Sorry for lag reply. Oh my god I've just spent the last 30 min scribbling furiously away at my desk because I was getting different answers from 2 different methods and didn't know why. When I did your way I got it wrong because the algebra was more complicated. So YEAH I really recommend just...
Can I assume that the question means the area above the x axis?
Anyway, firstly you must draw the graph. Are you trying to find the volume of the cylinder then subtract the volume not included in the region? If so isn't the x=1 the outer radius? And so then rather the inner radius should then...
Haha, all righty, first of all 595 - 325 = 270!
http://img26.imageshack.us/img26/2167/screenshotdocument1micr.png
I sometimes find drawing a picture simplifies things. The first picture on the left shows the amount of water (LIGHT BLUE WATER) that will be displaced AFTER the screw is dropped...
No need to make it so complicated; most projectile questions can simply be solved using the 3 equations.
v = u + at
s = ut + 1/2 at^2
v^2 = u^2 + 2as
All you need is to think about the initial condition, and final condition.
- To find the velocity at the very peak (i.e. right before impact)...
What is Heisenberg's Uncertainty Principle??
Homework Statement
Hey everyone,
Until very recently, I had always thought that Heisenberg's uncertainty principle was that
\Delta x \Delta p \geq \frac{h}{2\pi} (or \hbar)
However, I'm doing my final year of high school physics this year...
I think you're thinking of the potential of a single point charge - I think the formula for the potential of a dipole in an electric field given by
U = -\mathbf{p} \cdot \mathbf{E}
Without calculus =(
Well, in a dodgy series notation kind of thing...
I=\sum mr^2=m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 +...
In other words, the moment of inertia of a body is just the sum of all the individual moment of inertias of every point on it.
For example , if I had a rod of negligible...
There's a much neater way of doing this! Try using the difference of 2 squares and dividing the KE conservation equation by the momentum conservation equation
It's meant to just be the KE I think - there's no change in KE
There's only centripetal force in this problem - the "tangential force" you're talking about doesn't exist =\
remember the equations linking the rotational counterparts with their linear equivalents:
v=r\omega
Every point on the Earth has the same angular velocity \omega, but depending on the radius of the circular path which it traces, its tangential velocity will vary.
To have a tangential velocity...