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

L'Evil Fish
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(Original post by whyalwaysme??)
Oh wow thanks a lot man, makes a whole lot more sense now.
Lol was that sarcasm?
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whyalwaysme??
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(Original post by L'Evil Fish)
Lol was that sarcasm?
Loll no man.
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BrokenS0ulz
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Just got myself confused, does the strong nuclear force actually act between a neutron and a neutron and a neutron and a proton? They would have no electrostatic repulsion so surely there is no need for the strong nuclear force. Does it not only act between protons due to their repulsion?
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sagar448
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(Original post by BrokenS0ulz)
Just got myself confused, does the strong nuclear force actually act between a neutron and a neutron and a neutron and a proton? They would have no electrostatic repulsion so surely there is no need for the strong nuclear force. Does it not only act between protons due to their repulsion?
You are correct but if we go into detail the strong interaction force isn't really "just a force" it's actually exchange of mesons. A meson is a subatomic particle which is intermediate in mass between an electron and a proton and transmits the strong interaction that binds nucleons together in the atomic nucleus. So between a neutron and a neutron the meson will be transmitting the strong force by exchange.



EDIT: but as L'Evil Fish said, it also acts between quarks. :3
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BrokenS0ulz
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(Original post by sagar448)
You are correct but if we go into detail the strong interaction force isn't really "just a force" it's actually exchange of mesons. A meson is a subatomic particle which is intermediate in mass between an electron and a proton and transmits the strong interaction that binds nucleons together in the atomic nucleus. So between a neutron and a neutron the meson will be transmitting the strong force by exchange.



EDIT: but as L'Evil Fish said, it also acts between quarks. :3
Thanks
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sagar448
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(Original post by BrokenS0ulz)
Thanks
No problem but I wouldn't recommend using this in the exam because the materials just teach us that the strong nuclear force is really "just a force". :3
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BecauseFP
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(Original post by sagar448)
Photoelectric effect has an electron emission there was no electron being emitted (I think we are talking about the same question) e.e
Great, it looks like I'll be reposting this again.
(Original post by BecauseFP)
I've got two things to ask, sadly one is a repost since this part of the forum seems quite dead.

1) "One of the points on the spec for G485 is this:
"explain the use of deflection of charged particles in the magnetic and electric fields of a mass spectrometer"
And I've been wondering, have they ever actually asked this? I can't find it on any of the current papers but I'm unsure as to the legacy, does anyone know what they can ask or what sort of answers they expect? I've heard that a very similar thing appears somewhere in OCR A2 Chemistry but I don't know if that would transfer to Physics."

2) Can someone explain the apparent discrepancy in the mark schemes here (sorry about the weird quality of this, I had to squeeze it all onto something I could print because I asked a guy about everything on here a couple of months ago)? One of them talks about photons releasing electrons while all the other make no mention of electrons, annoyingly I'd also argue that photons causing electrons implies the photoelectric effect which another of the papers has explicitly mentioned as being wrong.
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sagar448
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(Original post by BecauseFP)
Great, it looks like I'll be reposting this again.
What?! I thought we already went through this....
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Hilton184
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(Original post by BecauseFP)
Great, it looks like I'll be reposting this again.
We went through all of this.

Use the more recent mark scheme. Include point about phosphor screen though. Examiner that said Image Intensifier doesn't use photoelectric effect is no longer around after being laughed out the office by his fellow colleagues, as clearly an image intensifier is making use of the photoelectric effect, as the first mark scheme you have posted demonstrates with the fact electrons are emitted by incident visible light photons on the photocathode.
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Elcor
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(Original post by sagar448)
You are correct but if we go into detail the strong interaction force isn't really "just a force" it's actually exchange of mesons. A meson is a subatomic particle which is intermediate in mass between an electron and a proton and transmits the strong interaction that binds nucleons together in the atomic nucleus. So between a neutron and a neutron the meson will be transmitting the strong force by exchange.



EDIT: but as L'Evil Fish said, it also acts between quarks. :3
You sure mesons is the correct word? They're made of a quark and an antiquark (so they're like similar in size to nucleons - doubt there's space for them to be whizzing around in one).

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sagar448
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(Original post by Elcor)
You sure mesons is the correct word? They're made of a quark and an antiquark (so they're like similar in size to nucleons - doubt there's space for them to be whizzing around in one).

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Quite sure I've also researched it a bit. Heres a link if you want to read more about it --> http://aether.lbl.gov/elements/stell...ng/strong.html
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BrokenS0ulz
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Could somebody have a look at 4cii on this paper
http://freeexampapers.com/A-Level/Ph...tion_Paper.pdf

The answer is 5 volts, I thought things in parallel had the same pd though, so why isn't it 5000V?
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[email protected]
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(Original post by BrokenS0ulz)
Could somebody have a look at 4cii on this paper
http://freeexampapers.com/A-Level/Ph...tion_Paper.pdf

The answer is 5 volts, I thought things in parallel had the same pd though, so why isn't it 5000V?
cant view question maybe upload a screen shot
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sagar448
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(Original post by BrokenS0ulz)
Could somebody have a look at 4cii on this paper
http://freeexampapers.com/A-Level/Ph...tion_Paper.pdf

The answer is 5 volts, I thought things in parallel had the same pd though, so why isn't it 5000V?
Ahh no you are getting mixed up, the RESISTORS in parallel have the same pd but with capacitors it's different. The capacitors in series have the same CHARGE and capacitors in parallel have the same PD. Since the questions states that the ball has capacitance you treat it as a capacitor. So you calculate the charge on the ball by multiplying the capacitance with 5000V. This charge that is resulted is the charge on the ball AND the capacitor. Now you divide this charge by the capacitance of the capacitor and it gives you 5V.

EDIT: I made little oopsies. :3

I genuinely think that question is wrong because it's connected in series while the question says it's in parallel. But if you try to work it out using series it gives the correct answer.
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BrokenS0ulz
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(Original post by [email protected])
cant view question maybe upload a screen shot
Spoiler:
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I've attached the whole question because I think you may need info from other parts of the question.


(Original post by sagar448)
Ahh no you are getting mixed up, the RESISTORS in parallel have the same pd but with capacitors it's different. The capacitors in parallel have the same CHARGE and capacitors in series have the same PD. Since the questions states that the ball has capacitance you treat it as a capacitor. So you calculate the charge on the ball by multiplying the capacitance with 5000V. This charge that is resulted is the charge on the ball AND the capacitor. Now you divide this charge by the capacitance of the capacitor and it gives you 5V.
Thanks for replying, but I'm pretty sure that capacitors in parallel do have the same p.d and that its capacitors in series that have the same charge.
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BrokenS0ulz
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(Original post by [email protected])
cant view question maybe upload a screen shot
oops won't let me edit the right picture in, here it is
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sagar448
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(Original post by BrokenS0ulz)

Thanks for replying, but I'm pretty sure that capacitors in parallel do have the same p.d and that its capacitors in series that have the same charge.
Oh sorry yes I corrected my self. In EDIT above in my post. Not thinking straight. Anyways I did the question again and it works fine when you assume that the ball is connected in series. I get the answer 5V. But yes you are right it should be 5000V if it's connected in parallel. Since these are old questions maybe it's a mistake??
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Shabz12357876877
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Hi does anyone have the 2014 g485 mark scheme please? ( on that note I also need the g482 mark scheme too lol) thanks a lot
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[email protected]
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Can anyone explain how June 2012 Question 3 Aii works?
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Tiwa
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(Original post by [email protected])
Can anyone explain how June 2012 Question 3 Aii works?
 \phi = BAN , hence you need to pick a certain value of the flux linkage. The value you need to you is the one with the amplitude, hence  2.0\times 10^-^2 Wb-turns . Then just rearrange the equation to find the flux density. The question already gives the number of turns and the cross-sectional area.
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