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    (Original post by HarryW95)
    I got this. That question really through me off though with the 32MJ of energy per cubic meter. Did we need to use that or not? I got 217 without using it and it sounded reasonable so I moved on.

    And can't remember for the mass but I ended up just doing density x volume.
    They didn't give you the volume of the heated gas though did they? I was reluctant to using the entire volume of the room as the volume of the heated gas so I did it another way.


    From what I remember I did something along the lines of:

    5 x 10^5 (thermal energy answer found in the first part) / 39x10^6 (39MJ) multiplied by the 0.78kgm^3

    My logic was that the fraction of the energy required and the energy produced by the heater would give the volume of heated gas, multiplied this by 0.78 and I got the mass.

    I'm unsure if it is correct though, I'd need confirmation.
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    (Original post by HarryW95)
    I got this. That question really through me off though with the 32MJ of energy per cubic meter. Did we need to use that or not? I got 217 without using it and it sounded reasonable so I moved on.

    And can't remember for the mass but I ended up just doing density x volume.
    I think you needed the 32mj for the mass since it was 32MJ per cubic meter but we only needed 5 x 10 to the something Joules to heat the room, so only needed a fraction of that 32 Mj then you can use density = mass x volume
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    (Original post by Political Cake)
    I also really liked the special case of newton's second law question. I remember wondering in January if that would come up then or not.
    It was strange how that question was worth only 1 mark in our paper, but worth 2 marks in previous papers.
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    (Original post by Edmundith)
    It was strange how that question was worth only 1 mark in our paper, but worth 2 marks in previous papers.
    it was worth 2
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    I wrote down my answers but stupidly I didn't write the exact question numbers.
    Only calculated values as well:

    1)
    16
    4

    2)
    10.5
    8469.6
    8.25*10^28

    3)
    top and bottom box
    0.833
    0.012
    0.329

    4)
    500386
    217
    9.24*10^-3

    5)
    2.47*10^5
    9.24*10^-3

    6)
    2.1*10^-5
    5.85*10^-3
    1.5*10^7

    I have no idea if these are correct...
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    (Original post by Ryan Gregg)
    They didn't give you the volume of the heated gas though did they? I was reluctant to using the entire volume of the room as the volume of the heated gas so I did it another way.


    From what I remember I did something along the lines of:

    5 x 10^5 (thermal energy answer found in the first part) / 39x10^6 (39MJ) multiplied by the 0.78kgm^3

    My logic was that the fraction of the energy required and the energy produced by the heater would give the volume of heated gas, multiplied this by 0.78 and I got the mass.

    I'm unsure if it is correct though, I'd need confirmation.
    That's correct because you have figured out the require volume, since itbwas 32 or 39 MJ or wahtever per cubic meter and we only needed 1\78 of that to heat the room, then times by the 0.72 density
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    (Original post by robolew)
    I wrote down my answers but stupidly I didn't write the exact question numbers.
    Only calculated values as well:

    1)
    16
    4

    2)
    10.5
    8469.6
    8.25*10^28

    3)
    top and bottom box
    0.833
    0.012
    0.329

    4)
    500386
    217
    9.24*10^-3

    5)
    2.47*10^5
    9.24*10^-3

    6)
    2.1*10^-5
    5.85*10^-3
    1.5*10^7

    I have no idea if these are correct...
    i ticked the top one but not the last one (Q3). i think the last one was is acceleration max when velocity is max? which it isn't i don't think. I can't exactly remember what it said though
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    (Original post by robolew)
    4)
    500386
    217
    9.24*10^-3


    I have no idea if these are correct...
    Yes! We got the same question 4 answers.
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    (Original post by Supernova_)
    i ticked the top one but not the last one. i think the last one was is acceleration max when velocity is max? which it isn't i don't think.
    It was "acceleration increases as speed decreases" I believe
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    (Original post by robolew)
    It was "acceleration increases as speed decreases" I believe
    Speed decreases as displacement increases. Acceleration increases as displacement increases. Hence acceleration increases as speed decreases.
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    I ticked the max acceleration at max displacement box, which makes sense since at the top of a pendulum swing it is furthest away from equilibrium position and has max acceleration at this top point
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    (Original post by robolew)
    It was "acceleration increases as speed decreases" I believe
    ah I thought it said 'acceleration increases as speed increases' oh well.
    I ticked the max acceleration at max displacement box, plus the top one ( acceleration opposite to displacement.) might get a mark if these two are correct.
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    (Original post by Supernova_)
    ah I thought it said 'acceleration increases as speed increases' oh well.
    I ticked the max acceleration at max displacement box, plus the top one ( acceleration opposite to displacement.) might get a mark if these two are correct.

    I said exactly the same as you, so someone has read the question wrong!
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    I only ticked the first and last box.
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    I got 250 moles in question 6 and I still got 1.5x10^7, using the molar approach. I found the moles of the added gas to be 60 and then I added them both to get 310, then I put them into the gas equation pV=nRT
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    These r my calculation answers pls confirm:erm:

    Name:  uploadfromtaptalk1371726943639.jpg
Views: 188
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    Name:  uploadfromtaptalk1371726980937.jpg
Views: 122
Size:  46.1 KB

    And i ticked first an last box:
    It is in opposite direction and increases when speed decreases as is always in
    s.h.m
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    I dont know how to do the last question in queation 5. Does anyone know how to do?

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    (Original post by niallh1995)
    I got 250 moles in question 6 and I still got 1.5x10^7, using the molar approach. I found the moles of the added gas to be 60 and then I added them both to get 310, then I put them into the gas equation pV=nRT
    I did same

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    Will there be any unnofficial mark scheme

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    (Original post by sj_1995)
    I figured out how many moles of air was added using the data they gave you, added it to the moles we had worked out earlier in the question, they did pV=nRT with my new number of moles on the system
    That is exactly what I did, did you get 250 moles for the first bit?
 
 
 
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