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OCR PHYSICS B G494~ 11th June 2015 AM ~ A2 Physics

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Reply 40
Avatar for Alex .G.
Alex .G.
I've heard only 100-150 people actually take this A2 - has anyone else heard this?
Original post by Alex .G.
I've heard only 100-150 people actually take this A2 - has anyone else heard this?


I'm only in AS right now and I didn't know it was that low, but now I see why. :frown:

Edit: where did you heard that btw? xD

Edit 2:
You can find last years ocr result statistics and it says around 4000 people were examined for this a2 course but yeah even then, 4000 isnt that high :tongue:
(edited 8 years ago)
Original post by Alex .G.
I've heard only 100-150 people actually take this A2 - has anyone else heard this?


It's about 4000


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How are people revising? I feel so unprepared for this exam...
Reply 44
Avatar for Alex .G.
Alex .G.
Original post by SheLikeTheMango
I'm only in AS right now and I didn't know it was that low, but now I see why. :frown:

Edit: where did you heard that btw? xD

Edit 2:
You can find last years ocr result statistics and it says around 4000 people were examined for this a2 course but yeah even then, 4000 isnt that high :tongue:


Im sure I heard it from somewhere - forgot where now! Think I saw it on some centre note of stat thing - it seemed silly, so I thought I'd check.
What on Earth is a worldline?
Reply 46
Avatar for smerkz
smerkz
1
Could someone please post a good solution for this question? The mark scheme is messy and unclear.

qqq.png
Original post by smerkz
Could someone please post a good solution for this question? The mark scheme is messy and unclear.

qqq.png


p=FAp = \dfrac{F}{A}

p=mv22r×12πr2p = \dfrac{mv^2}{2r} \times \dfrac{1}{2\pi r^2}

p=mv24πr3p = \dfrac{mv^2}{4\pi r^3}

But 3V=4πr33V = 4\pi r^3, and there are NN particles. Hence:

p=Nmv23Vp = \dfrac{Nmv^2}{3V}
Reply 48
Avatar for Robbo54
Robbo54
2
Original post by lizard54142
What on Earth is a worldline?


Essentially the axis are swapped so it's a time-distance curve
Original post by lizard54142
What on Earth is a worldline?


I like what you did there.
Original post by DomStaff
I like what you did there.


#physicsbanter
ImageUploadedByStudent Room1433964747.626672.jpg
I'm sure the C-13 showed a carbonyl group

But I have the wrong Benzene splitting


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Anyone ever actually looked at the section in the book on brightness of stars to calculate distances?
Reply 53
Avatar for smerkz
smerkz
1
Original post by DomStaff
Anyone ever actually looked at the section in the book on brightness of stars to calculate distances?


Yeah, but i'm not really sure what they could ask? The process? Maybe provide a formula and you stick some numbers in?
Original post by smerkz
Yeah, but i'm not really sure what they could ask? The process? Maybe provide a formula and you stick some numbers in?


Think it's a load of *******s. Give us some proper physics plsssssss
Original post by DomStaff
Anyone ever actually looked at the section in the book on brightness of stars to calculate distances?


Tbh I completely forgot that was on the course. You can probably wing it, that section is just common sense.
Original post by Mutleybm1996
ImageUploadedByStudent Room1433964747.626672.jpg
I'm sure the C-13 showed a carbonyl group

But I have the wrong Benzene splitting


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That's obviously wrong.
Original post by DomStaff
That's obviously wrong.


Sorry, wrong thread
do you do chemistry?
I'm pretty sure there was a carbonyl C13 one, others have said that too...
Original post by Mutleybm1996
Sorry, wrong thread
do you do chemistry?
I'm pretty sure there was a carbonyl C13 one, others have said that too...


No, I don't have a clue. Just waiting for a paper to print off and was bored.
Original post by DomStaff
No, I don't have a clue. Just waiting for a paper to print off and was bored.


which paper?
how are you feeling about this exam?
any advice/tips?

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