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OCR Physics A G485 - Frontiers of Physics - 18th June 2015

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I am so ****ed for this exam.
Reply 161
Original post by Smug Life
I am so ****ed for this exam.


Same, more so for G482 and maths C4 :frown:

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We have a whole month left till this exam, it's a glorified memory test. Keep calm and revise and you'll all be fine.
Original post by randlemcmurphy
1) If you have the official A2 guide, I would just remember everything about mass spectrometry the book offers. Alternatively, this may be useful.

2) I have come across the same issues as you. I know the exact question you are talking about, it is about a students suggestion. I think I have the reason my teacher gave me as to why in my notes. I will try and find it for you.

Any luck with #2?
could you please tell me the link of june 2014 physics g485 paper?
Do we need that r=mE/QBB equation for the exam ?
(edited 8 years ago)
Reply 166
Original post by TheGeneral95
Do we need that r=mE/QBB equation for the exam ?


It's not necessary to memorise that because there are marks for deriving it:

In a magnetic field the force due to the field (F=Bqv) causes the centripetal force (F=mv^2/r)

Equating the two and cancelling gives Bqr = mv

Therefore r = mv/Bq
Reply 167
Original post by Mr_Cupcakes
We have a whole month left till this exam, it's a glorified memory test. Keep calm and revise and you'll all be fine.


If you don't have 12 papers prior to it that is :/
Reply 168
Could someone please explain to me why the mass of a nucleus is less than the mass of it constituent nucleons?
Original post by Tiwa
Could someone please explain to me why the mass of a nucleus is less than the mass of it constituent nucleons?


To split the nucleus into its constituent nucleons, energy is required to overcome the strong nuclear force between its nucleons. As a result this energy is the work done on the constituent nucleons. Since energy and mass are equivalent according to E=mc2, this energy (work done), has an equivalent amount of mass that is added to the mass of the constituent nucleons because work was done on the nucleons.


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Reply 170
Original post by Mehrdad jafari
To split the nucleus into its constituent nucleons, energy is required to overcome the strong nuclear force between its nucleons. As a result this energy is the work done on the constituent nucleons. Since energy and mass are equivalent according to E=mc2, this energy (work done), has an equivalent amount of mass that is added to the mass of the constituent nucleons because work was done on the nucleons.


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Thank you!
Original post by MO2898
If you don't have 12 papers prior to it that is :/
Hahaha touché, who knows maybe my post was me just trying to convince myself that I'll be okay :redface:

By the way I've been finding these power points useful as additional information to the Physics 2 textbook my school gave us, might help someone else out too: http://www.st-ambrosecollege.org.uk/1273/academic/science---physics/physics-a2-level/
Reply 172
Original post by Mr_Cupcakes
Hahaha touché, who knows maybe my post was me just trying to convince myself that I'll be okay :redface:

By the way I've been finding these power points useful as additional information to the Physics 2 textbook my school gave us, might help someone else out too: http://www.st-ambrosecollege.org.uk/1273/academic/science---physics/physics-a2-level/


If it works, why not!
When the coil is parallel to the field how on earth is theta 90 degrees ? What have I missed :erm:
Reply 174
Original post by TheGeneral95
When the coil is parallel to the field how on earth is theta 90 degrees ? What have I missed :erm:


theta is from the normal of the coil

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Original post by Elcor
theta is from the normal of the coil

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Someone can correct me if I'm wrong because I'm not sure, but EMF= the rate of change of magnetic flux linkage, the rate of change of is greatest when the coil is parallel to the field.
(edited 8 years ago)
Reply 176
Original post by Mr_Cupcakes
Someone can correct me if I'm wrong because I'm not sure, but EMF= the rate of change of magnetic flux linkage, the rate of change of is greatest when the coil is parallel to the field.


Sorry ignore my previous reply, that's correct. If you imagine a graph of flux linkage against time (sinosudal wave) the gradient will be maximum when flux linkage is zero I.e. when cosθ=0 i.e when coil and field are parallel
(edited 8 years ago)
Anyone got the 2014 paper?
Reply 178
hey, can anyone send me the g484 and g485 2014 papers and mark schemes, please? I'm self teaching so I have no access to these papers :frown: your help would be much appreciated! hows everyone feeling?
Original post by pnaidu
hey, can anyone send me the g484 and g485 2014 papers and mark schemes, please? I'm self teaching so I have no access to these papers :frown: your help would be much appreciated! hows everyone feeling?


They were uploaded on the second page of this thread, post 37.


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