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    (Original post by ThatPerson2)
    0.018m is 1.8cm - think of the size of a bulb and the filament inside of it seems likely. I got 0.018m and rounded it to 0.02 (I generally do 2dp or 3sf) do you think this will count against me? I had 0.018m in the working.
    It might count against you, but I doubt it: it depends on the examiner and mark scheme.
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    (Original post by ThatPerson2)
    0.018m is 1.8cm - think of the size of a bulb and the filament inside of it seems likely. I got 0.018m and rounded it to 0.02 (I generally do 2dp or 3sf) do you think this will count against me? I had 0.018m in the working.
    They usually like the answers to be to at least 2 significant figures
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    (Original post by Alex .G.)
    Let's start the mark scheme?
    Yh go on !


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    Q1 Units

    Coulomb - As
    Conductance - AV^-1

    Q2

    3x10^8/1.7

    Q3

    3.2 or 4.8 ?

    Q4

    Q5

    Q6

    Q7

    Q8

    Q9
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    (Original post by jpetersgill)
    Yh go on !


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    Ok for
    Qu1: Power - AV
    Conductance: AV-1
    Charge: As

    Question 2 anyone? Also D, C, B for graphs as C had largest area using 1/2b*h for tensile strength
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    (Original post by Alex .G.)
    Q1 Units

    Coulomb - As
    Conductance - AV^-1

    Q2

    3x10^8/1.7

    Q3

    3.2 or 4.8 ?

    Q4

    Q5

    Q6

    Q7

    Q8

    Q9
    The other unit was power - IV


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    (Original post by HennersPD)
    Ok for
    Qu1: Power - AV
    Conductance: AV-1
    Charge: As

    Question 2 anyone? Also D, C, B for graphs as C had largest area using 1/2b*h for tensile strength
    Surely tensile strength is just A because it had the highest stress?


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    (Original post by Alex .G.)
    Q1 Units

    Coulomb - As
    Conductance - AV^-1

    Q2

    3x10^8/1.7

    Q3

    3.2 or 4.8 ?

    Q4

    Q5

    Q6

    Q7

    Q8

    Q9
    Q3 is 4.8 as it was 4-(-0.8) = 4+0.8 = 4.8D
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    (Original post by Alex .G.)
    Q1 Units

    Coulomb - As
    Conductance - AV^-1

    Q2

    3x10^8/1.7

    Q3

    3.2 or 4.8 ?

    Q4

    Q5

    Q6

    Q7

    Q8

    Q9
    Q3 was like input * 40/280 I think?


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    (Original post by HennersPD)
    Q3 is 4.8 as it was 4-(-0.8) = 4+0.8 = 4.8D
    Q3 was a potential divider I'm pretty sure.. But the lens one was 3.2D


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    Gerrow's Unofficial mark scheme.

    Add to it yourselves - I Can't remember the paper!


    Q1 Units

    Power AV
    Coulomb - As
    Conductance - AV^-1

    Q2

    3x10^8/1.7

    Q3

    3.2 or 4.8 ?

    Q4

    Q5

    Q6

    Q7

    Q8

    Q9
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    (Original post by HennersPD)
    Q3 is 4.8 as it was 4-(-0.8) = 4+0.8 = 4.8D
    Q3 = 3.2 D as 4.0 D is total curvature needed to be added to the incoming wavefronts and the lens adds 3.2 D and the curvature added by the eye is 0.8 D
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    Q4. 40/280 * 4.5V (not sure about the emf of the cell)
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    (Original post by Alex .G.)
    Gerrow's Unofficial mark scheme.

    Add to it yourselves - I Can't remember the paper!


    Q1 Units

    Power AV
    Coulomb - As
    Conductance - AV^-1

    Q2

    3x10^8/1.7

    Q3

    3.2 or 4.8 ?

    Q4

    Q5

    Q6

    Q7

    Q8

    Q9
    Let me just say what I remember

    The lens one:
    i) 1/1.25 = 0.8D
    ii) power = 3.2D

    The bits one:
    i) show that Q, 2^4 = 16
    ii) 500 x 300 x 4 / 8 = 75,000
    iii) 75,000 x 90 x 5 = 33,750,000

    The Scanning Tunnelling Microscope one:
    i) sensitivity = 580 pA nm^-1
    ii) diagram; one bump then two bumps next to each other
    iii) explanation: it'd scan the whole surface and produce a graph of all of it, wherever a "bump" on the graph was, an atom was

    The Sounds one:
    i) It's logarithmic because as you go to the right it increases with a constant factor of 10
    ii) frequency of highest amplitude in orchestra = 350 Hz (I think)
    iii) bandwidth = 9900 Hz

    The Potential Divider one:
    4.5 x 40/280 = 0.64 V
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    (Original post by Zain.chishty)
    Q4. 40/280 * 4.5V (not sure about the emf of the cell)
    Wouldnt it ve 40\280+40#4.5
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    (Original post by Zain.chishty)
    Q3 = 3.2 D as 4.0 D is total curvature needed to be added to the incoming wavefronts and the lens adds 3.2 D and the curvature added by the eye is 0.8 D
    But since u was negative so 1/f-0.8=4 so you add 0.8 = 4.8 surely?
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    (Original post by HennersPD)
    But since u was negative so 1/f-0.8=4 so you add 0.8 = 4.8 surely?
    Nah, -0.8 and -4 were both negative because they were on the same side of the lens in her eye
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    (Original post by LJflowrie)
    It probably was to be honest, hopefully i might get one method mark? I was rushing the first page because our invigilators hadn't given us formulae booklets before they started the exam so i didn't have one for the first 5 minutes which was really annoying...
    So did you get 4.8D for the power as u was negative?
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    (Original post by StrangeBanana)
    Let me just say what I remember

    The lens one:
    i) 1/1.25 = 0.8D
    ii) power = 3.2D

    The bits one:
    i) show that Q, 2^4 = 16
    ii) 500 x 300 x 4 / 8 = 75,000
    iii) 75,000 x 90 x 5 = 33,750,000

    The Scanning Tunnelling Microscope one:
    i) sensitivity = 580 pA nm^-1
    ii) diagram; one bump then two bumps next to each other
    iii) explanation: it'd scan the whole surface and produce a graph of all of it, wherever a "bump" on the graph was, an atom was

    The Sounds one:
    i) It's logarithmic because as you go to the right it increases with a constant factor of 10
    ii) frequency of highest amplitude in orchestra = 350 Hz (I think)
    iii) bandwidth = 9900 Hz

    The Potential Divider one:
    4.5 x 40/280 = 0.64 V
    For the bandwidth it was just largest - smallest?
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    Gerrow's Unofficial mark scheme.

    Add to it yourselves - I Can't remember the paper!


    Q1 Units

    Power AV
    Coulomb - As
    Conductance - AV^-1

    Q2

    3x10^8/1.7

    Q3

    1/1.25 = - 0.8D
    ii) power = -0.8 + (ans) = 4. So =4.8D


    Q4It's logarithmic because as you go to the right it increases with aconstant factor of 10
    ii) frequency of highest amplitude in orchestra =350 Hz (I think)
    iii) bandwidth = 9900 Hz

    (probably accepts 9500-10500)

    Q5 Voltage Resistance Power
    small voltage 3.14
    6 - 20.05

    The Potential Divider one:
    4.5 x 40/280 = 0.64 V

    Q6 (ish)The bits one:
    i) show that Q, 2^4 = 16
    ii) 500 x 300 x 4 / 8 = 75,000
    iii) 75,000 x 90 x 5 = 33,750,000


    Q7

    Q8

    Q9
    The Scanning Tunnelling Microscope one:
    i) sensitivity = 260 pA nm^-1
    ii) diagram; one bump then two bumps next to eachother
    iii) explanation: it'd scan the whole surface andproduce a graph of all of it, wherever a "bump" on the graph was, anatom was



    The Sounds one:


 
 
 
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