Recent content by Briggs

Homework Statement Find the general solution to the differential equation in implicit form. http://www.texify.com/img/%5Clarge%5C%21%5Cfrac%7Bdy%7D%7Bdx%7D%3D%28cosx-sinx%29e%5E%7Bcosx%2Bsinx%2By%7D.gif Homework Equations...

I think you have the values of T and t_0 mixed up. Remember time dilation would mean that slightly less time passes for the probe.

You are right, this is a conceptual question. Try to think about the energy at different points in the motion. What types of energy will be present when the bowl is at the very top of its swing? What happens to this energy as the ball progresses?

Looks like you might have made an arithmetic error when calculating in cents. 1017.864*0.099 = 100.768

v = \frac{2{\pi}r}{T} Should help you in the second part if you equate it to another equation with known variables.

Are you sure the values of time and distance are accurate? Given those two masses g should be equal to 3a.

If the velocity is perpendicular to the magnetic field you can find the radius of the arc by mv / |q|B Not sure if this is the right approach but it might help.

I think Twilight might scoop a few Oscars.

This was something I found helpful and thought others might want to know. I'm new to the whole LaTeX thing so I apologise if the topic is redundant.

OOoLatex Extension for Open Office The extension can be installed in Open Office by way of Tools -> Extension Manager -> Add. You will also need the following dependencies to carry out the behind the scenes LaTeX makery. Dependencies: MikText Ghostscript MinSYS It is probably best to...

The towel trick is only temporary and purposely causing it to overheat multiple times will eventually damage other components on the motherboard and ruin your 360 for good. Opening up the console and having a whack at the X-Clamp replacement is the only fix that addresses the root of the problem.

Sounds like you're talking about a Railgun

Have you played around with the equations of motion? (SUVAT)

Ah that was a mistake, the x values should be the other way around. So would \int_{\pi/8}^{\pi/4}cot^2(2x)dx= \int_{\pi/8}^{\pi/4}\frac{cos^2(2x)}{sin^2(2x)}dx integrate to \frac{\frac{1}{2}sin^2(2x)}{\frac{-1}{2}cos^2(2x)} ? Which I assume would simplify to -tan^2_(2x)

I have the question Evaluate \int_{Pi/4}^{Pi/8}_cot^2{2x}dx So integrating this should (I hope) give [\frac{-1}{2}cot{2x}-x] for those limits. But I have never evaluated the cot integral before, I know that cotx=1/tanx. Do I substitute this identity in and work from there?