Recent content by jeroen

Virtualization doesn't conjure 'computing power' out of thin air. It merely presents existing 'computing power' in a different way. And the conversion comes with some overhead.

Virtualization doesn't create computing power. A virtual machine is just software that emulates hardware. And computing power is not energy.

Expansion doesn't matter here. When there is a closed chamber then there will be an inside pressure and an outside pressure. The outside pressure will apply force to hold the chamber together. The inside pressure will apply force to take the chamber apart. Wether the chamber will come...

First you need to understand how the cycle 'cycles'. What sequence of pressures will you encounter as you go through the cycle? Just name them high, low and medium. Plot this sequence on a piece of paper. Also draw the situation described in your post and plot th two distance-pressure...

All you need to do is solve this equation: RL = 1/(1/R + 1/(2*R+RL)) And if R = 1: RL = 1/(1 + 1/(2+RL))

http://local.wasp.uwa.edu.au/~pbourke/geometry/lineline3d/

http://www.astro.uu.nl/~strous/AA/en/reken/grootcirkel.html

Measuring a star's distance by parallax basically is solving a triangle: The base and the two angles at the base are known and the height of the triangle needs to be calculated. However, the angles aren't known exactly, there is a certain error. So a measured angle of 60 degrees might...

Isn't the red light refracted less than the blue light? Blue light hitting the atmosphere is refracted downwards while the red light can follow a straighter path.

Delta means change. So delta velocity is the change of velocity. There is one equation that calculates force. The next calculation uses force to calculate accelleration. You can make one equation out of the two. The aim is to come up with two equations: v1 = v0 something dt r1 = r0...

t1 is the time dt is delta time r1 is range (or distance) f is force a is accelleration v1 is velocity dv is delta velocity dr is delta range r0 is the initial range v0 is the initial velocity t0 is the initial time m1 is the mass of the movable block m2 is th mass of the big block For simple...

You'd need to: -add a very small dt to t0 -calculate f from r -calculate a from f (or go straight from r to a) -calculate dv over dt and add it to v0 -calculate dr over dt and add it to r0 And then repeat this.

Assuming the thread is not attached to the tube, definitely! The weight will pull the thread one way while the ball will pull it the other way. If either of them pulls harder then the thread will move that way. You can use the lower ball to tell if the ball has more, less or the same...

A spring would be sufficient. Just measure the amount it is extended while you swig the weight and how much it extends per amount of force. What kind of equipment does your school have? With a spring, a sliding potentiometer, a multimeter, an oscilloscope or an electric graph you could put...

You might want to rewrite that sentence. ;)