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

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Unofficial markscheme anyone?
Could we have said that for G the measuring equipment and devices have improved precision


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Original post by Mutleybm1996
Could we have said that for G the measuring equipment and devices have improved precision


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Maybe, but I think what they were getting at was that the value of density used was way off. Besides, what values would you have improved the precision of?


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Original post by Rhetorical Hips
Maybe, but I think what they were getting at was that the value of density used was way off. Besides, what values would you have improved the precision of?


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Timing perhaps


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Ok for the natural log questions I divided the two values so you cancel the two constants, and re-arranged to get Epsilon, anyone do a similar method to this?
Reply 185
Original post by Robbo54
Ok for the natural log questions I divided the two values so you cancel the two constants, and re-arranged to get Epsilon, anyone do a similar method to this?


Exactly that.
Reply 186
Original post by ashysmithy105
Oh. My. God. I'm such an idiot. I put 9.4x10-7 = 94uA not 0.94uA :rolleyes: I wrote 9.4x10-7 above the answer section, do you think I'll only lose one mark then?


You might not even lose any, they're often lenient with these kind of things. At absolute most you'll lose one.
Original post by STATER
Exactly that.


I still don't see what I did wrong then because I got something ^-3. How many marks do you think I would drop?
Reply 188
How many marks was the question where we had to state why our value of G was wrong?
Original post by ryanWales
I don't think it can be sound because it was in space so there are no air particle. I put heat :smile:


There would still be sound as the sound would go through the solid spacecraft.
Reply 190
So what did everyone get for the frequency of lift oscillations? I got 0.45hz which I'm reasonably sure is wrong, I saw a few people mention they got 1.4hz?
My error was not subbing k=EA/L into the formula for T, i just calculated k and then subbed the value in but I must have got the value wrong.
Reply 191
Original post by Robbo54
I still don't see what I did wrong then because I got something ^-3. How many marks do you think I would drop?

Likely youll only lose 1
Reply 192
For the logs question, I did, my answer was 3.86 * 10^-20.

Anyone get this, rather than 3.96? I've checked it again just now, so I'm sure it's right unless I made a silly mistake.

my answer was epsilon = ln10 / [(1/(300*k))-(1/(400*k))] which gives 3.86 * 10^-20
Reply 193
Also, is 'all the particles have the given average velocity of about 500ms^-1' a correct assumption for the rate of collision question.
Reply 194
Original post by nair39
For the logs question, I did, my answer was 3.86 * 10^-20.

Anyone get this, rather than 3.96? I've checked it again just now, so I'm sure it's right unless I made a silly mistake.

my answer was epsilon = ln10 / [(1/(300*k))-(1/(400*k))] which gives 3.86 * 10^-20


Yep doing it now on my calculator I get the same. Hopefully I just remember my answer wrong!
Original post by urz13
Yep doing it now on my calculator I get the same. Hopefully I just remember my answer wrong!


I think I got 3.86 actually because thats what I get on my calcuator now. 3.96 is probably be remembering wrong. I doubt I would have written something different to what I had on my calculator.
Reply 196
Ok thanks, I'm sure you both just remembered it wrongly but got it right.

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Original post by nair39
For the logs question, I did, my answer was 3.86 * 10^-20.

Anyone get this, rather than 3.96? I've checked it again just now, so I'm sure it's right unless I made a silly mistake.

my answer was epsilon = ln10 / [(1/(300*k))-(1/(400*k))] which gives 3.86 * 10^-20


I think it was 2.something
I also think it was less than ln(10) too.
I got (ln(9.something) x k)/(1/300-1/400)



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(edited 8 years ago)
Reply 198
Original post by Mutleybm1996
I think it was 2.something
I also think it was less than ln(10) too.
I got (ln(9.something) x k)/(1/300-1/400)



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definitely ln10 because it was just 10/1=e^...
For the question with the table about the SHM, I thought the x in the equation for the change in velocity was a multiplication sign and not x for displacement. I am so annoyed! Why couldn't they have used a curly x or put it at the start/end of the equation? I couldn't work out why my answers were wrong but I knew they were.

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