Recent content by Peppino

Possibly, but on the various objects I have tested (a marker, a textbook, a can of Dr Pepper, a sliced cucumber) the 45 degree rule seems to uphold. If only I had a large range of various sized cylinders could this be tested. And the bullet is direct towards the very top of the cylinder, and...

I've encountered a problem that I do not believe I am able to answer using my very basic knowledge of classical and calculus-based physics, dealing with knocking objects over. Say we had a cylinder of length L and radius R and mass M. And suppose we shot a bullet of mass m at the very top of...

What is the correct spelling of the word? I've always used equilibriant but I've now heard it should be equilibrant. Any insight into this dilemma?

Oh, I see it. That's interesting. So when using different methods to find an integral, the value of C (or, what's included in C) would change?

Yes, I realized that. Any guess as to why my circuitous method didn't work?

Doing the non-moronic way I get the same answer as my calculator. I'll stick with that for now but any comments will still be appreciated

The table of values are completely different... on my calculator's answer, x = 1 yields -1, while on mine x = 1 yields -1.125. Is it not a function? Because an infinite number of different values of C could lead to a function with a derivative of the original function

Right... I meant to put the 1/8 in there. But when I graph the two expressions I get two completely different tables of values. Are you sure they are the same? As for the unnecessary U-Substitution... that's just me being a moron I guess

I have to find the integral of (4-x)x^{-3}. My TI-89 says it should be \frac{x-2}{x^{2}}+C but I can't seem to get it myself.I rearranged it to get (4x^{-1}-1)x^{-2} and then I used U-Substitution. And set U = 4x^{-1}-1 so that dU = -4x^{-2}dxThen I rewrote the integral as -\frac{1}{4}\int Udu...

Impact Speed: The speed at which a falling object hits the ground Terminal Velocity: The fastest speed an object can go as its acceleration is slowly decreased over time due to air resistance (air resistance works against gravity; eventually, the force of air resistance = the force of gravity)

...yield? Would it be useful for anything? I understand that it would yield measurements in ms (meter-seconds). Is there a term for it?

You would need to know the amount of friction and coefficient of restitution of the collision.

That would be 3.64 N, so there is a 3.64 N frictional force? Where would that come from?

But since there is no friction on the floor, wouldn't applying any force accelerate the big block?

1-kg Block 1 sits on 10-kg Block 2. Friction between the blocks is 0.4 static, 0.3 kinetic while the floor is frictionless. A 4-N force is exerted on Block 1. What is the acceleration of Block 2? I know Block 1 won't move relative to Block 2 since it is canceled out by the frictional force...