Recent content by PonderingMick

Can someone please tell me how to calculate this? My notes are such a mess. Is it as simple as Q = (Mx-My)c2 using atomic masses? Cheers, Mick

Just checked the answer and its actually 17.8 ms-1, will try and work through it again later.

t = (-66√662-4 * 4.9 * -15) /9.8 So I use -4.9 instead t = -66 + or - ( √662 -4 * -4.9 * -15) /9.8 t = -66 + or - ( √4356 - 294) /9.8 t = -66 + or - (66-294) /9.8 t = -66 + or - -23 t = -66+-23 = -89 or -43 t must be +ve?

Yes, the answer is in my textbook

Homework Statement Motorbike crossing a river using a ramp. River is 40m wide, the bank on the opposite side is 15m lower. The ramp angle is 53 degress. What is the speed needed at take off? Homework Equations I am using s = ut for the horizontal and s = ut x 1/2 at2 I think because the bank...

Can someone please tell me what applied uses surds have in Science? Apart from Pythagoras? Cheers, Mick

Help, How do I get X a/b = y/x I get to: x = y / (a/b) but this seems silly? Cheers, Mick

Ok, I get it now: \frac{1}{kg^{-1}} = kg for the same reason that 1 10-1 = 10 Thanks

Yes that's right, this means that my calculation is correct because the answer is a mass. But I still don't understand why the answer to this is kg: \frac{1}{kg^{-1}} = kg

OK, I must be wrong before I get that far. I have: m x (ms-1)2 N m2 kg-2 So on the top I get: m x m2s-2 and on the bottom I get m kg s-2 m2 kg-2 all the m and the s cancel so i just get the kg left

Can someone please explain how kg x kg-2 (in superscript) can equal kg? I hope it does or my calc is wrong! Anywhere I can revise this kind of thing? Mick