AQA Physics Unit 1 PHYA1 20th May 2014 OFFICIAL Watch

AreiMo1997
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#301
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#301
(Original post by Alatair97)
Someone help please?! I did the resistivity question wrong then corrected it at the LAST SECOND. I put 1.1x10-7 or something... was this correct?
yes. Thats the only question on electicity i gto right
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Gardnerr
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#302
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#302
(Original post by CharleyJay)
What did people put for the power question about when internal resistance is negligible, what would change?
I put that the current would increase, and because P=I^2R, power dissipated would be really high for 0.5 ohms, then become lower as the resistance increased up to 6.5 ohms.
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Randomer96
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#303
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(Original post by Ryejd)
You can say electrostatic, not sure if just ''photons'' will suffice though?
I saw on a markscheme it said (virtual) photons, which implied to me virtual wasn't necessary.
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cudders96
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#304
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#304
Yeah you can't use P=IV because there is internal resistance, and so we don't know V. So I^2R was the way to go.
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Ryejd
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(Original post by Randomer96)
I saw on a markscheme it said (virtual) photons, which implied to me virtual wasn't necessary.
There you go then
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Razzamataz179
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#306
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(Original post by JackTeh96)
I thought it is strong force?? by gluons??
Electromagnetic. The protons repel eachother in the nucleus (by the exchange of virtual photons) but because the strong nuclear force is so much stronger than the electromagnetic force (within a range), the nucleus isn't unstable.
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deadmau_5
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(Original post by Razzamataz179)
Because the experiment is to plot a graph of voltage against tempereature. So you put the circuit in a room, take a reading of the voltage, then using your graph, find the temperature.
We're talking about different things here. I'm talking about q.5 part a)
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ChordaeTendineae
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#308
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(Original post by CharleyJay)
What did people put for the power question about when internal resistance is negligible, what would change?
no lost volts, terminal pd=emf and i said there less resistance so current increases, so using P=IV power increases and i linked it to the graph
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Ryejd
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(Original post by Razzamataz179)
Because the experiment is to plot a graph of voltage against tempereature. So you put the circuit in a room, take a reading of the voltage, then using your graph, find the temperature.
You're talking about 2 different questions
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Me123456789
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(Original post by Randomer96)
I went for electrostatic and photons.
I did weak interaction and w boson... Is that still right?
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Ryejd
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(Original post by Me123456789)
I did weak interaction and w boson... Is that still right?
Nope
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ChordaeTendineae
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(Original post by Randomer96)
Ffs I crossed this out. I was hoping for a mark for the general statement that the power would still increase
ugh, I made a general statement :/
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LxH
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#313
For interaction between protons I said repulsion, virtual photons
For particle produce in the decay I said neutrino (conserves charge and number of mesons)
Got 17 Ohms for internal resistance
Missed out question on speed of muons
What the **** happened to me in that exam? I'm ****ing ruined. No way I'm getting into Leeds now having handed in this absolute **** stain of a paper. I can ****ing forget chemical engineering. A resit is not optional. I got a C in the ****ing ISA aswell. I can't scrape this back with mechanics either. Highest I'll get this year is a B and that's insufficient as ****.
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RowanChat32
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(Original post by team-punishment)
resistance would be proportional to power dissipated, i.e straight line
This is incorrect, power would decrease as resistance increases, and it would not be linear.
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JPeters
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#315
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#315
Did anybody get approx. 4.32eV after converting from Joules? I might have got the previous question wrong.
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TheSmartOne
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#316
That paper did not go well for me. I didn't know whether to put 'weak' with 'W+' or 'electromagnetic' with 'photon', so I went for weak! FGS ajhgaoifbnehobn

It was so hard. I was getting A's and the occasional B on past papers as well, I don't know what happened
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Rohan97
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(Original post by cudders96)
Yeah you can't use P=IV because there is internal resistance, and so we don't know V. So I^2R was the way to go.
You can use P=IV, since there's no internal resistance, it means that the voltage is no longer shared, therefore the voltage across the other resistor would be greater...

You could also mention that the circuit current increases so the power increases...
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jonathanyyt
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#318
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Thankfully I do Electronics for A-level, so to have a 6-marker on that was really helpful to me.

I got internal resistance as 3 ohms, which I think is correct because in a circuit like this the power is at is maximum when internal resistance equals external resistance, the graph supports this and the calculations as well.

If internal resistance was negligible, the power should decrease as the resistance goes up. Because without internal resistance, the voltage across the resistor is always the same. So since V=IR, as R increases, I decreases. And since P=IV, as I decreases and V stays the same, P decreases.

Think of it this way, connect a 10K resistor to a stabilised power supply of 9V you wouldn't feel a thing. Put your finger on a 1 ohm resistor (small power rating) connected to 9V after a while and it'll feel hot and smoke will come out, cos' there's more power.

For the interaction particle, it should be virtual photon, but I had a brain freeze and put gamma ray instead, hope they'll accept that in the mark scheme.
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sneakbo2
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(Original post by TheRAG)
For the 6 marker, did anyone else say that to measure room temp they had to plot a graph, measure pd without a water bath, then read off the graph?
Yep from the graph you would plot in the previous 6 marker, that's definitely the only liable answer. The rest would just be assuming, I think that is definitely correct
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Me123456789
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#320
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#320
For question 4 did you put the photons are in UV?
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