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    (Original post by particlestudent)
    Max 4, I think you need to mention potential for the last 2 marks.
    In what way?
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    (Original post by Dann_a)
    Yes - whenever you have to talk about the direction of a charge in the magnetic field you can say "Because of Flemings left hand rule it will in go in X direction (or the opposite if its negatively charged"
    Right hand rule he meant, that's left :') don't think iv ever seen right come up in an exam tbh
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    (Original post by alevelstresss)
    if anyone wants, here are my PHYA4 notes
    Hey, really useful notes thanks for these. You see in the capcitor section, it says E=Vc + Vr. What does the E(more of an epsilon i think) mean?
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    (Original post by boyyo)
    Hey, really useful notes thanks for these. You see in the capcitor section, it says E=Vc + Vr. What does the E(more of an epsilon i think) mean?
    thats the emf of the cell
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    (Original post by alevelstresss)
    Find the speed at which the water moves

    speed is equivalent to a length per second, so speed = volume / area

    mass = density x volume

    momentum = mass x velocity
    Still getting it wrong. Is volume cross-sectional area x the rate per second? so 7.2x10^-4 x 2x10^-4 = 1.44x10^-7 ?
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    (Original post by SirRaza97)
    Like this. Sorry its in landscape.

    Would it look this is if the right one was positive and the left one was negatively charged?
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    (Original post by Ayaz789)
    So it'll look like this?
    You haven't attached a picture .
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    Can someone show me their working for why the answer is B? Thanks
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    Attached Images
     
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    (Original post by particlestudent)
    They are both positive.


    Check page 22, I kind of explained it there .
    https://kaiserscience.files.wordpres...e-patterns.gif
    Would it not be the top right image?
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    (Original post by xMillnsy)
    Still getting it wrong. Is volume cross-sectional area x the rate per second? so 7.2x10^-4 x 2x10^-4 = 1.44x10^-7 ?
    length = volume / area
    length = (2.0x10-4)/(7.2x10-4)
    length = (5/18) m

    length per second is equivalent to a velocity

    v = (5/18) ms-1

    mass = volume x density
    mass = 1000(2.0x10-4)
    mass = 0.2 kg

    m = (1/5) kg



    therefore using p = mv

    p = (1/5) x (5/18)
    p = 1/18
    p = 0.056 kg ms-1
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    (Original post by boyyo)
    Hey, really useful notes thanks for these. You see in the capcitor section, it says E=Vc + Vr. What does the E(more of an epsilon i think) mean?
    EMF of the power source in the circuit.
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    (Original post by PiTheta97)
    Can someone show me their working for why the answer is B? Thanks
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    Final flux linkage-initial

    BAN-BANcos(50)
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    (Original post by particlestudent)
    Final flux linkage-initial

    BAN-BANcos(50)
    Made the mistake of just calculating BANcos50. Cheers

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    (Original post by PiTheta97)
    Can someone show me their working for why the answer is B? Thanks
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    Flux linkage is greatest when the coil is perpendicular to the field lines.

    When theta=50, Flux Linkage= BAN cos(theta) = 3.8x10^-3
    When theta =0, Flux linkage = 5.9x10^-3

    5.9x10^-3 - 3.8x10^-3 = 2.1 x10^-3 Wb Turns
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    (Original post by alevelstresss)
    length = volume / area
    length = (2.0x10-4)/(7.2x10-4)
    length = (5/18) m

    length per second is equivalent to a velocity

    v = (5/18) ms-1

    mass = volume x density
    mass = 1000(2.0x10-4)
    mass = 0.2 kg

    m = (1/5) kg



    therefore using p = mv

    p = (1/5) x (5/18)
    p = 1/18
    p = 0.056 kg ms-1
    ahh I get it now, thanks for your help
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    (Original post by Ayaz789)
    Haha its fine, draw it again and show me?
    Like this? The repulsion area is closer to the "smaller" charge.

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    (Original post by particlestudent)
    You haven't attached a picture .
    Well is the picture you attached on page 22?
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    hey physics people if I have two planets and want to work out the potential at a point between the two planets what do i do? Calculate the potentials due to each planet separately and add right? Also in the formula book gravitational potential is negative, do you sum the magnitude, or sum the negatives to get an even more negative value. And if the point is in the centre between the two planets?
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    (Original post by abro1089)
    hey physics people if I have two planets and want to work out the potential at a point between the two planets what do i do? Calculate the potentials due to each planet separately and add right? Also in the formula book gravitational potential is negative, do you sum the magnitude, or sum the negatives to get an even more negative value. And if the point is in the centre between the two planets?
    Been trying to get an answer for this....Im still at a loss lol
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    (Original post by SirRaza97)
    Like this? The repulsion area is closer to the "smaller" charge.

    Ahh okay thanks
 
 
 
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