Recent content by Gregie666

oops. looks like i read the data wrong! sorry 'bout that :)

from what i understand you took m=80+40. but the man is not accerlerating, only the platform. so m should be 40. so the equations should look like this: 1) F=ma 2) F=T-mg from here it's easy: ma=T-mg 40*2.5=T-10*40 T=40*12.5 notice that i used a capital T to denote the tention in the rope. i...

hi. can anyone push me in the right direction with the followin problem, please? Homework Statement a cart is moving on a frictionless surface at a speed of V_0 the mass of the cart is M. it suddenly starts to rain at time t=0. the rain is dropping vertically at a rate of q gk per...

as for the first question: calculate how long its going to take the rancher to drop from the level of the branch to the level of the horse (remember that he's falling at a constant accleration of g). then calculate how far the horse will gallop during that time. let's call that distance X...

ok.. i think you are right. so using conservation of momentum i can approach it like this: in an infinitesimally small ammonut of time, the following equation holds true: VM = (V + dV)M + (V - V_g )dm VM = (V + dV)M + (V - V_g )APdx VM = VM + Mdv + (V - V_g )APdx {{(V - V_g...

http://img186.imageshack.us/img186/2536/graphuq4.th.jpg take a look at the picture. you need to find the area enclosed between the two curves and the y axis. let's call this area S. first find the x value of the point of intersection of your two functions. let's call it b. then find the...

take a look at this picture: http://img143.imageshack.us/img143/3936/graphfq9.th.jpg it shows the function and the derivative. the picture you posted is wrong. nether the function nor the derivetive change direction at 0 or 1. when you want to find min/max points do the following: 1...

can you explain how you got a maximum at 1? you get a min point in 3/4 because the of the first derivative changes sign there. then you check x=0 and x=1 to see if there are maximum points there. f(x) is bigger then f(1) so there is a max at 0 but not at 1.

hi... a rocket of mass M is flying through a dust cloud the cloud has a density of P. the rocket's cross section is A. every dusticle the rocket colides with becomes permenantly attached. the rocket is ejecting material (as a propellant) at the same rate that it assimilates it. the speed of...

AH! ofcourse... i get it now. thatnk you very much.

i think so. but either i do it incorrectly or its just wrong. $\int {{{gyM} \over L}dy} = {{gM} \over {2L}}y^2 $ but its supposed to be 0.5MgH

the calculus in the example is too complex for me to follow beyound the initial few steps(this is only my first semester in college). but still it implies that: du=gydm=gyM/Ldy (when the differential element of length is dy) can anyone tell what i am doing wrong?

theres obviously an integral involved.. i think that the integral is supposed to be dy (when y is the height above 0). in every slice of rope the potential energy is y*dy/M*g. but if i integrate it this way it gives (0.5/M)gy^2. which is wrong.

Hi. i need to calculate the potential energy of a rope hanging from the cieling. the rope is uniform. the mass of the rope is M. the length of the rope is H. the bottom tip of the rope is the origin of the coordinate system. i'm *not* supposed to do it by treating the rope as a dimensionless...