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Edexcel Physics Unit 2 "Physics at work" June 2013 watch

  • View Poll Results: The last question - Does resistance increase or decrease?
    It increases ( using V=IR or some other method)
    70.73%
    It decreases using the 'lattice vibrations' theory
    29.27%

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    (Original post by Gunner121)
    Thank you so much and can please explain "Q 12 (c)"
    http://www.edexcel.com/migrationdocu...e_20090115.pdf
    Ignore what I said, I thought it was a standing wave.

    I'm off for the night guys. Good luck tomorrow!

    ~Charlie
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    In definition of polarized light the mark scheme says the vibrations are perpendicular to direction of wave? what does this mean ? does this not mean a transverse wave?
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    (Original post by jollygood)
    In definition of polarized light the mark scheme says the vibrations are perpendicular to direction of wave? what does this mean ? does this not mean a transverse wave?
    I think it means a transverse wave, as a transverse wave is defined as the particles move perpendicular to the direction of travel. But polarised light is when they vibrate in one plane only.


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    Good luck tomorrow everyone
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    2009 May question 15, part b. anyone? i dont get how is the current = 6/30? from where we get 6V? :/

    please anyone, help!
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    From part a), you know that the pd across 5ohm resistor is 3V?
    Therefore, the P.d. across 30ohm resistor has to be 6V (9-6) from Kirchoff's law.

    Therefore from ohm's law, I=V/R so I=6/30 = 0.2A
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    so what about the resistor R ? :/
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    Can someone kindly send me the link to January 2013 mark scheme. cant find it at the moment. that would be a great help. please
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    (Original post by jollygood)
    Can someone kindly send me the link to January 2013 mark scheme. cant find it at the moment. that would be a great help. please
    http://www.mediafire.com/?m7bm1p7951lsb
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    (Original post by arshia)
    so what about the resistor R ? :/
    Current across R: 0.6A-0.2A = 0.4A

    R=V/I so R = 6V/0.4A = 15ohm
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    (Original post by jollygood)
    Can someone kindly send me the link to January 2013 mark scheme. cant find it at the moment. that would be a great help. please
    hi
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    Can someone please tell me how to work out january 2011 question 2? please help!
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    Good luck for everyone


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    (Original post by Water_fall)
    Can someone please tell me how to work out january 2011 question 2? please help!
    Total resistance in the parallel combination :

    (1/(200 x 10^3)) + (1/(200 x 10^3)) = 100000 ohm

    Total resistance in the circuit : 100000 + 200000 = 300000 ohm

    Current in the circuit : I= V/R : 12/300000 = (4 x 10^-5) A

    Then you can calculate PD in the parallel combination : V=IR :

    (4 x 10^-5) x100000 = 4V

    Hope you can understand
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    jan 2013 paper?
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    (Original post by Freddy-Francis)
    jan 2013 paper?
    6PH02_01_que_20130118.pdf6PH02_01_rms_20130307.pdf
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    (Original post by Water_fall)
    Can someone please tell me how to work out january 2011 question 2? please help!
    Let the resistance of the single resistor be R1, and the resistance of the parallel combination be R2.

    R_1 = 200000 \ \Omega

    R_2 = \left( \dfrac{1}{200000} + \dfrac{1}{200000} \right)^{-1} = 100000 \ \Omega

    Now, there are two ways you could do this.

    You could either find the current using:

    V = IR

    V = I(R_1 + R_2)

    12 = I(200000+100000)

    I = 4 \times 10^{-5} \ \mathrm{A}

    Then, (V2 is the pd across the parallel combination)

    V_2 = IR_2

    V_2 = 4 \times 10^{-5} \times 100000 = 4 \ \mathrm{V}



    Or, alternatively, you could use the fact that:

    V \propto R

    Which means that,

    V_1 : V_2 = R_1 : R_2

    R_1 : R_2 = V_1 : V_2 = 200000:100000 = 2:1

    Using knowledge of basic ratios, and the fact that V1+V2=12, you can easily arrive at the correct answer here.
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    good luck everyone
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    good luck every1!! lets smash this paper. praying its better than unit 1!!
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    (Original post by lipgloss9)
    Total resistance in the parallel combination :

    (1/(200 x 10^3)) + (1/(200 x 10^3)) = 100000 ohm

    Total resistance in the circuit : 100000 + 200000 = 300000 ohm

    Current in the circuit : I= V/R : 12/300000 = (4 x 10^-5) A

    Then you can calculate PD in the parallel combination : V=IR :

    (4 x 10^-5) x100000 = 4V

    Hope you can understand
    Thank you so much!!
 
 
 
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