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    A pendulum clock loses 1.5 minutes per day. If the pendulum is 0.85m long, by how much must its length be changed to make the clock run on time?

    Kinda confused really, in this case what would the period (T) be and the length, would it be O.85+-X?? Im confused.
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    Shouldn't the period be one optimally, because it should swing once per second?
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    ye... what else can you tell me..
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    T=2pi x root (L/g)

    So for the pendulum currently, the time period is 1.8495...

    In one day, there are (24x60x60)/T swings = 46715 swings

    In a day, 1.5 minutes are lost so 90/46715 seconds are lost per swing = 1.93x10^-3

    So, for each swing, the T must be changed by 1.93x10^-3 seconds to make it run on time.

    Using the formula, you end up with L=0.851772... or 0.84822... depending on whether you added or subtracted the time (it doesn't matter which)

    So you need to change the length by 0.001772m

    I'm pretty sure this is right but if anyone has any other ideas, feel free to say.
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    T=2pi x root (L/g)
    what equation is it i havnt seen this in my book
    doing OCR physics a
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    LaTeX, people.
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    huh?
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    http://www.thestudentroom.co.uk/wiki/LaTex

    It makes it much easier to understand what you're asking.

    edit: Or answering with.
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    (Original post by D-Day)
    http://www.thestudentroom.co.uk/wiki/LaTex

    It makes it much easier to understand what you're asking.

    edit: Or answering with.
    There's really not much need for LaTeX in this situation, it's quite clear what the quantities are :p:

    Purist :p:
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    (Original post by ashy)
    There's really not much need for LaTeX in this situation, it's quite clear what the quantities are :p:

    Purist :p:
    If posting in LaTeX becomes habit for more people, I'll not have to decipher "y=x^2/3+9x-5/2x+17."
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    (Original post by D-Day)
    If posting in LaTeX becomes habit for more people, I'll not have to decipher "y=x^2/3+9x-5/2x+17."
    Good point.

    You just don't need to use latex to type out y = 2x though :p:
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    T = 2 \pi \times \sqrt{ \frac{L}{g} }
    is this something i missed in my book? couldnt find it
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    (Original post by ashy)
    Good point.

    You just don't need to use latex to type out y = 2x though :p:
    Habit. I hardly notice myself doing it anymore :o:
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    (Original post by D-Day)
    Habit. I hardly notice myself doing it anymore :o:
    You don't notice yourself typing out [latex]blah[/latex] every time? :eek:

    It's so irritating.
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    lol my thread has been hijacked, jus jokin, thanx for all the help
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    (Original post by SSze)
    T = 2 \pi \times \sqrt{ \frac{L}{g} }
    is this something i missed in my book? couldnt find it
    Yes :p:
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    (Original post by D-Day)
    Just testing something unrelated.

    \int^{20,000}_0 e^{−2x} − {x − 10 000}^2 + 100000000
    Woah, your latex is totally narly duude.
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    (Original post by Toxic)
    lol my thread has been hijacked, jus jokin, thanx for all the help
    Yeah, sorry. Get a couple of geeks together and this is the result...
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    :five:
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    (Original post by D-Day)
    Yes :p:
    checked twice, no such equation:confused:
 
 
 
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