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AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread] Watch

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    Would love the Bridge SHM one
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    If an alternating current just keeps goes backwards and forwards then wouldn't the force just keep going backwards and forwards perpendicular to it?


    I can sort of see that the field would still go in the same direction using LHR but how does it induce an alternating voltage if it's not changing direction?
    how does the voltage size change from one side of the transformer to the other using transformer and where does the difference go??

    Is the alternating current actually alternating or do they mean going left on one side of the coil and right in the other?? Then surely it's not alternating it's just a continuous flow in the same direction
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    (Original post by Kennethm)
    I dont think saying that dampening changes the natural frequency is correct? Someone tell me if Im wrong. The rest looks pretty good here is what I wrote:

    if a question went something like - 'bridges blah blah resonance blah blah dampening blah blah, answer must include blah blah'

    bridges must be fitted with dampening systems because bridges are exposed to many external forces acting upon it at any given time. Without dampening these forces may set the bridge in violent (or damaging) motion when the natural frequency of the bridge matches the frequency of the periodic force applied to the bridge - for example people walking on the bridge with a matched frequency (soldiers marching.)

    Pretty sure people could add to this. Hope it helps too.
    Sounds good. What I would add: The bridge can resonate when driven by wind at one of the natural frequencies of the bridge (note: yes, bridges, as well as things like musical instruments, have multiple natural frequencies). This will cause the bridge to vibrate with rapidly increasing amplitude and get destroyed. Soldiers should be asked to break step when they march across the bridge because if they mark in step with each other they can cause the bridge to resonate. etc.
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    Do we have to use our rounded values from the previous part of a question, or can we use the exact value.
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    (Original post by EEngel95)
    Do we have to use our rounded values from the previous part of a question, or can we use the exact value.
    Use the exact unrounded value in all intermediate calculations. Round up to an appropriate number of sig figs when stating answers.
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    (Original post by Kennethm)
    Thats already been done. I know they've done transformers for 6 markers like 3 times now but there is a lot of content on transformers where as there isnt much to say about EMI. There might be a question on a rotating coil and how/when/where the magnetic flux is a maximum compared to when the emf is a maximum. That could confuse a lot of students as some people don't explicitly know the difference.
    And could you explain the answer to that please??
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    Wtf? I think it could be down to the wording of the question...
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    (Original post by Sbarron)
    And could you explain the answer to that please??
    emf induced is at a pi/2 phase difference between flux linkage. When magnitude of flux linkage is maximum emf induced is 0, and vice versa. The emf induced is maximum when the coil is parallel to the field lines because the coil cuts the flux lines perpendicularly at that instant. Flux linkage is easier to guess intuitively as it becomes maximal when the coil is perpendicular to the field.
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    (Original post by NEWT0N)
    emf induced is at a pi/2 phase difference between flux linkage. When magnitude of flux linkage is maximum emf induced is 0, and vice versa. The emf induced is maximum when the coil is parallel to the field lines because the coil cuts the flux lines perpendicularly at that instant. Flux linkage is easier to guess intuitively as it becomes maximal when the coil is perpendicular to the field.
    Spot on, and thank you newton for contributing to that SHM I posted.
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    they might be useful
    https://onedrive.live.com/redir?resi...hint=folder%2c
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    Just looked over my paper from last year and damn was I marked harshly! I didn't get one mark on the six marker!
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    What happens in a uniform electric field, when both plates are positive and also when :

    1) Both plates have the same PD

    2) One plate has a higher positive PD then the other positive PD

    And also when both are negative

    1) of same PD

    2) One plate is more negative then the other
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    (Original post by AR_95)
    What happens in a uniform electric field, when both plates are positive and also when :

    1) Both plates have the same PD

    2) One plate has a higher positive PD then the other positive PD

    And also when both are negative

    1) of same PD

    2) One plate is more negative then the other
    1) If both plates have the same PD, there is no electric field.
    2) There is an electric field in the direction of the higher value PD to the lower value PD. V is the difference between the values.

    1) Same as above
    2) Same as above, will be in the direction of the least negative to the more negative value.

    Anyone feel free to correct me if I'm wrong because I didn't look this up. I think it's correct though
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    (Original post by AR_95)
    What happens in a uniform electric field, when both plates are positive and also when :

    1) Both plates have the same PD

    2) One plate has a higher positive PD then the other positive PD

    And also when both are negative

    1) of same PD

    2) One plate is more negative then the other
    When one plate has a higher PD than another plate, the direction of the electric field lines go from higher PD -> Lower PD. This is vice versa for two negatively charged plates, but in this instance the electric field lines will go from less negative -> More negative.


    When two plates have the same PD, whether it be negative or positive, the net effect of the electric field at a midpoint between them will be zero.
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    (Original post by JJBinn)
    1) If both plates have the same PD, there is no electric field.
    2) There is an electric field in the direction of the higher value PD to the lower value PD. V is the difference between the values.

    1) Same as above
    2) Same as above, will be in the direction of the least negative to the more negative value.

    Anyone feel free to correct me if I'm wrong because I didn't look this up. I think it's correct though
    Yep, electric field lines are the paths that a small positive charge would follow if placed in the field
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    Great video on resonance

    https://www.youtube.com/watch?v=csYUGJLtZnw
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    (Original post by NEWT0N)
    emf induced is at a pi/2 phase difference between flux linkage. When magnitude of flux linkage is maximum emf induced is 0, and vice versa. The emf induced is maximum when the coil is parallel to the field lines because the coil cuts the flux lines perpendicularly at that instant. Flux linkage is easier to guess intuitively as it becomes maximal when the coil is perpendicular to the field.
    Something like this?
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    (Original post by ShutUpLegs)
    Just looked over my paper from last year and damn was I marked harshly! I didn't get one mark on the six marker!
    I just did the written, you really didn't need to even touch the 6 marker to get an A*.

    Sometimes its better to retain or gain marks elsewhere as opposed to spending 10 minutes on 6 marks. The way I see 6 markers are if they are this difficult and demanding then don't bother unless you've done everything else as perfectly as possible and you have the time for it.

    Some 6 markers aren't demanding at all and its easy to get 4/5 ish marks from them at first glance which is fine.
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    (Original post by palaseum9)
    Something like this?
    Yup
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    (Original post by huniibehi)
    Does anybody get how they concluded 38s for this question?

    Posted from TSR Mobile
    I hate those type of questions. Basically, the pendulum of period 1.9 seconds completes more oscillations within a given time than the pendulum of 2.0 seconds.

    One way in calculating the time when they both come back into phase with each other is by having n as the number of oscillations completed by the 2.0 second pendulum... because you know the 1.9 second pendulum will always complete more oscillations than the 2,0 second pendulum you can formulate it...

    (n+1)1.9=2.0n

    If you re-arrange to find n it should give you n=19, so 2.0 second pendulum completes 19 oscillations before they both come back in phase and if you find the product of the number of oscillations and time period it gives 38 seconds.
 
 
 
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