AQA Physics Unit 1 PHYA1 20th May 2014 OFFICIAL Watch

hltorrance
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#141
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#141
(Original post by Monkey246)
How do you know which force an interaction involves; strong or weak???
eg. from a past exam paper (june 13 unit1)
X → π– + p you were meant to know that this was the weak interaction involved in this decay. How???
Because strange particles ie the pion are formed through weak interaction i think
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Maricamac
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#142
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#142
(Original post by hltorrance)
Because strange particles ie the pion are formed through weak interaction i think
Most of the time it is a weak interaction, the only time you will ever interact by strange interactive force is when strangeness IS conserved (there has to be a value for strangeness though)
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corsilo96
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#143
An a is around 75% and full ums is around 64-68 out of 70
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particlestudent
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#144
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(Original post by hltorrance)
Because strange particles ie the pion are formed through weak interaction i think
No, strange particles can only feel the strong force, not the weak interaction
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hltorrance
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(Original post by particlestudent)
Don't worry about memorising all the quark structures for Kaons and Pions. Just memorise K+ and K- will simply be the opposite
You can work out quark structures using the data on the formula structure there is not need to memorise them
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__Adam__
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#146
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(Original post by Maricamac)
can I just ask how do we work out specific charge, I always get it wrong.
Usually i put electron charge x no of electrons over the (electron mass x electron no) x (neutron mass x neutron number)

Specific charge is Overall charge of nucleus or ion in coulombs divided by the Total mass of the nucleus or ion in kilograms.

Units C/kg
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hltorrance
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(Original post by aLeXaNdRa08)
Oh , so how would you get E=VIt with those equations?
E=VQ. Q=IT so E=VIT
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Boop.
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#148
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#148
Does anyone know how to do 2bii?

PA01 JUNE 2005

http://www.egsphysics.co.uk/files/a_...W-QP-JUN05.PDF

http://www.egsphysics.co.uk/files/a_...W-MS-JUN05.PDF
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rhiam
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#149
(Original post by Monkey246)
How do you know which force an interaction involves; strong or weak???
eg. from a past exam paper (june 13 unit1)
X → π– + p you were meant to know that this was the weak interaction involved in this decay. How???
the weak interaction changes quark type
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dathtom
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(Original post by particlestudent)
No, strange particles can only feel the strong force, not the weak interaction
No, strangeness is only conserved by strong interaction, all decays are weak.

Unless the mark scheme, textbook and AQA are wrong, which is likely but for the exam I'd go with what they are saying
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Kieranw5
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#151
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#151
(Original post by hltorrance)
Because strange particles ie the pion are formed through weak interaction i think
Whenever it's a decay it's a weak interaction
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jjpneed1
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(Original post by dathtom)
It's asking you for the specific charge.
read his post again

I don't really understand, they say '..for B and C respectively' yet B should have the higher wavelength because a lower energy photon is needed to go from B->A compared to C->A. But it doesn't really affect anything because, upon realising this, you can swap the answers round.

Do 2 x 10^-18 - (energy of photon), where the energy of the photon is what you worked out in part i. Then give the one that comes out as 1.5 x ... to B, other one to C.
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Usey11
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#153
If youve done part bi you know the energy of the photons.
Than using that and the equation for energy levels (hf =E1-E2) you can find them out, E2 = -2*10^-18, so use your energy of photons and that in the rearranged formula (hf + E2 = E1) to find it out. So its your hf -2*10^-18.
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simon105
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#154
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I can't find a decent summary of variable resistors anywhere...

So if you change the resistance with the variable resister, what does it affect - the current or the voltage? Or both?
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jjpneed1
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(Original post by simon105)
I can't find a decent summary of variable resistors anywhere...

So if you change the resistance with the variable resister, what does it affect - the current or the voltage? Or both?
It changes the current and hence changes the voltage of some other component.

I suppose you could argue that the potential difference across the other component because the potential difference changes across the variable resistor (and total pd is constant in series), but again current still decreases because of higher resistance
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Fifa97
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#156
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#156
Have diodes ever come up before? I have a feeling it might..
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simon105
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(Original post by jjpneed1)
It changes the current and hence changes the voltage of some other component.

I suppose you could argue that the potential difference across the other component because the potential difference changes across the variable resistor (and total pd is constant in series), but again current still decreases because of higher resistance
So if you increase the resistance of the variable resistor, it decreases the current flowing in the circuit, and the decrease in current decreases the p.d. of other components due to V=IR? Seems to make sense
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jjpneed1
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(Original post by simon105)
So if you increase the resistance of the variable resistor, it decreases the current flowing in the circuit, and the decrease in current decreases the p.d. of other components due to V=IR? Seems to make sense
That's right, assuming their resistance is constant
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GullyyXD
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#159
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#159
it isn't conserved. muon+ is an anti particle so the lepton number is -1 and the antimuon neutrino is also an anti particle therefore -1 lepton no. K+ isnt a leptop so the lepton no. is 0 meaning it isnt conserved. 0 ---> -1-1
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toddmcnugget
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#160
Does anyone have a list of the experiments we're supposed to know such as the resistivity one?
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