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    How did people find the ISAs on Capacitance and SHM?
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    (Original post by henry.philips)
    How did people find the ISAs on Capacitance and SHM?
    Not fantastic, got 41 on SHM one, hbu?


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    Is unit 5 synoptic at all?
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    anyone doing the unit 1 exam on Tuesday?
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    (Original post by Exams987)
    anyone doing the unit 1 exam on Tuesday?
    i am!!!!
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    (Original post by kother2015)
    i am!!!!
    i have made a thread here

    http://www.thestudentroom.co.uk/show...7#post64958147
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    [QUOTE=Exams987;64960117]i have made a thread here

    http://www.thestudentroom.co.uk/showthread.php?t=4087081&p=64958 147#post64958147[/QUOTE]


    thanx

    r u re sitting unit 2 as well?
    out of interest what ums did you get last year ? and what u predicted?
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    in
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    Following. Doing Turning Points here, let's hope they don't accelerate a potato through a fluorescent tube this time.
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    For those who are taking Turning Points, I'm making flash cards as I go along my revisions and past papers. Hope you'll find it useful.
    http://www.cram.com/flashcards/phy5a-7323575
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    Can anyone help with this q?
    Three ice cubes each of mass 25 g are dropped into a glass containing 330 g of water at a temperature of 22 °C. The ice is at 0 °C and melts so that the temperature of the water decreases.Calculate the final temperature of the water when all of the ice has melted. Assume that no heat is lost to the glass or the surroundings.specific latent heat of fusion of ice = 3.3 × 10^5 J kg^1 specific heat capacity of water = 4.2 × 10^3 J kg^1 K^1
    Thanks
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    I keep answering questions with the wrong number of sig figs...

    Is it true that you can only lose 1 mark in the paper to wrong number of significant figures?? or am i getting confused
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    (Original post by Lili1998)
    Can anyone help with this q?
    Three ice cubes each of mass 25 g are dropped into a glass containing 330 g of water at a temperature of 22 °C. The ice is at 0 °C and melts so that the temperature of the water decreases.Calculate the final temperature of the water when all of the ice has melted. Assume that no heat is lost to the glass or the surroundings.specific latent heat of fusion of ice = 3.3 × 10^5 J kg^1 specific heat capacity of water = 4.2 × 10^3 J kg^1 K^1
    Thanks
    QIce = (0.025 × 3) × 330000 = 24750 J
    QIce = 0.075 × 4200 × ΔT = 315ΔT
    QWater = 0.330 × 4200 × ( 22 - ΔT) = 1386 × ( 22 - ΔT)

    QIce = QWater therefore

    24750 + 315ΔT = 1386 × ( 22 - ΔT)

    Rearrange to find ΔT, giving a ΔT of 3.37 ºC

    Hope this helps.
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    (Original post by Lili1998)
    Can anyone help with this q?
    Three ice cubes each of mass 25 g are dropped into a glass containing 330 g of water at a temperature of 22 °C. The ice is at 0 °C and melts so that the temperature of the water decreases.Calculate the final temperature of the water when all of the ice has melted. Assume that no heat is lost to the glass or the surroundings.specific latent heat of fusion of ice = 3.3 × 10^5 J kg^1 specific heat capacity of water = 4.2 × 10^3 J kg^1 K^1
    Thanks
    Hey I had a go but i'm still in the process of revising this so don't take my word for it hahah..

    Tf (final temp)

    heat energy supplied by water = energy to melt ice + heat energy gained by the melted ice

    0.330 x 4200 x (22-Tf) = 0.025 x 3.3x10^5 + 0.025 x 4200 x Tf

    which rearranges to..

    Tf = 14.9ºC

    not 100% sure but the answer seems fairly reasonable ahah
    is there a mark scheme?
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    (Original post by Calchioni)
    Hey I had a go but i'm still in the process of revising this so don't take my word for it hahah..

    Tf (final temp)

    heat energy supplied by water = energy to melt ice + heat energy gained by the melted ice

    0.330 x 4200 x (22-Tf) = 0.025 x 3.3x10^5 + 0.025 x 4200 x Tf

    which rearranges to..

    Tf = 14.9ºC

    not 100% sure but the answer seems fairly reasonable ahah
    is there a mark scheme?

    WELLLL i can already see where i've gone wrong, I didn't multiple by the number of ice cubes...
    This is a perfect example of where i screw up my exams -.-
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    (Original post by scrlk)
    QIce = (0.025 × 3) × 330000 = 24750 J
    QIce = 0.075 × 4200 × ΔT = 315ΔT
    QWater = 0.330 × 4200 × ( 22 - ΔT) = 1386 × ( 22 - ΔT)

    QIce = QWater therefore

    24750 + 315ΔT = 1386 × ( 22 - ΔT)

    Rearrange to find ΔT, giving a ΔT of 3.37 ºC

    Hope this helps.
    Thanks... that makes sense
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    How's everyone feeling about the exam?


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    What is the differrence between a cell and a high voltage unit?

    Because they are both in a thermionic emission circuit (Turning points module), so what do they both do?

    Any input appreciated
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    HELP:An isotope of technetium Tc which is in a metastable state, decays emitting only γ rays. When the isotope is placed 20 cm from a γ ray detector the count rate is 25 counts per second. The background count rate is 120 counts per minute. Calculate the count rate, in counts per second, when the detector is placed 30 cm from the isotope.
    I keep getting the answer 9.1, but the mark scheme says the answer is 12.2, how do they do this?
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    (Original post by Alby1234)
    HELP:An isotope of technetium Tc which is in a metastable state, decays emitting only γ rays. When the isotope is placed 20 cm from a γ ray detector the count rate is 25 counts per second. The background count rate is 120 counts per minute. Calculate the count rate, in counts per second, when the detector is placed 30 cm from the isotope.
    I keep getting the answer 9.1, but the mark scheme says the answer is 12.2, how do they do this?
    Which textbook is this?

    I would have thought it was 25/1.5^2 = 11.1

    So I don't get that answer either for the corrected count rate
 
 
 
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