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

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    (Original post by Mehrdad jafari)
    ive got one today at 2 and tomorrow morning another
    1 tomorrow, 1 on Monday, 1 on Tuesday THEN phya4


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    My biggest worry is that I'm really prone to making silly mistakes that I'd otherwise avoid. I drop around 3-5 marks alone on these mistakes or just by not reading the question properly

    Also I'd advise getting the written paper out of the way first, because with the questions you either know it or you don't, compared to the multiple choice in which you may need more time to work out possible answers.
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    Is anyone revising AS stuff?


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    (Original post by CD223)
    Is anyone revising AS stuff?


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    Not really - a huge amount of the synoptic content that comes up in phya4 is on the data sheet, I've yet to come across anything that I couldn't remember so it doesn't seem worth it!
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    (Original post by Lau14)
    Not really - a huge amount of the synoptic content that comes up in phya4 is on the data sheet, I've yet to come across anything that I couldn't remember so it doesn't seem worth it!
    That's true - there was a question on fundamental nodes the other year though :/


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    I keep getting these ones wrong so if someone could help




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    (Original post by AR_95)
    I keep getting these ones wrong so if someone could help




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    If you're running out of time remember transformers are quite efficient but not 100% so it's usually the value under 98%


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    (Original post by gcsestuff)
    If you're running out of time remember transformers are quite efficient but not 100% so it's usually the value under 98%


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    Lol yeah you say that then AQA read this and stick an efficiency of 7% in this year's paper


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    How did you get Is = 0.5?
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    (Original post by AR_95)
    How did you get Is = 0.5?
    It gave values for the current in the question


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    (Original post by CD223)
    It gave values for the current in the question


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    ****ing hell like I said before I lose easy marks because i dont read the question

    I didnt even read the part about 0.5 , just stopped at 0.26
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    (Original post by CD223)
    Lol yeah you say that then AQA read this and stick an efficiency of 7% in this year's paper


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    Yeah someone somewhere will have a need for a inefficient transformer and they will base it on that


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    (Original post by AR_95)
    ****ing hell like I said before I lose easy marks because i dont read the question

    I didnt even read the part about 0.5 , just stopped at 0.26
    Just let that be a lesson for next week

    (Original post by gcsestuff)
    Yeah someone somewhere will have a need for a inefficient transformer and they will base it on that
    Haha, true!


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    part 2? Sometimes its the easy questions that get to me
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    (Original post by AR_95)


    part 2? Sometimes its the easy questions that get to me
    It's a uniform electric field so the potential varies linearly.

    

E = \dfrac{\Delta V}{\Delta d} = \dfrac{2.0V}{0.6m}

    

\Rightarrow E = 3.33 Vm^{-1}\ \text{(3 sf)}


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    Hi, can someone please shed some light on Jun14 MC QUESTION 24. Why is the answer A?
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    (Original post by TALAKHAN)
    Hi, can someone please shed some light on Jun14 MC QUESTION 24. Why is the answer A?
    The general equation for the induced emf is:

    

\displaystyle \epsilon = BAN \omega \sin (\omega t)

    Due to the nature of the sinusoidal variation in emf (when \displaystyle \sin (\omega t) = 1), its peak value is:

    

\displaystyle \epsilon_0 = BAN \omega

    As \displaystyle \omega = 2 \pi f, and flux linkage is equivalent to \displaystyle = BAN:

    Dividing both sides by omega gives the max flux linkage equal to:

    

\displaystyle BAN_{Max} = \dfrac{\epsilon_0}{\omega}

    

\displaystyle \therefore BAN_{Max} = \dfrac{\epsilon_0}{2 \pi f}


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    (Original post by CD223)
    The general equation for the induced emf is:

    

\displaystyle \epsilon = BAN \omega \sin (\omega t)

    Due to the nature of the sinusoidal variation in emf (when \displaystyle \sin (\omega t) = 1), its peak value is:

    

\displaystyle \epsilon_0 = BAN \omega

    As \displaystyle \omega = 2 \pi f, and flux linkage is equivalent to \displaystyle = BAN:

    Dividing both sides by omega gives the max flux linkage equal to:

    

\displaystyle BAN_{Max} = \dfrac{\epsilon_0}{\omega}

    

\displaystyle \therefore BAN_{Max} = \dfrac{\epsilon_0}{2 \pi f}


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    thanks
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    (Original post by MsFahima)
    thanks
    No problem! Good luck for next week


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