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    (Original post by Stonebridge)
    If you find k using 66N, and then find k using 19N, you get two slightly different answers. The difference between them is the uncertainty in the answer.
    How did you find the average when you did this initially?
    Okay so K will just equal force/area so as you say, i did the force for 19N now so, 19/2*10^-5 = 9.5*10^5

    for 66N, k = 66/2*10^-5 = 3.3*10^6

    average value of k = 2.12*10^6 N m^-2

    First of all is that correct? if so, could you please show me how to do the percentage uncertainty.
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    If it was me i'd do this.

    1) work out k for all 3 values using  k = \frac{2\pi F}{A \times \alpha}

    This gives

     k_1 = 9.5\times10^5

     k_2 = 1\times10^6

     k_3 = 1.65\times10^6

    Now the average is

     k_{av} = \frac{k_1 + k_2 + k_3}{3} = 1.2\times10^6

    The error on this can be found using

     err = \frac{1}{N} \sum |k_i - k_{av}|

    so in this case

     err = \frac{1}{3} (|k_1 - k_{av}| + |k_2 - k_{av}| + |k_3 - k_{av}|)

    Which gives

     err = \frac{1}{3} (|9.5\times10^5 - 1.2\times10^6| + |1\times10^6 - 1.2\times10^6| + |1.65\times10^6 - 1.2\times10^6|) = 3\times10^5

    As a percentage error:

     \% err = 100\times\frac{err}{k_{av}} = 100\times\frac{3\times10^5}{1.2\  times10^6} = 25\%

    So we get 25%. Not sure how your markscheme gets 10%...
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    (Original post by spread_logic_not_hate)
    If it was me i'd do this.

    1) work out k for all 3 values using  k = \frac{2\pi F}{A \times \alpha}

    This gives

     k_1 = 9.5\times10^5

     k_2 = 1\times10^6

     k_3 = 1.65\times10^6

    Now the average is

     k_{av} = \frac{k_1 + k_2 + k_3}{3} = 1.2\times10^6

    The error on this can be found using

     err = \frac{1}{N} \sum |k_i - k_{av}|

    so in this case

     err = \frac{1}{3} (|k_1 - k_{av}| + |k_2 - k_{av}| + |k_3 - k_{av}|)

    Which gives

     err = \frac{1}{3} (|9.5\times10^5 - 1.2\times10^6| + |1\times10^6 - 1.2\times10^6| + |1.65\times10^6 - 1.2\times10^6|) = 3\times10^5

    As a percentage error:

     \% err = 100\times\frac{err}{k_{av}} = 100\times\frac{3\times10^5}{1.2\  times10^6} = 25\%

    So we get 25%. Not sure how your markscheme gets 10%...
    Yeah, this makes much more sense than the mark scheme for percentage uncertainty.... i will go with this. Thanks for that mate, i have given you +rep
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    http://www.edexcel.com/migrationdocu...e_20080617.pdf

    question 2(a) ------ how do i figure out what the precision is????? and for the graph it's a power law isn't it, so a log-log graph will do? this basic part gets me, what goes on the y-axis and what goes on the x-axis????
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    Yeah, this makes much more sense than the mark scheme for percentage uncertainty.... i will go with this. Thanks for that mate, i have given you +rep
    Lol just realised I made a stupid numerical mistake! Should be

     k_3 = 1.1 \times 10^6

    which gives

     k_{av} = 1.02\times10^6

    and

     \% err = \frac{1.83\times 10^5}{1.02\times10^6} = 18\%

    Sorry about that
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    (Original post by spread_logic_not_hate)
    Lol just realised I made a stupid numerical mistake! Should be

     k_3 = 1.1 \times 10^6

    which gives

     k_{av} = 1.02\times10^6

    and

     \% err = \frac{1.83\times 10^5}{1.02\times10^6} = 18\%

    Sorry about that
    Don't worry, you've still got your rep :p: if you could take a look at my next question, that would be greatly appreciated

    oh wait, to one of the questions above, does the frequency go on the x-axis cause that is the input variable and controls the current right?
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    question 2(a) ------ how do i figure out what the precision is????? and for the graph it's a power law isn't it, so a log-log graph will do? this basic part gets me, what goes on the y-axis and what goes on the x-axis????
    I think that the precision of the current readings could be described as 'how close to the true value they actually are'. In this case we see that the readings are all quoted to 2s.f, so the precision of each is (presumably)  \pm 0.005 mA . I'm not 100% about this kind of thing myself, but that is how I think it goes!

    For the graph bit, whenever I get these type of questions I do always do this.

    Consider that the gradient of any graph is dy/dx. This means that if I plot a graph like the one in the question, I have a gradient found like this:

     \frac{dy}{dx} = \frac{\frac{1}{I^2}}{\frac{1}{f^  2}} = \frac{f^2}{I^2}

    I then look to get the equation I have started with into this form. In this case

     \frac{f^2}{I^2} = k + (\frac{fR}{V})^2

    Now I can see what a graph of these variables will give me.

     grad = k + (\frac{fR}{V})^2

    This means that I should have  grad \alpha (\frac{fR}{V})^2

    where  \alpha means 'is proportional to'.

    This is something I can test, meaning that this graph will be usable to verify that this equation is describing my results.

    I hope that makes sense..?
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    oh wait, to one of the questions above, does the frequency go on the x-axis cause that is the input variable and controls the current right?
    The independant variable goes on the x axis - this is the one that is being changed by the experimenter. The dependent variable goes on the y axis - this is the one that changes because the experimenter changed the dependent variable, i.e. the experimenter does not directly change this variable.

    As you rightly say, here frequency is on the x axis because it is the independant variable as the experimenter is setting it himself.
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    (Original post by spread_logic_not_hate)
    I think that the precision of the current readings could be described as 'how close to the true value they actually are'. In this case we see that the readings are all quoted to 2s.f, so the precision of each is (presumably)  \pm 0.005 mA . I'm not 100% about this kind of thing myself, but that is how I think it goes!

    For the graph bit, whenever I get these type of questions I do always do this.

    Consider that the gradient of any graph is dy/dx. This means that if I plot a graph like the one in the question, I have a gradient found like this:

     \frac{dy}{dx} = \frac{\frac{1}{I^2}}{\frac{1}{f^  2}} = \frac{f^2}{I^2}

    I then look to get the equation I have started with into this form. In this case

     \frac{f^2}{I^2} = k + (\frac{fR}{V})^2

    Now I can see what a graph of these variables will give me.

     grad = k + (\frac{fR}{V})^2

    This means that I should have  grad \alpha (\frac{fR}{V})^2

    where  \alpha means 'is proportional to'.

    This is something I can test, meaning that this graph will be usable to verify that this equation is describing my results.

    I hope that makes sense..?

    Yeah that is going in depth, can't we just use y = mx + c???? since frequency is the independent variable that will be on the x-axis and therefore k will have to be the gradient wouldn't it?
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    Yeah that is going in depth, can't we just use y = mx + c???? since frequency is the independent variable that will be on the x-axis and therefore k will have to be the gradient wouldn't it?
    I get you, as in

     y = \frac{1}{I^2}

    and

     x = \frac{1}{f^2}

    so identifying with y = mx+c gives m = k, i.e. the gradient is k. So the student should look for a linear graph to confirm the relationship (as k is constant, the gradient should be constant - i.e linear)

    That sounds like you're doing it right and that I might have looked into this a bit too deeply :facepalm:
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    also next part of question

    "Discuss whether this suggests the teacher has too much fat."

    lmao.
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    (Original post by spread_logic_not_hate)
    also next part of question

    "Discuss whether this suggests the teacher has too much fat."

    lmao.
    Lol yeah, that question cracked me up :p: just because it's high resistance don't mean the teacher is fat, could also be tall as well lol...... you know for the precision for the current, it's rounded to 2 d.p like 0.30, so if it was 0.300, would it be then +/- 0.0005 mA?

    edit - now i am confused how you calculate the resistance????
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    you know for the precision for the current, it's rounded to 2 d.p like 0.30, so if it was 0.300, would it be then +/- 0.0005 mA?
    Yeah that's right.

    edit - now i am confused how you calculate the resistance????
    Looking at y = mx+c, the axis intercept c will be equal to

     c = (\frac{R}{V})^2

    so from the y-intercept on your graph you can find c, hence find R.

    And yeah lets hope that teacher is tall
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    (Original post by spread_logic_not_hate)
    Yeah that's right.



    Looking at y = mx+c, the axis intercept c will be equal to

     c = (\frac{R}{V})^2

    so from the y-intercept on your graph you can find c, hence find R.

    And yeah lets hope that teacher is tall
    Lol, yeah let's hope that he/she is...... thank you so much for this mate, it has helped me along a lot on the graph questions.... this is worth a lot of marks in the exam, so don't wanna throw away easy marks lol.... that's all i need for now, i think i have everything else covered, thanks once again mate and will be giving you +rep for this again
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    Lol, yeah let's hope that he/she is...... thank you so much for this mate, it has helped me along a lot on the graph questions.... this is worth a lot of marks in the exam, so don't wanna throw away easy marks lol.... that's all i need for now, i think i have everything else covered, thanks once again mate and will be giving you +rep for this again
    no probs, certainly don't be wasting those easy marks! Thanks for the +rep also
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    (Original post by spread_logic_not_hate)
    no probs, certainly don't be wasting those easy marks! Thanks for the +rep also
    Hey there mate, this is another graph question from a different paper that is confusing me. Here it goes, i will just type it all out.

    "Element: Boron, Silicon, Strontium, Tin, Uranium
    Nucleon number: 10, 28, 88, 120, 238
    Nuclear radius R/ 10^-15m: 2.69, 3.93, 5.34, 5.99, 7.74

    It is predicted that for most nuclei R = RoA^1/3 where Ro is a constant.

    Plot a suitable graph to verify this relationship."

    So we use y = mx + c cause we want it in that,

    logR = LogRo + 1/3Loga-----> gradient equals 1/3 and Ro = c(intercept)..... in the question they only give you ONE extra column but wouldn't you need two extra columns cause it's LogA and LogR...... Also what value do you get for R cause mine are coming out as -14 for all of them and the mark scheme has the values

    (2.15/3.04/4.45/4.93/6.20) that has confused me
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    this is really confusing me how they do this and the mark scheme doesn't show how they do it.... so if someone could answer with the answer asap please!!!
 
 
 
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