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

Smug Life
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#161
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#161
I am so ****ed for this exam.
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MO2898
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#162
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#162
(Original post by Smug Life)
I am so ****ed for this exam.
Same, more so for G482 and maths C4

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Mr_Cupcakes
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#163
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#163
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.
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BecauseFP
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#164
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#164
(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?
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q1147996060
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#165
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#165
could you please tell me the link of june 2014 physics g485 paper?
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TheGeneral95
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#166
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#166
Do we need that r=mE/QBB equation for the exam ?
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MO2898
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#167
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(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
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MO2898
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#168
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#168
(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 :/
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Tiwa
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#169
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#169
Could someone please explain to me why the mass of a nucleus is less than the mass of it constituent nucleons?
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Absent Agent
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#170
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(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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Tiwa
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(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!
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Mr_Cupcakes
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#172
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(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

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/...sics-a2-level/
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MO2898
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#173
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#173
(Original post by Mr_Cupcakes)
Hahaha touché, who knows maybe my post was me just trying to convince myself that I'll be okay

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/...sics-a2-level/
If it works, why not!
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TheGeneral95
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#174
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#174
When the coil is parallel to the field how on earth is theta 90 degrees ? What have I missed :erm:
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Elcor
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(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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Mr_Cupcakes
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#176
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#176
(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.
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MO2898
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#177
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(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
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Theman69
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#178
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#178
Anyone got the 2014 paper?
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pnaidu
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#179
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#179
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 your help would be much appreciated! hows everyone feeling?
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Absent Agent
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#180
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(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 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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