} is a function of x only, then the solution to...

it seems like the units don't work out you want work, which is N*m/s or just joule/s, ds/dt is also velocity, is it not? so you have: change in work per time = force*velocity*velocity? so...W = N * m^2/s^2 = N*m/s <--nopealso it's a partial differential equation, but i suppose you...

Torque = r x F it is a cross product and r x F is not the same as F x r "The cross product is anticommutative, a × b = −b × a " http://en.wikipedia.org/wiki/Cross_product

Bernoulli's equation (in that form) has to be applied between two points in a steady flow (with the exception of obtaining an approximate answer for 'quasisteady' flows) , I don't believe that any point in the jet besides the very base of it can be considered steady (since it diffuses). So you...

although it does seem pointless, I've seen it used and I believe N*s is acceptable

area velocity relation? the continuity equation? It can only be applied to incompressible flow, the density must remain constant.

well, since you defined P1 to be absolute pressure, then so should you define P2 you could have just taken gauge pressure, it's up to you but being consistent is the most important thing

hmm... it would seem that as m1 and m2 approach the same value then the acceleration of the system will get closer to 0 since the forces on both sides of the pulley would be balanced that's fine, but you're not actually setting the masses exactly equal, and so you should have some...

maybe YOU don't ...but... Maple does :cool: let's see if my newbie latex skills can do it justice > a:=exp(-x^2)- 2*x^2; a := e^{-x^2} - 2x^2 > int(a,x); int/indef1: first-stage indefinite integration int/indef1...