LCMATHS wrote: » I was going to say air resistance, but isn't air resistance just a type of friction?
AtomicKoala wrote: » Ok, now that everyone has said the paper was easy, did anyone else think it was actually quite.. difficult?
FatRat wrote: » I didn't even do the mechanics question but I always though that to get max height all you need to do was find the height when the velocity = 0 ?
Solid_Shepard wrote: » Question three was standard, except for that absolutely bizarre final part (which I certainly got wrong), where it was asking you to talk about graphing/getting the slope (as the y-intercept = -.03d) without actually asking about it directly that.
FatRat wrote: » What's this? Are you sure that's the answer? For that part I took the difference in length between the first resonance and the second resonance. This is the same length as an anti-node to the next anti-node i.e. half a wavelength. Therefore multiply the length by two and you have your wavelength. Then use the formula c= f x lambda ??
FatRat wrote: » Anyone know if gravity was a correct force in the last part in q1? I said gravity and I levelled the air track to avoid the force.
Overheal wrote: » Technically gravity is not the force, gravity is an acceleration. Weight is the force of mass accelerated by gravity, F=ma
2thousand14 wrote: » Isn't gravity one of the fundamental forces of nature?
(iv) Name the type of acceleration that the ISS experiences as it travels in a circular orbit around the earth. centripetal / gravitational What force provides this acceleration? gravitational (do not accept “gravity”)
TheBegotten wrote: » Anyone else think that last part of the particle physics question to be rather sneaky? And the Beta - decay, I don't even think that's on the course!
MegGustaa wrote: » I essentially did the same - S(y) = u(y)t - gt^2, differentiate to get V(y) and put it equal to zero to get time of max height, then sub that back in. The angle was 15 degrees. I had u as 63.6.
CianDowd wrote: » Beta decay is on the course but I have no idea what the last part was. I just went through the log table sticking in pieces of equations to do with what they asked for... :$ XD
EoghanIRL wrote: » All you had to do was let f=qvB equals to mv^2/r Mv is equal to momentum right ? Mass times velocity . Divide across by v Mv= qBr Did anyone else do something like this or am I completely wrong ? Hoping this is ok as I couldn't see a different way to do it.
CianDowd wrote: » I think that you're mostly right but in my answer I used: S=ut+(1/2)at^2 So: S(y)=u(y)t-(g/2)t^2 Whether you meant to type it I'm not sure but I think that g had to be halved.