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AQA Physics Unit 1 PHYA1 20th May 2013

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Reply 1060
Avatar for Zakee
Zakee
There was a question which involved two resistors in parallel and one was given to you as 2.0 ohms, and the other (after working out)
was either 4.2 ohms or 4.22 ohms. The next question was 'What is the total resistance of the circuit'.

Is the total resistance just 1/2 + 1/4.22 = 1/R(total) or was there another step involved?
Am I the only one that got 30 000 ohms for the very last question?
Can someone go over the marking points for the whole why do emitted electrons have varied Ek up to a maximum or whatever it was..
I think I got it but just want to check
Reply 1063
Avatar for Surf
Surf
15
Original post by pilotluke1
What was the answer to 2A?


Do you mean the proton one?
A = Neutron
B = W+ boson
C = Positron
D = Electron Neutrino
Reply 1064
Avatar for GeneralOJB
GeneralOJB
OP
Original post by Surf
Do you mean the proton one?
A = Neutron
B = W+ boson
C = Positron
D = Electron Neutrino


A was a d quark, not a neutron.
Reply 1065
Avatar for Jatt.B
Jatt.B
2
I thought this was the easiest paper yet! I just didn't get the last question! I thought the answer is 3500 ohms but everyone I have spoken to is getting a range of different answers
Reply 1066
Avatar for Surf
Surf
15
Original post by Zakee
There was a question which involved two resistors in parallel and one was given to you as 2.0 ohms, and the other (after working out)
was either 4.2 ohms or 4.22 ohms. The next question was 'What is the total resistance of the circuit'.

Is the total resistance just 1/2 + 1/4.22 = 1/R(total) or was there another step involved?


Didn't you have to add the internal resistance of the cell to that value? Not sure - I can't remember the question exactly... Wasn't it about 2.85 ohms or something?
Reply 1067
Avatar for GeneralOJB
GeneralOJB
OP
Original post by Zakee
There was a question which involved two resistors in parallel and one was given to you as 2.0 ohms, and the other (after working out)
was either 4.2 ohms or 4.22 ohms. The next question was 'What is the total resistance of the circuit'.

Is the total resistance just 1/2 + 1/4.22 = 1/R(total) or was there another step involved?


You had to add the internal resistance as well.
Reply 1068
Avatar for amar96
amar96
8
Original post by Surf
Didn't you have to add the internal resistance of the cell to that value? Not sure - I can't remember the question exactly... Wasn't it about 2.85 ohms or something?


you dont include the internal resistance
Original post by BayHarborButcher
Am I the only one that got 30 000 ohms for the very last question?


Got the same answer.
Original post by Zakee
There was a question which involved two resistors in parallel and one was given to you as 2.0 ohms, and the other (after working out)
was either 4.2 ohms or 4.22 ohms. The next question was 'What is the total resistance of the circuit'.

Is the total resistance just 1/2 + 1/4.22 = 1/R(total) or was there another step involved?


That's what I did, but I included an unrounded 4.22, as it was recurring.
Whether that makes much of a difference I don't know

Oh and I added the 1.5 internal resistance to the value you get from the other 2

got 2.86 or something like that
Original post by davon806
For the line of dc voltage,how many line should I draw?I only drew one on 45V!I am not sure whether I have to draw a line on -45V


same! I drew a line at 45V too. :smile:
Original post by amar96
you dont include the internal resistance


Yes you do! :smile:


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Original post by Surf
Do you mean the proton one?
A = Neutron
B = W+ boson
C = Positron
D = Electron Neutrino

Yeah that question for A I put a down quark.
Original post by SamHedges
I thought the reason it was maximum kinetic energy was because the ones further from the metal's surface would need to absorb higher frequency photons to escape so the ones with maximum KE are near surface. Surely decreasing frequency meant the ones in the metal deeper in would need more energy to escape and so their emission would be more frequent for higher frequencies?


I wrote that if you decreased the frequency of the photon, the kinetic energy of the photon would decrease up until you reach there threshold frequency, beyond which no electrons would be emitted. This is because photons consist of energy E=HF. Therefore its energy depends on its frequency.

Doubling the intensity would double the number of electrons released because double the amount of photons are striking the metal surface, and one photon releases one electron.

Posted from TSR Mobile
Reply 1075
Avatar for GeneralOJB
GeneralOJB
OP
Original post by amar96
you dont include the internal resistance


Who told you that?
Reply 1076
Avatar for SamuelJ
SamuelJ
:mad: I did the specific charge for the nucleus and not the ion

Will I lose 1 or 2 marks?

And also, on the Feynman diagram does it matter which lepton goes on which arrow on the RHS?
Original post by pilotluke1
Yeah that question for A I put a down quark.


I wrote it as "d quark" do you think I'll still get the mark?

Posted from TSR Mobile
Reply 1078
Avatar for Surf
Surf
15
Original post by pilotluke1
Yeah that question for A I put a down quark.


yeah, my bad - I meant to put d quark.... absent minded.


As for the total resistance - you do have to add the internal resistance.
Reply 1079
Avatar for GeneralOJB
GeneralOJB
OP
Original post by SamuelJ
:mad: I did the specific charge for the nucleus and not the ion

Will I lose 1 or 2 marks?

And also, on the Feynman diagram does it matter which lepton goes on which arrow on the RHS?


No it doesn't matter which lepton goes on which arrow. If you did the nucleus instead of the ion you'll lose both marks.

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