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Size:  269.2 KB Thanks, I have no idea how to do question 4 and 5, could someone explain it to me, thanks.
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    (Original post by JerryG)
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Size:  269.2 KB Thanks, I have no idea how to do question 4 and 5, could someone explain it to me, thanks.
    I think Pythagoras can give you the magnitude and trig the direction for Q4
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    (Original post by 123Master321)
    I think Pythagoras can give you the magnitude and trig the direction for Q4
    There are no distances on the Axed though, just speeds
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    (Original post by JerryG)
    There are no distances on the Axed though, just speeds
    Yh but its the same if instead of velocity you used force. Imagine the combined effect of going 50m/s north and 30m/s east if that helps
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    (Original post by 123Master321)
    Yh but its the same if instead of velocity you used force. Imagine the combined effect of going 50m/s north and 30m/s east if that helps
    So how would I use that to work out direction?
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    (Original post by JerryG)
    So how would I use that to work out direction?
    tan(theta) = 5/3
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    Direction is the angle, once you find out all 3 sides of the right angles triangle by using pythagoraus then you can use tan x = o/h to find the angle
    O = side opposite angle
    H = hypotneuse
    Then do inverse tan x to find x

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    (Original post by 123Master321)
    tan(theta) = 5/3

    Right thanks, any idea on the next question?
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    (Original post by JerryG)
    Right thanks, any idea on the next question?
    What is N and N0
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    (Original post by 123Master321)
    What is N and N0
    I have no idea, I've just been set this work in prep for starting a level.
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    (Original post by JerryG)
    I have no idea, I've just been set this work in prep for starting a level.
    Heard of logarithms before and I think it was from a-level maths. So probably go on exam solutions and go to the index and find logs, then see if you can turn what it's saying into a straight line equation then you u should be good from there
    my bridging work was just some advanced gcse stuffy and a few easy
    a-level stuff like suvat


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    (Original post by JerryG)
    Right thanks, any idea on the next question?
    To start with, log both sides to get:

    log(N) = log(N0)-hg

    See what that gives you in terms of y = mx+c
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    (Original post by Flame Alchemist)
    To start with, log both sides to get:

    log(N) = -hg log(N0)

    See what that gives you in terms of y = mx+c
    Instead of log, you want ln right?
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    (Original post by 123Master321)
    Instead of log, you want ln right?
    Assuming natural base, yeah. It's all used confusingly interchangeably in different texts. But you're right that's it's usually best to specify.
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    Logs in physics. This looks like more of a maths question. I did experimental data questions like this in maths.


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    I looked online, but can only find logs with numbers, never things like N and NO
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    (Original post by JerryG)
    I looked online, but can only find logs with numbers, never things like N and NO
    Okay, since you're struggling to gain a footing, I'll explain the answer to you.

    Logarithms are essentially the inverse of exponentiation, so we can use them to construct direct, linear proportionalities from exponential relations. Know that the natural logarithm (ln) is log of base e, such that ln(e) = 1.

    In this equation, N is the variable to track, whereas N0 is the initial value of N. It's not clear which of h and g is variable or constant, but let's assume g is variable.

    ln(N) = ln(N0e^-hg)

    ln(N) = ln(N0) + ln(e^-hg)

    ln(N) = ln(N0) - hg ln(e)

    ln(N) = ln(N0) - hg

    It is evident that:
    ln(N) ∝ -g
    Thus, we can plot a straight line of ln(N) against -g, with gradient h and y-intercept ln(N0), allowing to easily determine those values if necessary.
 
 
 
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