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    (Original post by Stickyelmo)
    I just count the number of squares
    Attachment 228288

    Ok, i now understand why I was confused, when the mark scheme and you mentioned counting squares, i presumed you meant the small ones and thought it was ridiculous to count all of them!! But I have (finally) caught on that the larger squares were meant! I've done the question now and got it right.. was having a very stupid moment. Thanks for your help x
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    Any idea what experiment questions may arise? I've revised SHC for a solid and liquid and an SHM one came up in Jan so i'm thinking maybe damping or Centripetal force?
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    (Original post by JustFacts)
    Any idea what experiment questions may arise? I've revised SHC for a solid and liquid and an SHM one came up in Jan so i'm thinking maybe damping or Centripetal force?
    what are the answers to the experiments you've revised? would you mind writing them out?
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    Yeahh Would be great if anyone has a list of all the experiments and their answers
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    (Original post by Converse girl)
    When do we use a=v^2/r?
    One example I can think of is when an object is fly in a vertical circle, at the top weight provides all the centripetal force. So to work out the velocity or radius you'd use a=v^2/r a being free fall 9.81
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    (Original post by Kreayshawn)
    what are the answers to the experiments you've revised? would you mind writing them out?

    (Original post by Exotica)
    Yeahh Would be great if anyone has a list of all the experiments and their answers

    The thing is i doubt they'd ask the same experiment twice except for SHC.
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    ShC experiment is just where you placed a FIXED mass of solid or liquid inside an insulating material and heat it to about 10k . This is done by a heating coil if you are using a liquid and an electric heater if its a solid. You measure the change in temperature and the energy by connecting your heat source to a circuit with a ammeter,voltmeter, resistor as E equals current times voltage times time. Rearrange the equation to find the SHC. A limit would be its not a it would be difficult to create a thermal equilibrm so heat may be lost. Brownian Where smoke particles in still air appear to move haphazardly because they are being stuck by 'invisible' air molecules which shows that the air molecules are hitting the smoke particles from every side and are extremely small. The continuous movement shows that they are being hit at constant intervals.
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    Hi guys,

    Is 'derivation of 1/3 pv squared' and 'derivation of 3/2 k t' part of the OCR G484 Physics syllabus?? There hasn't really been a question on this so would I need to know this??

    Cheers
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    (Original post by spanishovenfork)
    Hi guys,

    Is 'derivation of 1/3 pv squared' and 'derivation of 3/2 k t' part of the OCR G484 Physics syllabus?? There hasn't really been a question on this so would I need to know this??

    Cheers
    The only equation you have to derive(according to the spec) is Keplars third law. That question has been asked before once a few papers back so I would make sure you're able to do that.
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    (Original post by JustFacts)
    ShC experiment is just where you placed a FIXED mass of solid or liquid inside an insulating material and heat it to about 10k . This is done by a heating coil if you are using a liquid and an electric heater if its a solid. You measure the change in temperature and the energy by connecting your heat source to a circuit with a ammeter,voltmeter, resistor as E equals current times voltage times time. Rearrange the equation to find the SHC. A limit would be its not a it would be difficult to create a thermal equilibrm so heat may be lost. Brownian Where smoke particles in still air appear to move haphazardly because they are being stuck by 'invisible' air molecules which shows that the air molecules are hitting the smoke particles from every side and are extremely small. The continuous movement shows that they are being hit at constant intervals.
    Just to add to the Brownian Motion part - the observations support kinetic theory of gases by showing that particles move in a rapid random motion (you could also argue that collision time is neglible)

    And for SHC an improvement you could find the SHC in a vacuum
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    (Original post by martynsteel)
    Just to add to the Brownian Motion part - the observations support kinetic theory of gases by showing that particles move in a rapid random motion (you could also argue that collision time is neglible)

    And for SHC an improvement you could find the SHC in a vacuum
    Thanks for that Brownian motion point. I'll make sure I memorise that . And yeh I know you could do it in a vacuum but they always ask for a limit to the experiment but having it done in a vacuum would be an improvement. Also if they ask for resonance or damping experiment what would you say?
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    Guys I am incredibly stuck on June 2011 Q 5 b ii. How do I calculate the number of collisions? I don't understand the mark scheme either. ):

    Nvm, read the question wrong xD
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    Can anyone help me with June 2012 Q3bi please. Really confuzed by how the markscheme tells to calculate force F as the weight isnt equal to the cetripetal force. ~_~ thanks anyone please
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    do we ever actually have to use
    x = A sin(2pft)
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    (Original post by Kreayshawn)
    do we ever actually have to use
    x = A sin(2pft)
    If it's on the specification, probably. But they've never asked a question using it
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    (Original post by 9876543211)
    Can anyone help me with June 2012 Q3bi please. Really confuzed by how the markscheme tells to calculate force F as the weight isnt equal to the cetripetal force. ~_~ thanks anyone please
    I think I can see your confusion, but in fact the 'weight' is the centripetal force.

    The mass they are using in F=mg is the mass of the slotted masses which are providing the centripetal force (take another look at the diagram), NOT the weight of the bung - as indicated by the mark scheme in the guidance comments.

    So the weight of the slotted masses provides tension in the nylon thread, and this tension is the centripetal force.


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    (Original post by Rhodopsin94)
    I think I can see your confusion, but in fact the 'weight' is the centripetal force.

    The mass they are using in F=mg is the mass of the slotted masses which are providing the centripetal force (take another look at the diagram), NOT the weight of the bung - as indicated by the mark scheme in the guidance comments.

    So the weight of the slotted masses provides tension in the nylon thread, and this tension is the centripetal force.


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    OHhhhhhh thanks very much, yeah i never noticed that on the diagram :L. yeah that makes perfect sense now thhhaaannkkk yooouu
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    (Original post by Kreayshawn)
    do we ever actually have to use
    x = A sin(2pft)
    It depends on the starting point of the oscillations.

    Usually, x = Acos(2pift) is used since a mass is displaced from it's rest position, and then undergoes simple harmonic motion. Notice in this situation, you start off with your 'maximum' displacement (the maximum displacement is the amplitude you started with) so at t=0, there is max displacement therefore a cosine curve.

    Technically it's impossible to start simple harmonic motion with x = Asin(2pift) because if it is already in it's rest position then it won't move. But if you released a mass, and then timed oscillations from the point where the mass passes it's equilibrium position, then you could use x = Asin(2pift).

    If given a graph, check whether it is a sine or cosine curve.


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    (Original post by eggfriedrice)
    One example I can think of is when an object is fly in a vertical circle, at the top weight provides all the centripetal force. So to work out the velocity or radius you'd use a=v^2/r a being free fall 9.81
    okay i found some questions so i am going over them thanks
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    Am I right in saying, that on a temperature-time graph for ice melting, when in a solid state the specific heat capacity is less so the gradient of that point in the graph is steeper, but when it becomes a liquid the specific heat capacity is more so the gradient of that point on the graph is less steep?
 
 
 
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