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    (Original post by ArchieL)
    Haha, you know it! Geoff's retiring this year
    :afraid:

    Also the section C notes were Tim's work this year, he rustled them out as soon as the prerelease came out! The physics dept. at csfc are so hyped on caffeine it's unreal.
    :elefant:
    I know, it sucks!
    Ok awesome! Are there any answers for the section C notes around?
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    (Original post by ArchieL)
    There are some great YT vids on the subject, but..
    Uncertainty = http://www.thestudentroom.co.uk/showthread.php?t=227989
    % Uncertainty = https://www.youtube.com/watch?v=1dTn2pt5PuA
    Resolution = Smallest observable change: In the case of a digital display, the amount of d.p, in a graph it's how many units 1 box represents.
    Thankyou very much
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    liomeeringca,

    Read the MS again, it says 169m^2s^-2 and the same for 900.

    I just used v^2 = u^2 +2as, then rearrange for s.
    I apologise, the reason they have those units is because they've been squared:
    (m/s)^2 = (m^2/s^2)
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    (Original post by AndItsOllie)
    I know, it sucks!
    Ok awesome! Are there any answers for the section C notes around?
    I'm afraid not mate, I've got a few if you need them! Just ask
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    (Original post by ArchieL)
    I apologise, the reason they have those units is because they've been squared:
    (m/s)^2 = (m^2/s^2)
    I understand now, thank you for your help
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    (Original post by marseille_h)
    This might seem straightforward but I'm a bit confused:

    A car has an acceleration of 0.86g. Calculate the time it takes to reach a velocity of 27m.s-1.
    g=9.81

    v=u+at
    so t= (v-u)/a

    is g=acceleration in this case and the acceleration of the car they give we take it as u and v=27 ?


    Acceleration would be 0.86 x g, so 0.86 x 9.81. I assume you would take u as 0 and v as 27 and then use v=u+at to find t.
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    (Original post by jacobfun)
    Acceleration would be 0.86 x g, so 0.86 x 9.81. I assume you would take u as 0 and v as 27 and then use v=u+at to find t.
    which formula do we use to get acceleration and do 0.86 x g ?
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    (Original post by marseille_h)
    which formula do we use to get acceleration and do 0.86 x g ?
    No formula. 'g' is shorthand for the acceleration due to gravity (on earth) ~ 9.81ms-2
    => an acceleration of 0.86g = (0.86)(9.81) = 8.4366 ~ 8.44
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    (Original post by ArchieL)
    No formula. 'g' is shorthand for the acceleration due to gravity (on earth) ~ 9.81ms-2
    => an acceleration of 0.86g = (0.86)(9.81) = 8.4366 ~ 8.44
    Oh i understand i thought it was a wierd unit. THANKS
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    How does everyone feel about potential dividers, thermisters and general electricity questions then? I'm bricking myself
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    (Original post by ArchieL)
    How does everyone feel about potential dividers, thermisters and general electricity questions then? I'm bricking myself
    None of this in unit 2 i think..
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    (Original post by liomeeringca)
    How would one calculate gravity from a set of results like that?
    V^2=U^2+2as. F=ma, and the component for force in the direction of the ramp is gsin(theta). Therefore you can rearrange that to a=gsin(theta)
    V^2=U^2 + 2gsin(theta)s. U=0, so you can rearrange it to g=V^2 divided by 2sin(theta)s
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    I saw someone had already answered this but I'm still unsure how to determine the value for g from v2= u2+2as

    Can someone explain step by step please?

    I hate all these maths based questions
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    (Original post by ArchieL)
    How does everyone feel about potential dividers, thermisters and general electricity questions then? I'm bricking myself
    Wasn't this G491?
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    (Original post by marseille_h)
    None of this in unit 2 i think..
    As far as I was aware, section C is a synopsis unit (of sorts?)
    Also I've got a question on it on some prelrelease questions set by the department at college, so I'd guess there might be a little on it? Don't get me wrong, I'd be over the moon if there wasn't! Haha
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    (Original post by King Hotpie)
    V^2=U^2+2as. F=ma, and the component for force in the direction of the ramp is gsin(theta). Therefore you can rearrange that to a=gsin(theta)
    V^2=U^2 + 2gsin(theta)s. U=0, so you can rearrange it to g=V^2 divided by 2sin(theta)s
    Thank you so so much,
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    (Original post by ArchieL)
    As far as I was aware, section C is a synopsis unit (of sorts?)
    Also I've got a question on it on some prelrelease questions set by the department at college, so I'd guess there might be a little on it? Don't get me wrong, I'd be over the moon if there wasn't! Haha
    I haven't looked a potential dividers in aaaaaaages! I really hope there's nothing on it, like you say D:
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    For the pre-release material on measuring the speed of light... why can it be assumed that the 'hotspot' where the chocolate melted is an antinode? :confused:
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    (Original post by seharnaznaz)
    I saw someone had already answered this but I'm still unsure how to determine the value for g from v2= u2+2as

    Can someone explain step by step please?

    I hate all these maths based questions
    With the trolley on the ramp, it has an acceleration, g, acting directly downwards. Construct a triangle with this force, another force c acting ϴ degrees to it originating from the floor (so they meet). Then add another from the top of grunning parallel to the slope. This cuts the line with c and forms a triangle. From the cosine rule, cosϴ=opp/hyp, so =a/g. Rearrange to get a=gsinϴ. (It's best to draw a diagram of all component forces, stripped down).

    If we then substitute this into the equation v2=u2+2as with u=0, we get that result (v2 =2.g.l.sinϴ). Also it's probably worth noting that plotting a graph with v^2 against sinϴ gets us with a gradient of 2.g.l
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    (Original post by seharnaznaz)
    For the pre-release material on measuring the speed of light... why can it be assumed that the 'hotspot' where the chocolate melted is an antinode? :confused:
    Antinodes is where the intestity is the highest compared to nodes where it is =0
 
 
 
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