OCR Physics B G495 Field and Particle Pictures June 21st 2011 Exam Thread

is it actually worth while doing any of the section C of the past g495 papers? i dont know how it will benefit doing the section C of this one

Reply 181

readyforresults

Original post by Rogercbinboy

Can someone confirm this order with transformers:

The alternating current in the primary coil creates a changing flux in the iron core, this travels through the secondary core and induces an alternating voltage in the secondary coil.

Then with the eddy currents:

The alternating current in the primary coil induces a changing flux in the iron core, this creates an emf in the core since it's a good conductor of electricity and that causes currents in the core that create their own flux, and according to Lenz' law these flux lines oppose the original change of flux decreasing the overall flux.

That sound right/in the right order?

!!

Reply 182

Helsy

12

Original post by Summerdays

Don't worry, it's because I have done ALL of the past papers. How many have you done so far?

Umm.... about 8. Papers from 2004 to 2008, and then in the next couple of days I'm going to do the rest up until the present day. I left the 2001 papers etc for when I run out of newer ones to do.

I'm still pretty much screwed. :frown:

Reply 183

Rogercbinboy

11

Original post by a93

is it actually worth while doing any of the section C of the past g495 papers? i dont know how it will benefit doing the section C of this one

I haven't bothered myself, though I guess you could find out what sort of questions they ask spending time reading around our section C would probably be more useful.

EDIT: Does anyone has a clear reason why the earth bulges at the equator by the way?
I think it's something to do with the centripetal acceleration using up some of the energy of the gravitational acceleration or something but I'm not sure how to put it into words for an exam question.

(edited 12 years ago)

Reply 184

Mikkels88

yeh my teachers explanation on those notice article answers isnt exactly easy to follow... my understanding was that for the circular motion you need a resultant force to the centre of the earth, and he explains it using pressure (didnt really get this bit) where by the earth exerts an outward pressure on an element of its volume, and so to maintain a resultant force to the centre you gotta move that element (just a point some distance from the centre) further out, so that the outward pressure force is decreased and the centripetal force is larger than it, giving a resultant and circular motion.

Do not quote me on that, im not really sure what im saying :wink:

Reply 185

Ollie901

1

I'd just say that the outsides of the earth (in the direction of the spin) have the biggest linear velocity so will have a greater circular acceleration towards the centre of the earth (v=romega, a=v^2/r) and will therefore feel a bigger centrefugal force, hence the bulge.

(edited 12 years ago)

Reply 186

Rogercbinboy

11

Cheers you two, gotta say I get the 2nd explanation a lot more though.

My teacher would have an aneurysm if he caught any of us using the word 'centrifugal' but gotta admit that it makes the definition so much easier.

Reply 187

mashman9

Hi guys really not looking forward to this exam...

Ive got some mock questions for section C that our teacher gave us...

"the distance from the north pole to the south pole is 20014km, how long is one nautical mile?"

when it says the distance from north to south does it mean along a meridan or directvly through the earth like the diameter?

And can someone explain one minute of arc?

Thanks :smile:

Reply 188

Rogercbinboy

11

Original post by mashman9

Hi guys really not looking forward to this exam...

Ive got some mock questions for section C that our teacher gave us...

"the distance from the north pole to the south pole is 20014km, how long is one nautical mile?"

when it says the distance from north to south does it mean along a meridan or directvly through the earth like the diameter?

And can someone explain one minute of arc?

Thanks :smile:

I think it probably means the diameter of the earth through the earth, if so then divide that by 2 to get r and do (pi x radius)/180 or (2pi x radius)/360 (makes no difference but since the meridian is 180 degrees might get kudos for that) so get 1 degree of the meridian.

A minute is a 60th of a degree so divide the answer by 60 to get 1 minute of the arc.

Reply 189

41jms

1

anyone have the feb 2011 markscheme?

Reply 190

41jms

1

oh nevermind, found it :smile:

Reply 191

bear54

Just wondering in questions where you use your answer to a previous question in a calculation, how many of you use the value they give in the question if it was a 'show that'? They won't start penalising you for it will they?

Reply 192

stuffedfigs

Original post by bear54

Just wondering in questions where you use your answer to a previous question in a calculation, how many of you use the value they give in the question if it was a 'show that'? They won't start penalising you for it will they?

My physics teachers always told me to use your value if you believe it is the correct one. I don't think you'll be penalised though.

Reply 193

JoeCarr

Original post by bear54

Just wondering in questions where you use your answer to a previous question in a calculation, how many of you use the value they give in the question if it was a 'show that'? They won't start penalising you for it will they?

Yeh you definitely are not penalised for that, if anything it's safer because you know that is the 'correct' value and although you can still get marks for ecf on the next question, I think unless I am 100% sure because it was a very basic question, that using the given value is safer. On the mark schemes they say 'using such and such value, the answer is...' on questions like that, so overall no they do not penalise you for that. :smile:

Reply 194

41jms

1

http://www.johnbright.conwy.sch.uk/depts/science/Content/Past_Papers/Y13/Physics/G495Jan11.pdf

could someone explain 10c to me please ? :smile:

Reply 195

JoeCarr

Original post by 41jms

http://www.johnbright.conwy.sch.uk/depts/science/Content/Past_Papers/Y13/Physics/G495Jan11.pdf

could someone explain 10c to me please ? :smile:

I can try, but I can't guarantee i'll do a good job. Also, at the risk of sounding presumptuous, since this 'nucler/atomic ratio' came up in january this year, I don't think it will be in this paper, especially since it is rarely if at all found in any of the other past papers.

Right, basically the nucleus is obviously a component of the atom, however the density of 'gold' in this example is for the whole atom. The ratio is as it says on the paper, of the radii of the nucleus over the whole atom. The nucleus is 6*10^-5 times smaller radially speaking. Now to work out density, we need volume, and so this ratio of radii is cubed to get volume (4/3pi is irrelevant as we have a ratio so they cancel). We now have a ratio of volumes of 2.16*10^-13. This says the volume of the nucleus is 2.16*10^-13 times smaller, so we can either do 1/2.16*10^-13 as a density ratio, multiplied then by our original density (as volume to density is an inverse m/v relationship), or we can simple divide the density value by 2.16*10^-13. :smile:

This assumes that all the mass is found in the nucleus, as otherwise we couldn't simply use the given density value times by 1/2.16*10^-13 as not all our mass would be in the nucleus.

Sorry it is all a bit wordy, hope it helped.

(edited 12 years ago)

Reply 196

jimmeh

Original post by 41jms

http://www.johnbright.conwy.sch.uk/depts/science/Content/Past_Papers/Y13/Physics/G495Jan11.pdf

could someone explain 10c to me please ? :smile:

Volume is proportional to

r^3

(Don't worry about the shape of a sphere and the whole 4/3 pi, as it all cancels out.

Sooo, cube the radius's (radii?!) and the ratio, and you get a ratio of volumes.

Then divide the density of a whole gold atom (which you're given) by that ratio to get the density of a gold nucleus.

And for the assumption I'd put that there's no gaps between the gold nuclei...

It's an absolute bastard of a question though...

Edit: JoeCarr (above me) has done a far better job of explaining! Read that one :tongue:

(edited 12 years ago)

Reply 197

Ceri25

Can somebody please help me with uncertainties? I really get confused with absolute and percentage and whatever. Are there set formaulas that can be learnt for them?
Any examples using the pre release would be super appreciated as well!! I tried some questions from the earlier posted section c questions and I was unable to do all the uncertainty questions :frown:

Reply 198

41jms

1

Original post by JoeCarr

I can try, but I can't guarantee i'll do a good job. Also, at the risk of sounding presumptuous, since this 'nucler/atomic ratio' came up in january this year, I don't think it will be in this paper, especially since it is rarely if at all found in any of the other past papers.

Right, basically the nucleus is obviously a component of the atom, however the density of 'gold' in this example is for the whole atom. The ratio is as it says on the paper, of the radii of the nucleus over the whole atom. The nucleus is 6*10^-5 times smaller radially speaking. Now to work out density, we need volume, and so this ratio of radii is cubed to get volume (4/3pi is irrelevant as we have a ratio so they cancel). We now have a ratio of volumes of 2.16*10^-13. This says the volume of the nucleus is 2.16*10^-13 times smaller, so we can either do 1/2.16*10^-13 as a density ratio, multiplied then by our original density (as volume to density is an inverse m/v relationship), or we can simple divide the density value by 2.16*10^-13. :smile:

This assumes that all the mass is found in the nucleus, as otherwise we couldn't simply use the given density value times by 1/2.16*10^-13 as not all our mass would be in the nucleus.

Sorry it is all a bit wordy, hope it helped.