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CIE PHYSICS P4 QUESTIONS

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falafail:
DO NOT POST ANYTHING HERE EXCEPT QUESTIONS RELATED TO PHYSICS PAPER 4 OR I WILL MURDER YOU  >:( >:( >:(

okay, so:

1) june 02 Q 4b
2) nov 05 Q 4b  in the markscheme it says "endpoints of line correctly labelled" what are we supposed to label exactly???
3) june 06 Q 4c
4) nov 02 Q 1a how do we do (i)? and for (ii) why do we use 7 minutes and not 9?
5) june 04 Q 2b(ii) and for Q6 the last row is like + + 0, but shouldn't it have been 0 + -, 'cause the particles move so they're doing work??
6) june 06 Q 2b(ii)
7) how do you determine the amplitude from a ke/pe graph?

HELP :(

AndrewCedric:
1) The EP graph NEVER GOES NEGATIVE. So A t=0 it is max.. At t= 0.025 approx, it is 0.. It reaches Max again at at t=0.05

2)Label The maximum and minimum displacement.

3)Max Speed--> Horizontal ( right / left ) part of the circle          Max Accel--> The uppermost vertical part of circle

4) Use the formula Power x Time ( 2min ) = mass x specific heat x ( 100 - 20)
FOr part 2, We use 7, cause only at that time the water starts boiling. Only for the last 7 min that the temperature is constant. So we can apply Q=mL

5)Use formula p=1/3 (density) ( Mean square speed )
NO. FOr the box thing. THERE IS NO WORK DONE. W=pdV. No change in volume. So no work done.
We know Q must be +.. SO leaving U must be +. Alternatively, working from U. No change in Temperature. But Their state changes, thus their EP increases. REMEMBER THIS IS NOT IDEAL GAS, SO EP CANNOT BE NEGLECTED.

6)Use pV=nRT for the balloon too. And Compare No of moles with part a.

7) KE/PE graph against what?? Usually we do displacement. So, Just look for the Max EP and down to the x axis. Thats the amplitude.

HOPE IT HELPS!!!!! ;D

ifraha18:
cn someoen xlain the graph of q7 part 1 june 2005.??

Onion:

--- Quote from: ifraha18 on May 23, 2010, 05:12:11 am ---cn someoen xlain the graph of q7 part 1 june 2005.??

--- End quote ---

Half-life is the time for undecayed atoms to be reduced by half. So the graph would be an exponential growth, starting from origin. At t = 2.6, mass of Fe-56 would be 1/2(1.4) = 0.7 micrograms

Mass of Fe-56 would increase over time, so at another 2.6  hours (t = 5.2), mass = (0.7) + 1/2(0.7) = 1.05 micrograms and so on ...

ifraha18:
Thanks...:)

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