Recent content by StheevilH

then it would equals to mg / cm^2... which is the unit of Q. where did you get mg/cm^3? oh wait... damn it... you changed L into cm^3... thank you so much

I think that should equals to just "cm" If you look at the attachment, equation 24, that is what I am after and the units were taken from other sources. But to me, equation makes sense but units don't from what I did. Is there a problem with my algebra skill?

Okay I found some references and the equation I am after is Q = 2 * C * (D * t / ∏)^(1/2) where Q is the weight of drug released per unit area (hence unit is mg/cm^2) C is the initial drug concentration D is the diffusion coefficient (unit of cm^2 min^-1) and t is release time in...

Oh... Thanks Borek :) worked it out

Um.. I know of Archimedes... and his principle... but I do not know that principle in depth.. but is that principle why it is fair to assume that the identical pressures being applied inside out?

oh... after rearranging to solve for n, subsituting n = m / M and M can be found on the web (yea... periodic table... should have one... somewhere...) and substitude the values in? I also am curious that the assumption I made, "the pressure inside the balloon equals the...

Homework Statement The question states "What mass can the balloon lift at sea level where the density of air is 1.22 kg / m^3".. Extra information given are, use of ideal gas equation is needed, which is pV = nRT where p is pressure, V is volume, n is number of moles, R is constant...

Roger, copy that (Y)

Oh..lol question... That was pointing at the Vo being equal to V... So... we can ignore about the directions in the calculations?? The magnitudes are same...so we can just put them as same? and...did I actually get it that way.. @_@

Ah...I worked out...I think... lol..yeah I thought PE was meant to be in projectile formula.. Eh...from what I got...is KE = PE... 1/2 * K * x^2 = 1/2 * m * v^2 Hence, v^2 = K * x^2 / m .. Using range formula; R = Vo^2 * Sin 2 (*) / g.. R = K * x^2 / m * g...Sin 2 (45) = 1...so...

Wait.. don't worry about the above one I wrote... I found out potential energy of the spring is 1/2 * K * x^2.. but this doesn't match with any projectile formula I got.. is x = Vx? or V itself? not in component form? Gah...I'm not good at this... Waiting for a reply...

Eh... write out as many equations for projectile motion? I learned only 4 types of them though... and none of them matchs.. Before that...what is the equation of conservation of energy?? I don't get the whole thing..to begin with...so..would you be able to be more detail?? please?? =]

R = Kx^2 / mg <-- anybody reconise this? Well.. I had experiment about springs... I got this R = K x^2 / m g.. formula to explain... where R = range (displacement of x) K = constant of spring (elastic) x = extension of spring m = mass g = gravity Can anybody prove this...