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PHYA1 - Physics Unit 1 Exam - 24th May 2011

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Original post by jackegan93

Original post by jackegan93
That's actually what I said.. will I still get marks?


I'm pretty sure we're right lol.

I'm sure the bottom equation actually worked fine. But some people are saying muon has no lepton number now... or it has -1, which I don't get...

Unless both of them didn't work and they just wanted a reason for the one you chose?
Original post by TooEasy123
I'm pretty sure we're right lol.

I'm sure the bottom equation actually worked fine. But some people are saying muon has no lepton number now... or it has -1, which I don't get...

Unless both of them didn't work and they just wanted a reason for the one you chose?


muon*+ has a lepton number of -1.
Reply 362
Original post by jackegan93
Yeah anyway, anyone remember the Argon question? Like what the atom before was?



think it was potassium, K!
Original post by TooEasy123
I'm pretty sure we're right lol.

I'm sure the bottom equation actually worked fine. But some people are saying muon has no lepton number now... or it has -1, which I don't get...

Unless both of them didn't work and they just wanted a reason for the one you chose?


I'm confident that the bottom one didn't work now too though, more confident than I am with my own choice, so it's probably wrong.

Say, do you remember the question with Argon in it and what it said?
Reply 364
For the proof question i proved that the resistance of the resistors was 3, using my answer of 5 ohms from the previous question - would i still get marks for that?
i.e. i did 1/2r + 1/r = 1/rt
rt = 2r/3

2r/3 + r = 5
5r = 15
r = 3
Reply 365
has anyone got an unofficial mark scheme?
Original post by jackegan93

Original post by jackegan93
I'm confident that the bottom one didn't work now too though, more confident than I am with my own choice, so it's probably wrong.

Say, do you remember the question with Argon in it and what it said?


Wtf... I checked it sooo many times, I'm pretty sure it was a normal muon lol.

Was it the one where it's like Ar + e- = something + Ve
Reply 367
What do u think the grade boundaries are ???????????
anyone???
Original post by TooEasy123
Wtf... I checked it sooo many times, I'm pretty sure it was a normal muon lol.

Was it the one where it's like Ar + e- = something + Ve


Not sure I can't remember much about it, I'm hoping someone will jog my memory. Ah well
it was a positive muon so it had a lepton number of -1
Reply 370
Original post by Spas_13
Thats wrong mate, unlucky. Good luck in the retake :wink:


LOL ! :colondollar:
Original post by matthew769
this would be correct had the question asked for the current value where there was more resistance, but it asked for the current value through the part where there was only one resistor :smile: i imagine you would get a mark for your working as it is all correct, just the answer is incorrect


:frown: Aw well just 2 marks right
Reply 372
I really think the mark boundries will be quite low. Does anyone thing ill get any marks for this. This is for the calculating the internal resistance "So for some reason i used 1.5 as the emf (1 mark lost) and again for some reason i calculated the lost volts as :1.5v-1.4V = 0.1V. I then read this value off the graph and I got A Current value of 0.3. R=V/I so R=0.1/0.3 so i got an answer around 0.33 ohms, will i get anything?
Reply 373
Was the definition question 'ionisation energy' ?
can anyone remember the question at the end of the emf graph asking for energy lost in watts (or joules per second).
On the question that asked you to convert eV to joules, was it MeV or eV?
Reply 376
ev so multiply 13.6 by 1.6 x 10^-19
Original post by anil10100
Doing a good job in my eyes though haha


I agree :wink:
Reply 378
50/70 for an A ? what do people think ?
Original post by A level Az
How did you get to that answer? Did you take away the last point from the first and divide it by the total current?


Yep, just drew a triangle and found the gradient.
Firstly along the y axis which was 1.52 from the highest point (of the triangle) subtracted by 0.08 which was the lowest point. Then divided by the highest and lowest on the x axis which was just 3.20 because the lower was 0.00 :P

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