Recent content by Shaye

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    SHM - Pendulum and Mass on a spring

    I have only be able to write something like: 2x(2π√(l/g)) = 2π√(m/k) 2π is a constant therefore; 2x(√(l/g)) = √(m/k) You could square both sides; 2^2x(l/g) = (m/k) But now I'm lost as to how to proceed. PS- Book answer is B Thanks
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    Calculate the charge on an oil droplet

    Yeah I should probably start to do this going forward @PeroK Good tip!
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    Calculate the charge on an oil droplet

    The book gives an answer of Q = 3.2 x 10^-19C I get an answer of 6.67 x 10^-19C. Workings below:
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    Hurricane forces — Comparing the force from a 60 mph wind to a 120 mph wind

    Thanks @Orodruin and @haruspex. After much hand holding I eventually got it now lol
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    Hurricane forces — Comparing the force from a 60 mph wind to a 120 mph wind

    The only thing I can think of is p=m/v and m=pv and ke = 1/2(pv)*v^2...but I'm still at a loss
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    Hurricane forces — Comparing the force from a 60 mph wind to a 120 mph wind

    The marker wrote that the answer is 4 and it's because m and v double. I don't understand how m doubles??
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    Momentum Conservation: How to Reconcile a Negative Value?

    Thanks everyone! A DOH! moment for me. Keeping track of the vectors is very useful
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    Momentum Conservation: How to Reconcile a Negative Value?

    Maybe a silly question but on the above question using the conservation of momentum: momentum before firing (0) = momentum after firing (55*35)+(M*2.5) If I re-range the above it's M = -(55*35)/2.5 = -770kg. I can I reconcile that minus sign (basically get rid of it)? Thanks
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    Calculating moments about a support point

    I mean what I found similar was the fact that you could take moments about any point on the ruler example e.g. at [0.2,0.8], [0.4.0.6] etc. and when you add the moments up they should always equal the couple...
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    Calculating moments about a support point

    The books solution says to take moments about x2 to find the force of x1 as 110N to 2 significant figures. Then by equilibrium: X1 + X2 = 250N therefore x2 = 250N - 110N = 140N. I understood this to be (4/9)*250 = 110 2 s.f. Similar to the reasoning of a couple i.e. (Couple = F*d) for...
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    Calculating moments about a support point

    Thanks Delta2. Sorry I got confused. I didn't think about the CoM/CoG where all the weight is acting. But now it's fine :-) The sum of the moments about the CoM should total 250N
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    Calculating moments about a support point

    Moment about X2 to calculate force at X1: x1 * 9 = (250 * 2) Therefore, x1 = 500/9 = 55.5N The book however gives force at x1 as 110N. So I figured I have not understood a concept somewhere
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    What is the maximum frictional force?

    Phew! Thanks. I thought I was going mad
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    What is the maximum frictional force?

    Advance Physics for You https://www.amazon.com/dp/1408527375/?tag=pfamazon01-20
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