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Edexcel A2 Physics Unit 5 'Physics from Creation to Collapse' watch

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    thanks
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    (Original post by havinghoops)
    The answers to the papers in the attachments above.

    ALSO, can anyone explain to me the multichoice question in the specimen paper, starting:


    QUESTION 3: A child is playing on a swing. The graph shows how the displacement of the child varies with time.

    I know the axis are wrong, got that. But to find the answer, I tediously found the gradient at zero, so get something near to pi. Is there a simplier way of doing it. Both my physics teachers are drawing a blank
    Thanks for these, there really helpful
    And as for Q3)
    We will use the following equations:
    v=Aωsin(ωt)
    T=2π/ω

    It wants us to find the maximum velocity, and this will occur when sin(ωt)=1.
    So when sin(ωt)=1 , v=Aω

    Rearrange T=2π/ω
    into ω=2π/T

    From the graph we can see that T is 3 seconds.
    ω=2π/3
    and A is 1.5
    v=1.5ω

    Sub in ω for v=1.5ω
    v=1.5x2π/3
    v=π
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    (Original post by havinghoops)
    The answers to the papers in the attachments above.

    ALSO, can anyone explain to me the multichoice question in the specimen paper, starting:


    QUESTION 3: A child is playing on a swing. The graph shows how the displacement of the child varies with time.

    I know the axis are wrong, got that. But to find the answer, I tediously found the gradient at zero, so get something near to pi. Is there a simplier way of doing it. Both my physics teachers are drawing a blank
    I'm not sure if my way was the 'right' way, but what I did was look at the graph: it's a cosine curve; therefore, the velocity graph would be a sine curve. The point at which a sine curve is 1 is it's max. point, and also, pi.
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    (Original post by Uberuvy)
    Thanks for these, there really helpful
    And as for Q3)
    We will use the following equations:
    v=Aωsin(ωt)
    T=2π/ω

    It wants us to find the maximum velocity, and this will occur when sin(ωt)=1.
    So when sin(ωt)=1 , v=Aω

    Rearrange T=2π/ω
    into ω=2π/T

    From the graph we can see that T is 3 seconds.
    ω=2π/3
    and A is 1.5
    v=1.5ω

    Sub in ω for v=1.5ω
    v=1.5x2π/3
    v=π
    It's a multi-choice question, I don't think it's absolutely necessary to do all that!


    or maybe the way I did it was cheating. :ninja:
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    (Original post by unamed)
    It's a multi-choice question, I don't think it's absolutely necessary to do all that!


    or maybe the way I did it was cheating. :ninja:
    The way you did do it was cheating :p:
    Because if the amplitude would have been higher e.g 2 and not 1.5 then you would have got it wrong. It was just lucky that in this question the max value on a cosine graph and the max velocity was the exact same.

    But its a specimin exam so i doubt they would really ask you something like that right at the begining for 1 mark. However its a fairly simple calculation once you remember it as its the same for all types of SHM questions.
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    (Original post by Uberuvy)
    The way you did do it was cheating :p:
    Because if the amplitude would have been higher e.g 2 and not 1.5 then you would have got it wrong. It was just lucky that in this question the max value on a cosine graph and the max velocity was the exact same.

    But its a specimin exam so i doubt they would really ask you something like that right at the begining for 1 mark. However its a fairly simple calculation once you remember it as its the same for all types of SHM questions.
    Well, now I have been corrected, I won't try to cheat again!
    It is pretty simple - and boring, which is why I try to find ways around it.
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    Good luck everyone. This is gonna be one hell of an exam.
    Thank you all for your attachments.

    ENGLAND GOT RAPED ! :P
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    (Original post by unamed)
    Well, now I have been corrected, I won't try to cheat again!
    It is pretty simple - and boring, which is why I try to find ways around it.
    Right with you on that one lol. I don't feel prepared for the exam but il be as glad as hell to get it over and done with.
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    (Original post by samydude)
    Good luck everyone. This is gonna be one hell of an exam.
    Thank you all for your attachments.

    ENGLAND GOT RAPED ! :P
    :woo: :woo: :woo: :woo: :woo: :woo: :yes: :yes: :eek3: :eek3:
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    (Original post by BPat)
    Sorry about the double post but can someone give me the link to that Hodder site where you needed a username and password?

    It was posted on this thread but i can't find it.

    Thanks.
    http://www.hodderplus.co.uk/edexcelp...n.html?unit-5/

    useredexcelphysicsa2
    passgravity1
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    no one really answered my question on what is standard candles? and parallex
    • PS Helper
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    (Original post by BPat)
    Sorry about the double post but can someone give me the link to that Hodder site where you needed a username and password?

    It was posted on this thread but i can't find it.

    Thanks.
    http://www.hodderplus.co.uk/edexcelp...tml?index.html
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    (Original post by lefneosan)
    http://www.hodderplus.co.uk/edexcelp...n.html?unit-5/

    useredexcelphysicsa2
    passgravity1
    Thank you both so much!! Really appreciate it!
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    how do you define temperature?

    Something like a measure of the total internal energy of a substance?
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    (Original post by dana.h)
    no one really answered my question on what is standard candles? and parallex
    Standard candles are stars (variable stars, supernovas etc) whose properties mean that its luminosity can easily be worked out from other measurements. If we know its luminosity(L) and we can measure its flux(F), we can then substitue the values into the equation:
    F = \dfrac{L}{4\pi d^2}
    to find the value of d.

    Parallax is another way to find distance of stars but can only be used for close by stars (pretty use it was somin like only 650 lightyears away). You observe the star from opposite sides of the earths orbit about the sun and calculate the angle it makes via the amount it appears to move against more background distant stars. Once you know the angle you can stick it in abit of trig to find the distance.


    Hopefully that helps
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    I have a question regarding the 'madeup test 1'; question 10, b
    I'm slightly confused with the explanation.
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    (Original post by Kameo)
    how do you define temperature?

    Something like a measure of the total internal energy of a substance?
    Average kinetic energy of the molecules within an object.
    Internal energy is just the idea of random distribution of potential and kinetic energy
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    (Original post by unamed)
    I have a question regarding the 'madeup test 1'; question 10, b
    I'm slightly confused with the explanation.
    So its abit like this

    Pearth------Gx------Gy
    From Gx to Pearth we know that the wavelength was shifted by 28nm (684-656 = 28).

    And we also know that the velocity a galaxy is receeding is proportional to the distance away from us.
    Seeing as Gy is twice as far from Pearth than Gx is (as you can see in the above diagram XD ) its shifted by a factor of 2 so it will be 56nm longer (28x2 = 56).
    Hense add 56 to the original wavelength and it will give us an answer of 712nm.
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    (Original post by Kameo)
    how do you define temperature?

    Something like a measure of the total internal energy of a substance?
    When an object is heated, it's kinetic energy determines what temperature it has. The potential energy determines it's state. So, say you're heating ice. When it is changing state from ice to water, it will almost stay at the same temperature for the time period it takes for it to change from ice to water, as all the heat energy is going into the change in potential energy to change its state.

    So, the actual temperature of a substance is more related to its kinetic energy than the potential energy.

    Remember: during times of increase of temperature the heat energy supplied is going into Kinetic Energy component of the internal energy.

    And during changes of state it's going to potential energy. Because Kinetic Energy isn't changing during these times, the temperature remains the same.

    Oh, and thank you Uberuvy and unamed for your help
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    (Original post by Uberuvy)
    So its abit like this

    Pearth------Gx------Gy
    From Gx to Pearth we know that the wavelength was shifted by 28nm (684-656 = 28).

    And we also know that the velocity a galaxy is receeding is proportional to the distance away from us.
    Seeing as its twice as far (as you can see in the above diagram XD ) its shifted by a factor of 2 so it will be 56nm longer (28x2 = 56).
    Hense add 56 to the original wavelength and it will give us an answer of 712nm.
    Ah. I understand (thanks to your wonderfully accurate diagram!) now, I got to the 56, but I just didn't know where to go from there. :facepalm:
 
 
 
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