Recent content by wanamaa

Homework Statement A sign is to be hung from the end of a thin pole, and the pole supported by a single cable. Your design firm brainstorms the six scenarios shown below. In scenarios A, B, and D, the cable is attached halfway between the midpoint and end of the pole. In C, the cable is...

oh ok, I guess I got potential energy and kinetic energy confused then, when I said there'd be no potential energy. What you said makes a lot of sense, actually. Thanks a bunch for your help! :)

Thank you! That makes so much more sense now. I didn't post the follow up question but here it is: "Assume spring constant k = 1000 n/M, mass m = 0.200 kg, initial spring compression x = 0.15 m, coeffiction of friction µk = 0.20, vertical height of incline h = 2.0 m, and angle T = 45...

Homework Statement The spring in the figure shown has a spring constant of k. It is compressed a distance of x meters, then launches a block of mass m kilograms. The horizontal surface is frictionless, but the coefficient of kinetic friction for the block on the incline is µk. The vertical...

Yay, I was close on that one! Ok, that is what I was trying to do in the beginning and kept running into the "can't take a log of a negative number" problem. But won't I run into the same problem with the newly made "-e^x - C" or would I just rearrange the variables, like "-C-e^x"? Sorry for...

Basically I am trying to solve for the separable differential equation for y, and I also was given an initial y value (which I forgot in my first post but posted it I think in my 3 or 4th post on this thread), which is y(-1)=0 for both equations. I know how to input initial values, I just got...

oh snap! I screwed up writing the notation for x^2/x. I meant x^2/2. Sorry, now I feel stupid :redface:. To solve the second equation, I am given an initial value of y(-1)=0. Forgot to mention that in the very beginning :blushing: oops. I just don't know if I need to refine the equation...

whoops, looks like it wouldn't let me copy the equations you wrote up above. Weird.

I wrote that because that's what the integral for x dx is. Unless my TI-89 is lying to me... I understand where I forgot the C, but with what you wrote, where did the 8 come from? Is it just some arbitrary number to help solve for y or does it have any significance? Maybe I completely...

Hi there, I was working on two of my homework problems for calculus and I'm stuck. First equation is: dy/dx=e^(x+y), solving for y So far here is what I have: dy/dx=(e^x)(e^y) therefore dy/dx=(e^x)/(1/e^y) INT(1/e^y)dy=INT(e^x)dx from the integration, my calculator comes up with...