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2012 Higher Physics Discussion

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    Anyone know the answer to mc Q6?
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    (Original post by KeithyDee)
    That is not tutor/teacher answers though, just what some people have said.
    well here you go then


    1E
    2A
    3C
    4C
    5C
    6D
    7B
    8D
    9B
    10C
    11B
    12E
    13D
    14E
    15D
    16A
    17A
    18D
    19B
    20B
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    (Original post by aroy45)
    Can someone explain what I'm doing wrong for 21a


    This was posted from The Student Room's iPhone/iPad App
    Your diagrams wrong, the angle should be 70 degrees
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    Ill put my answers up for people to correct or agree
    21a 15.9km at 157 degree
    ii) 3.53m/s
    B). 15.9km at 157
    ii) 2.94m/s

    22ai) 61.2m
    ii) 11.475m
    B). More likely as ball will not travel at far and reach max height earlier

    23) proof
    ii). 42318m/s
    B) 5.6x10^-3
    C) xenon as f=ma as mass is bigger acceleration the same. (very unsure)

    24). Proof p/Tk is a constant
    B). More collisions with each other and outer walls
    C). Stopper could be moved by different pressures acting could change volume. (wrong I think)

    25) 1.5A
    ii) 3v
    iii) 13.5W

    B) Internal resistance is lower

    26) graph with voltage starting at 0 and levelling out at 12V
    B) 5kohms (might be 6kohms)
    C)the voltage is the same (not sure)
    ii). The capacitance in this instance is the same as the voltage does not change (unsure)

    27) proof
    B) differential
    ii) 5.6
    B(a) 1.93v
    B(b) 55mm (think its 65)

    28) 51 degrees
    B) partial internal reflection
    ii) 48.8 degrees
    C). Sketch shows total internal reflection

    29a) 3.3x10^-6m
    B) 3.03x10^5
    C) yes as it falls within the uncertainty range

    30) decrease
    B) photoconductive
    ii). Current increases
    C) 6.75x10^-6

    31) 0.0125j (wrong?)
    B) alpha
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    (Original post by KeithyDee)
    The displacement was 15.7 +/- 0.2km.
    no, it was 16km and +- 10m cos you have to round of 15.7 to 16 and every answer 2 significant figures cos the question was in 2 sig fig, there was no decimal places.
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    [QUOTE=TheZorse94;37806292]Ill put my answers up for people to correct or agree
    21a 15.9km at 157 degree
    ii) 3.53m/s
    B). 15.9km at 157
    ii) 2.94m/s

    22ai) 61.2m
    ii) 11.475m
    B). More likely as ball will not travel at far and reach max height earlier

    23) proof
    ii). 42318m/s
    B) 5.6x10^-3
    C) xenon as f=ma as mass is bigger acceleration the same. (very unsure)

    24). Proof p/Tk is a constant
    B). More collisions with each other and outer walls
    C). Stopper could be moved by different pressures acting could change volume. (wrong I think)

    25) 1.5A
    ii) 3v
    iii) 13.5W

    B) Internal resistance is lower

    26) graph with voltage starting at 0 and levelling out at 12V
    B) 5kohms (might be 6kohms)
    C)the voltage is the same (not sure)
    ii). The capacitance in this instance is the same as the voltage does not change (unsure)

    27) proof
    B) differential
    ii) 5.6
    B(a) 1.93v
    B(b) 55mm (think its 65)

    28) 51 degrees
    B) partial internal reflection
    ii) 48.8 degrees
    C). Sketch shows total internal reflection

    29a) 3.3x10^-6m
    B) 3.03x10^5
    C) yes as it falls within the uncertainty range

    30) decrease
    B) photoconductive
    ii). Current increases
    C) 6.75x10^-6

    31) 0.0125j (wrong?)
    B) alpha

    agree with everything but tehcnically speaking 21) answers had to be in 2 sig figures since question was in 2 sig figures dont think that matters unless you put too many
  7. Offline

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    (Original post by DragonHeartstring)
    Me too-ish! I drew it going west instead of east :rolleyes: Just hope that what I got as displacement and bearing is right for what I drew, then I might still pick up marks.
    I also wrote that v=d/t for it, so I'm hoping I maybe pick up a few half marks for formula. And pretty sure I got the time right in the last question, just did everything else wrong!, so annoying as well when you're convinced it's perfect in the exam! :rolleyes:
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    (Original post by S119234)
    I also wrote that v=d/t for it, so I'm hoping I maybe pick up a few half marks for formula. And pretty sure I got the time right in the last question, just did everything else wrong!, so annoying as well when you're convinced it's perfect in the exam! :rolleyes:
    I know! I was hoping for a scale drawing in this exam, and then I go and mess it up! Typical :rolleyes:
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    I think things are looking fairly good for me, thanks guys
  10. Offline

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    Oh well, now the wait for August begins. I'm still undecided as to open it slowly showing each grade one at a time or just whip it out haha
  11. Offline

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    (Original post by Siwel95)
    Oh well, now the wait for August begins. I'm still undecided as to open it slowly showing each grade one at a time or just whip it out haha
    I'm just going to whip it out (my philosophy in life ) and see how it goes

    Just enjoy the rest of your summer !
  12. Offline

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    (Original post by KeithyDee)
    I'm just going to whip it out (my philosophy in life ) and see how it goes

    Just enjoy the rest of your summer !
    haha same goes for you
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    Could someone say what the answer to 31(a) is ?, I got 0.00002J using D=E/m, but not particularly sure.

    Also, for 28(b)(i), would it not be an acceptable answer to say that it's because the angle in air = 90degrees ?, cause that then leads to the critical angle equation.
  14. Offline

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    Does anyone know what the grade boundary for an A has been the last few years?
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    (Original post by S119234)
    Could someone say what the answer to 31(a) is ?, I got 0.00002J using D=E/m, but not particularly sure.

    Also, for 28(b)(i), would it not be an acceptable answer to say that it's because the angle in air = 90degrees ?, cause that then leads to the critical angle equation.
    I got some sort of variation of 0.00002
  16. Offline

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    (Original post by S119234)
    Could someone say what the answer to 31(a) is ?, I got 0.00002J using D=E/m, but not particularly sure.

    Also, for 28(b)(i), would it not be an acceptable answer to say that it's because the angle in air = 90degrees ?, cause that then leads to the critical angle equation.
    yes i got that for 31a) , as for question 28 i cannot remember the question haha
  17. Offline

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    (Original post by Herzschmerz)
    Does anyone know what the grade boundary for an A has been the last few years?
    2011: 64 marks (71%)
    2010: 67 marks (74%)
    2009: 68 marks (75%)
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    (Original post by Siwel95)
    yes i got that for 31a) , as for question 28 i cannot remember the question haha
    Thank goodness! , it was the one asking why the value of theta equals the critical angle for the light in water. Pretty sure that it's because angle in air is 90 degrees.
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    What has been the grade boundaries for a B over the years?
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    (Original post by S119234)
    Thank goodness! , it was the one asking why the value of theta equals the critical angle for the light in water. Pretty sure that it's because angle in air is 90 degrees.
    ah yes, i wrote that it is refracted at 90 degrees

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