Recent content by yougene

Thanks for your time. This is noted. Even with the rotated axis, T still has a vertical component. What he seems to be implying is that tension is only the horizontal component of force T. This I'm finding hard to digest.

Ok, so in the red box it says that the tension is obtained from the horizontal dimension. This seems very wrong to me. Could someone explain this to me? Isn't there tension in the vertical dimension? Homework Statement http://img27.imageshack.us/img27/9276/inclineplanesolution.jpg

Bah. If physics is a language of precision why go out of the way to play semantic games in word problems? No such problem, but you're right this problem doesn't involve any sort of "forces".

Homework Statement http://img198.imageshack.us/img198/3834/relativeboatvelocity.jpg The note says "( Wind relative to boat is irrelevant! )" Why is it irrelevant?

That's what I'm thinking as well, thanks guys. If anyone has a third opinion let me know.

The only features I can discern here, are two parallel lines( ceiling and base of triangle ), the sides of the incline, and the inclines angle Theta. Other than the parallel ceiling there is nothing that relates the rope to the incline. In my mind if I move the strings hanging point left or...

By what reasoning?

Homework Statement By what relation/reasoning are these two angles, theta, the same? http://img198.imageshack.us/img198/6273/inclinedplane.jpg

Figured it out. It's amazing how writing things out always puts it in perspective.

Homework Statement If you step on your car's brakes hard, the wheels stop turning (i.e., the wheels "lock") after 2.0 revolution. Part A. At the same constant acceleration, how many revolutions do the wheels make before stopping if your initial speed is twice as high? Homework...

Thanks, this is all coming back now. :P So then my equation would work only if C = 0. @Mark44 So if my C = -1 then saying Ax^2 + Bx = 1 would be a valid way to solve the equation? Or are you saying Ax^2 + Bx + C = 1 would be valid to solve for?

Homework Statement Lets say you have Ax + Bx^2 + C = 0 Usually I would use the Quadratic formula in this situation. But what if you do this x( A + Bx ) + C = 0 x( A + Bx ) = -C Couldn't you then say x = -C and A + Bx = -C x = (-C - A) / B ?

Thanks, that looks comprehensive.

Thanks, that sums it up nicely. What if an acid is being added to a basic solution? How does Kw play into that type of scenario?

Thanks for the reply The 7 was a typo. That doesn't address the fragmentation in my understanding though. Can I use Kw equilibrium with weak acids/bases? Why?