Recent content by Steven Thomas

That's a really clear and coherent answer thank you. I think I now know how I'll proceed. I'd like to quote it almost in my lab book as a rational for my choice, did you get that from any particular book or are you a lecturer or anything that I can put down?

Let's assume they are touching. We call the radius of the spheres r and the separation between the two spheres (centre to centre) R. When they touch we have R = 2r. You can substitute this into the equation for the force of gravity, and substitute for the mass in terms of density and volume...

Hi @RUber Thanks for your reply. Ihave tried your method and it gives me much better error percentages on the coefficients and it also reduces the SSE and RMSE and moves the Rsquare and adjusted Rsquare closer to 1, so that's good. I've since also tried using 1/percentage error squared, where...

@Sunsethorizon If you attempt to solve assuming that they are touching, do you get the correct answer? Otherwise, as you say before, I think you will have two unknowns, both the mass / radius of he lead spheres and their separation.

When it says placed next to each other, does that mean the spheres are touching one another? If so you can rewrite the one radii in terms of the other and you only have one unknown.

Hi all, I'm currently in the middle of performing an experiment for the final project of my MSc, and I have a question about how I should go about weighting the data when fitting a curve to it using the MATLAB fitting tool. Firstly, a bit of background about the problem. I am seeing how low...

Thanks for everyones help, I've managed to sort these and the other ones that I had a problem with because of the help y'all supplied. I'll also try that latex thing (I did wonder why it was asking if I'd like to preview some rubber :p) next time I post. Thanks all.

So then I'm pretty sure that the -1/a term goes to minus infinity and hence my whole integral does. As for my -x.e^-x, do I take it as f(x) is -x and let g(x) = e^x as then I'd have f(x)/g(x)? Differentiating those would give me -1/e^x so it would tend to 0?

My other one was integrating the product of x and e^-x between 0 and infinity. I can do the indefinite integral which gives me: -x.e^-x - e^-x between 0 and ∞, I just can't evaluate the first term when I use the infinity limit because I get negative infinity multiplied by zero...

I got 1/a - 1 As a-> infinity I get negative infinity. Is that the answer? Or is the answer that it doesn't diverge? Also, how do you insert the equation so nicely on this forum? I have the feeling I'm going to be posting a few like this. I do have another one actually if you'd be so kind as...

Sorry that is infinity minus 1, not infinity to the power of minus 1, which in turn equals infinity. Is this an acceptable complete integration, or should I apply L'Hopital's rule?

Thanks for your help here guys but I worked it out and it turned out to be real simple in the end. I rearranged to make V the subject for both and then used the fact that E2 = E1/4 to give me a simultaneous set of equations I could solve real easy (also using V2=V1-2. Thanks anyway.

Homework Statement Integral of ∫1/x^2 (or ∫x^-2) between 1 and 0.The Attempt at a Solution I can integrate it no problem to give me -1/x or x^-1, but when I put it between the limits of 1 and 0 I get ∞-1 which is just ∞. Is this right or do I need to use L'Hopital's rule. If so, how? I'm...

Surely there has to be a better way? Am I able to discard my second root of V=1.5 because 1.5-2 = -0.5 and I can't really have a negative voltage? Steven Thomas.

Hi there, Firstly I took the difference between the starting and finishing vectors to get a movement vector of (10, 0, -9). Then there is the vector of force that Jill exerts (0, 240, 0). Two vectors are perpendicular if their dot product is zero. So in this case the dot product is...