Recent content by FallenLeibniz

@Charles Link: I do thank you for your reply. It's much appreciated. I wasn't looking so much for the answer as the fact that setting the volumes equal could be just written down on paper enough. More or less, I was wondering the physical justification from behind the derivation. But, I do get...

Homework Statement The problem that I am having stems from a problem given in the following way:[/B] "a)Show that for a gas, the mean free path ##\lambda## between collisions is related to the mean distance between nearest neighbors ##r## by the approximate relation ##\lambda \approx...

Homework Statement The following is a small concept exercise from an. EM text (Electromagnetism: Pollack and Stump). I believe I have the explanations correct, but am just looking for "peer-review" as they seem "hand-wavy": Suppose a conductor has a cavity inside it, and there is a point...

@andrewkirk : Figured it out. I kept looking at something wrong, but have seen the blood clot.

I was under the impression in my work that the resistive force would never be greater in magnitude than the effect of gravity. In my setup, $$k{v}^2$$ is the resistive acceleration and is pointing up the plane (in the positive x-direction) and $$g\sin(\theta)$$ points down the plane (in the...

@andrewkirk I am not sure what you mean by "you have distance x pointing up the plane". I did say that only my notion of what a "positive distance" was is going up the incline. Can you elaborate? @Student100 : I am attaching my work on the problem here...

Homework Statement I am trying to prove what is asked in the following problem: A particle of mass m slides down an inclined plane under the influence of gravity. If the motion is resisted by a ##f=kmv^2##, show that the time required to move a distance d after starting...

Riiight. That makes more sense than the block jumping up in the air. Oops.

After that don't you just square that velocity then divide by two times 9.8?

Has anyone been able to get the answer that the OP says was correct? I have done this problem with both paradigms to try amd help them however I can not see how this answer of .35m is reached. Both ways give me that the mass hits at 3.597 m/s.

Would you like kind enough to say what was wrong?

Here is the corrected equations list. I also want to thank the posters who have replied so far. I had been stuck on this for a really long time.

OK, so I took a look at my equations and there is something incorrect. The sum should be ##\vec{T_y}=\vec{T_a}+\vec{T_b}##. Furthermore the magnitudes of ##\vec{T_a}## and ##\vec{T_b}## are the same as the Tension (I believe a poster mentioned this above, but I didn't see what was meant till...

Yes,that wa Yes I do apologize. ##\vec{T}=\vec{Ta-Tb}## should be ##\vec{Ty}=\vec{Ta-Tb}##. My bad. I will update the equation list when I get back to my high tech white board . The above mentioned physics is still correct though right?

The force pair adds up to result in a vertical upward resultant force that counterbalances the weight of the second mass at equilibrium.