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    (Original post by shloke123)
    Yh as carla said, I doubt it will come up. Just make sure you know your quality of measuremen,t and all of g492 and the exam should be simple enough
    What do we actually need to know about quality of measurement?

    And it's not worth worrying about all the information rate questions on the Matthew Arnold 1a questions then?

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    I don't know, everything you can find out tbh
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    For the first article surely the only stuff they can ask us about is frequency and amplitude??

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    (Original post by emfp21)
    For the first article surely the only stuff they can ask us about is frequency and amplitude??

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    And inertia, time period, data storage and resolution.


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    And quality of measurement!!!
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    And response time and sensitivity.


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    (Original post by emfp21)
    What do we actually need to know about quality of measurement?

    And it's not worth worrying about all the information rate questions on the Matthew Arnold 1a questions then?

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    Things like how to reduce uncertainty and being able to suggest systematic errors/ random errors in an experiment.

    No no I would concentrate on the wave section for the oscilloscope part of the advanced notice. It'll probably be calculating frequency/ period of a wave or something.
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    (Original post by Knowing)
    Yup, it says we'll be given a fresh copy.

    My school's organisation is crap though, they lost the formula booklets so we had to start 20 minutes late for G491

    Oh I see, thanks, my copy didn't have the front page
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    what is resolution?
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    (Original post by Hicko)
    what is resolution?
    The smallest change a measuring device can detect
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    In the equation d sin θ = nλ how you get rid of the sin??

    This question I'm doing now for example has this in the markscheme:

    First: They have rearranged the equation to
    sin θ = nλ / d
    Second: then they have put the numbers in: 2 x 5 x 10^ -7 / 1.6 x 10^ -6 which equals 0.625.

    Obviously this number (0.625) is what
    sin θ equals but the question wants just θ. The mark scheme then just says, and so θ = 39

    How did they get from
    sin θ = 0.625 to θ = 39 ??? I've found this problem in a few papers and still don't understand how it works!!
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    9rd? What you been smoking?
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    (Original post by liberata)
    In the equation d sin θ = nλ how you get rid of the sin??

    This question I'm doing now for example has this in the markscheme:

    First: They have rearranged the equation to
    sin θ = nλ / d
    Second: then they have put the numbers in: 2 x 5 x 10^ -7 / 1.6 x 10^ -6 which equals 0.625.

    Obviously this number (0.625) is what
    sin θ equals but the question wants just θ. The mark scheme then just says, and so θ = 39

    How did they get from
    sin θ = 0.625 to θ = 39 ??? I've found this problem in a few papers and still don't understand how it works!!
    You just arcsin it.


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    (Original post by liberata)
    In the equation d sin θ = nλ how you get rid of the sin??

    This question I'm doing now for example has this in the markscheme:

    First: They have rearranged the equation to
    sin θ = nλ / d
    Second: then they have put the numbers in: 2 x 5 x 10^ -7 / 1.6 x 10^ -6 which equals 0.625.

    Obviously this number (0.625) is what
    sin θ equals but the question wants just θ. The mark scheme then just says, and so θ = 39

    How did they get from
    sin θ = 0.625 to θ = 39 ??? I've found this problem in a few papers and still don't understand how it works!!
    Inverse sin

    (Sin^-1(theta))
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    (Original post by liberata)
    In the equation d sin θ = nλ how you get rid of the sin??

    This question I'm doing now for example has this in the markscheme:

    First: They have rearranged the equation to
    sin θ = nλ / d
    Second: then they have put the numbers in: 2 x 5 x 10^ -7 / 1.6 x 10^ -6 which equals 0.625.

    Obviously this number (0.625) is what
    sin θ equals but the question wants just θ. The mark scheme then just says, and so θ = 39

    How did they get from
    sin θ = 0.625 to θ = 39 ??? I've found this problem in a few papers and still don't understand how it works!!
    Press the sin^-1 button on your calculation ( shift sin) and then ans or type in 0.625 and then you should get the angle :3
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    (Original post by alicerose2)
    Press the sin^-1 button on your calculation ( shift sin) and then ans or type in 0.625 and then you should get the angle :3
    I've just typed in sin^-1 (using the shift key), then next to that (0.625) and when I've pressed enter it's given me 0.675131532?? Is my calculator in the wrong mode maybe? I just did it on a different calculator and got 39 so I've not typed it in wrong :/
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    (Original post by liberata)
    I've just typed in sin^-1 (using the shift key), then next to that (0.625) and when I've pressed enter it's given me 0.675131532?? Is my calculator in the wrong mode maybe? I just did it on a different calculator and got 39 so I've not typed it in wrong :/
    Its probably in radians
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    (Original post by Carla Huynh)
    Its probably in radians
    It's in normal mode I've just checked... maybe I just have a dodgy calculator
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    reset your calculator mate
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    what types of questions might come up for the 3rd section of the insert lads?
 
 
 
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