Recent content by wolfspirit

Hi, Thanks for that! I was using the wrong equation, i should have used $$ \frac{a X b}{|aXB|} $$

ah ok i thought one could enter latex code here...

Homework Statement Find a unit normal to the plane containing a and b. a=(3;-1; 2) and b=(1; 3;-2) Homework Equations \frac{V}{|V|} where V = a+b The Attempt at a Solution (3;-1; 2) +1; 3;-2)=(4,2,0)=V |V| = \sqrt{4^2+2^} = 2\sqrt{5} there fore the unit vector is (\frac{2}{\sqrt{5}}...

Thanks :)

I don't what the force is that's my point he can't stand on the moon since that is the pivot point, since we do not know where he is standing how can we know the force that he can exert on the lever to try and move the Earth..

Homework Statement given a lever and a place to stand, Archimedes claimed to be able to move the earth. amusing he could use the moon as a privet point, how long would the lever have to be? (the Earth moon distance is 382000km Homework Equations toque = (force)(lever arm) toque =I cross alpha...

ah brilliant much clearer thanks! :)

With the x/x you can use l'hopitals rule to get round the fact that you have somepthing that is tending to 0/0 but for the tanh(x)/x part you cant? so how do you evaluate it? and I am a bit confused as to how Because i thought ((e^x-e^-x)/e^x+e^-x))/x was the same as: ((e^x-e^-x)/e^x+e^-x))...

Hi, I'm trying to find the limit as x tends to zero of the function (tanhx-x)/x this is what i have but i have no idea if i am on the right lines? lim x-> 0 can be split up into two problems: limx->0 (tanhx)/x - limx->0 x/x limx->0 x/x = 1 limx->0 (tanhx)/x can be expressed as limx->0...

I keep getting the wrong answer when i try to differentiate cotx.. this is what i get: cotx = 1/tanx =cosx/sinx=cosx ⋅ sin^-1 so by the product and chain rule we have: sinx⋅(sin x)^-1+cos⋅(-1sin^2 x)^-1 ⋅(cosx)^-1 = sinx/sinx - cosx/cosx ⋅ sin^2x =1-1/sin^2 x where as the correct answer...

Many thanks for your reply! I have re read the book and it makes a lot more sense now however this is a few things I'm still not 100% sure about. so just to get this straight in my mind: to quote the book; "hadrons interact through strong interaction and through the electromagnetic...

Hi, I have been getting a little confused; the textbook (and from what i can see on-line) are saying that Hadrons only interact through the strong interaction (along with the electromagnetic force if charged Q) and yet K Mesons decay through the weak interaction. If my understanding is correct...

i did have a little look on the net but i came up empty thans Ryan

I have been thinking, weather balloons can go quite high (some up to 1000ft) so would it be possible to take 4 or 5 weather balloons and attach them to the corners of a frame. in the centre of the frame would be a rocket ready to launch at a set altitude. the only problem i can think of is...

thank you for your help :) it seams i have a bit of reading to do before i get back to the really interesting stuff