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# Official Thread For OCR Physics A G484 Jan 2011 watch

1. (Original post by Oh my Ms. Coffey)
That didnt go too well, I couldnt get the second radius for satellite to work.
What did you get for the first one? 6.4x10^6?
I think thats what you get using g=GM/r2, with g as 9.81?

However the value for g for the satellite is unkown so that can't be used.

After some rearranging of forumalae I got like 2.7x10^7.
2. (Original post by HRM-24)
Steep grad from 18 to 0, then less steep under 0 10 -18.
If the question said it that the shc is lower, I think then the line will be steeper (faster temperature change), vice versa. can anyone confirm what the question said (I dont' remember)?
3. (Original post by rishmiester)
6400km seems like the radius of the earth.
I got 2.0 10^7 m, or summin like that

True that
negative gradient straight line going from 18 to 0, then horizontal line for a bit, then steeper negative gradient line going down till -18. u?
2.0x10^7 was the one they gave you And my graph went Less steep, flat, steeper also
(Original post by rishmiester)
Phew!
velocity of oxygen molecule: 450?
Yes i got this too !
4. (Original post by In the looney bin)
2.0x10^7 was the one they gave you And my graph went Less steep, flat, steeper also
Yes i got this too !
YEAHHHH BOIII!
that or we're both wrong without realising it :P

2.0x10^7 was given in the question? I may not remember the correct question... what was the answer for that question then?
5. (Original post by rishmiester)
YEAHHHH BOIII!
that or we're both wrong without realising it :P

2.0x10^7 was given in the question? I may not remember the correct question... what was the answer for that question then?
You had to work out one radius, given only mass of Earth and velocity of sattelite ( which i got as 2.7 or 2.9 or something x10^7)

Then this satellite deccelerates and moves into a smaller orbit of radius 2.0x10^7.

You had to give a possible way of deceleration, and then work out the new velocity, which i got as four thousand and something m/s lol, cant remember
6. in the last question did you have to use the period of 100 days in the calculation?
7. Found this paper on the whole pretty good. - Can't really remember most of my answers lol :O - But I genuinely did not revise one bit for this exam. Couldnt care less, told my teacher not to enter me for it, wanted to do it in June. Oh well.

EDIT: Is it only me that all of a sudden feels as if about 90% of this papers questions is now a blur in my head? - I can only remember a couple of the worded ones LOL, and a few methods I used but hardly any numerical answers LOOL!!
8. I had less steep to 0 and then steeper to -18
9. (Original post by In the looney bin)
You had to work out one radius, given only mass of Earth and velocity of sattelite ( which i got as 2.7 or 2.9 or something x10^7)

Then this satellite deccelerates and moves into a smaller orbit of radius 2.0x10^7.

You had to give a possible way of deceleration, and then work out the new velocity, which i got as four thousand and something m/s lol, cant remember
Yeah, i definitely don't remember
the x10^7 part looks quite familiar though, as does the four thousand and something part.

I missed out the "abscence of any external forces" part on the very first question!

did you get the coalesced speed as 4/3 in the first question?
10. (Original post by In the looney bin)
2.0x10^7 was the one they gave you And my graph went Less steep, flat, steeper also

Yes i got this too !
i got the radius as 2.7 or 2.9 x10^7 and also got less steep flat steeper
11. (Original post by EmperorMustard)
I had less steep to 0 and then steeper to -18
Why? - It said the specific heat capacity of milk was much lower after 0 celcius. So it would lose heat at a slower rate in comparison to the shc above 0.
12. (Original post by rudebwoi93)
Why? - It said the specific heat capacity of milk was much lower after 0 celcius. So it would lose heat at a slower rate in comparison to the shc above 0.
SHC is the amount of energy needed to change per degree. With a low shc, this means it would take less energy per degree of temp changed. So, with a constant power (from the fridge, removing energy), it would take less time in total to remove enough energy to get to -18, hence steeper graph.
13. (Original post by rudebwoi93)
Why? - It said the specific heat capacity of milk was much lower after 0 celcius. So it would lose heat at a slower rate in comparison to the shc above 0.
Yeah I thought that too.
14. (Original post by rudebwoi93)
Why? - It said the specific heat capacity of milk was much lower after 0 celcius. So it would lose heat at a slower rate in comparison to the shc above 0.
Delta t = E/CM

C = less, E/M is the same.

Delta T increases at a greater rate.
15. (Original post by rishmiester)
Yeah, i definitely don't remember
the x10^7 part looks quite familiar though, as does the four thousand and something part.

I missed out the "abscence of any external forces" part on the very first question!

did you get the coalesced speed as 4/3 in the first question?
Yes i got 4/3, what is this external forces bit? Conservation of linear momentum?
16. (Original post by In the looney bin)
Yes i got 4/3, what is this external forces bit? Conservation of linear momentum?
The first part of the very first question, it asked you to state the law of the conservation of linear momentum.
17. what did everyone get for the last two questions, one asked to calculate the amount in mole, the other for the mass of nitrogen lost?
18. (Original post by HRM-24)
Steep grad from 18 to 0, then less steep under 0 10 -18.
Did the question require you to put temperature values on the side?
I thought it said it didn't...
19. That exam didn't go too well for me. I finished with a few minutes to spare but after the exam there were so many things I could think of that I forgot/did wrong. That'll probably be another retake in June then.
20. (Original post by sc0307)
what did everyone get for the last two questions, one asked to calculate the amount in mole, the other for the mass of nitrogen lost?
I think I got 8. something for the number of moles, and for the mass I got something around 0.023?
Sorry can't remember the exact values..

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