Recent content by Antineutron

No. Read again, this part. It must have done some work? What did the bullet do work on? What kind of collision is it when this occurred?

Although the sum of all the Q's are 0, make sure you put the correct sign for each Q. They are not all positive.

alpha's do matter, even if they are from a table. If they both had same materials, hence same alphas, that ring will never come off! Length initial of sphere equals .0005 times Length initial of ring added to length initial of ring. can you take it from here?

\frac{\partial f}{\partial z} = -3x^2z^2 = -3x^2z^2 + \frac{\partial h}{\partial z} \Rightarrow h(z) = 0 Im having trouble with latex, but your partial f over partial z which is -3x^2z^2 = -3x^2z^2 + partial h over partial z. so, h(z) = 0=g(y,x)

explain why you believe it would stay the same.

okay so, you start out with kinetic energy of the bullet then you hit the wood of certain mass, that slows down the bullet, but it slows down by raising the wood and bullet up some height. then after that the bullet comes out of the wood at lower kinetic energy thus comes out slower then...

I think you got this line wrong

what material is the sphere and what material is the ring? What does that tell you about alpha value for each one? also on started out at what length in regards to the other? The change plus the initial value for each material should equal each other right?

Okay think about you started out with this much energy then you had to spend that energy on other things then how much energy do you have left?

You can have a vector as long as it begins from a point and end at a another point right? Thats what I get from vector formulas. It is not required for a vector to start at the an origin right? Is that what you mean?

use alpha for both because the size of sphere is proportional to the diameter. as the diameter gets larger or smaller the smaller the sphere's surface area gets. right? so you can use alpha for the shere and the right and make the lengths equal each other like you said.

yes.

I don't know what to say... I agree, the limit of 1 messes it up. maybe the person that made up the problem made a mistake?:confused:

no because underneath the (x+1) is not an irreducible quadratic factor Here is one that is irreducible x^2+1 try to factor it

if you just try to simplify and do the limits the denomintor goes to 1 and the top alternates from -1 to 1 and back and forth.