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ANSWERS: OCR Physics B(Advancing Physics) G491~ 20th May 2014~ AS Physics Watch

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    OCR PHYSICS B(ADVANCING PHYSICS)~ Physics in Action!

    ANSWERS ARE ON POST 559


    Welcome to the OCR Physics B (G491) Exam Thread!

    Feel free to share notes/tips/materials/questions and thoughts on the exam.

    DATE OF EXAM: G491 - 1Hr - 20TH MAY 2014 - AM

    SYLLABUS:

    http://www.ocr.org.uk/Images/74952-specification.pdf

    - Scroll down to G491 ( PAGES 16-20)


    http://www.ocr.org.uk/qualifications...ics-h159-h559/

    ( Legacy OCR papers)

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    Hi there,

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    So i'm the only one sitting this exam?


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    Hey

    You're not the the only one
    How are you finding the electricity section i think its the part where i'm gna struggle in the exam
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    Oooh, I'm sitting this exam too.

    How are you finding revision?
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    (Original post by HappyCheesecake)
    Oooh, I'm sitting this exam too.

    How are you finding revision?
    Easier than G492!


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    (Original post by Mutleybm1996)
    Easier than G492!


    Posted from TSR Mobile
    Bits and pieces of it, definitely. But, as I'm also taking M1, G492 isn't as difficult.
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    (Original post by Mutleybm1996)
    Easier than G492!


    Posted from TSR Mobile
    I haven't started G491 yet but I've had a look at G492. Doing M1 helps with that.

    What do you think of the Advanced notice paper?
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    Help please! I'm stuck on question 5 of this paper:
    http://www.ocr.org.uk/Images/61672-q...-in-action.pdf

    The mark scheme answer is (at least 12 x 103 x 11 ) =1.3(2) x 105 (bit s-1) - But i don't know where to get the 103 from!

    Many thanks
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    (Original post by Kuna9613)
    Help please! I'm stuck on question 5 of this paper:
    http://www.ocr.org.uk/Images/61672-q...-in-action.pdf

    The mark scheme answer is (at least 12 x 103 x 11 ) =1.3(2) x 105 (bit s-1) - But i don't know where to get the 103 from!

    Many thanks
    Rate of transmission of a digital signal (bits per second)= sample per second x bits per sample

    Therefore it should be 12x10^3x11 which equals 132000, this is accepted by the mark scheme.

    I have no idea where the have got 103 from though
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    (Original post by ZahraP)
    Hey

    You're not the the only one
    How are you finding the electricity section i think its the part where i'm gna struggle in the exam
    Yep, thats the section im going to struggle in too. Its the most difficult. :confused:
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    http://www.ocr.org.uk/Images/144795-...-in-action.pdf

    can someone please explain how to do 10. c) i), the mark scheme is useless
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    (Original post by Kuna9613)
    Help please! I'm stuck on question 5 of this paper:
    http://www.ocr.org.uk/Images/61672-q...-in-action.pdf

    The mark scheme answer is (at least 12 x 103 x 11 ) =1.3(2) x 105 (bit s-1) - But i don't know where to get the 103 from!

    Many thanks
    (Original post by ZahraP)
    Rate of transmission of a digital signal (bits per second)= sample per second x bits per sample

    Therefore it should be 12x10^3x11 which equals 132000, this is accepted by the mark scheme.

    I have no idea where the have got 103 from though
    I think the markscheme means 10^3 (10 cubed = 1000) - because the sampling rate is 12kHz (12,000 Hz) - so thats where the "103" or "10^3" came from
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    (Original post by Knowing)
    I think the markscheme means 10^3 (10 cubed = 1000) - because the sampling rate is 12kHz (12,000 Hz) - so thats where the "103" or "10^3" came from
    That makes perfect sense; thank you
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    The grade boundaries are becoming ridiculous. In Jan 2010 the grade boundaries for an A was 36 and now it is like 45 for an A! I don't think the exams have got any easier through out the years so why have they increased the grade boundaries!?
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    Does anyone know how to do question 8c from the May 2011 paper? I don't understand what it says on the mark scheme or how to work it out.
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    (Original post by kobobo)
    Does anyone know how to do question 8c from the May 2011 paper? I don't understand what it says on the mark scheme or how to work it out.
    Since both shuttles are said to be identical, you can work out the ratio of how far they are away from the camera by measuring the size of the shuttle images.

    Size of Atlantis Image = 2.6cm
    Size of Endeavour Image = 0.7cm

    Since both shuttles are identical, we can say that the real object size for both = 1.

    The equation you need is Image size = Object size/Distance from camera.
    ^ Rearrange the equation, and input the values, and you get this:

    Distance of Atlantis from Camera = 1/2.6
    Distance of Endeavor from Camera = 1/0.7

    Now put these figures into the ratio equation they give you in the question:

    (1/0.7) / (1/2.6)

    This will give you ~3.7(14...etc.), which is the right answer

    Hope this makes sense
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    (Original post by Knowing)
    Since both shuttles are said to be identical, you can work out the ratio of how far they are away from the camera by measuring the size of the shuttle images.

    Size of Atlantis Image = 2.6cm
    Size of Endeavour Image = 0.7cm

    Since both shuttles are identical, we can say that the real object size for both = 1.

    The equation you need is Image size = Object size/Distance from camera.
    ^ Rearrange the equation, and input the values, and you get this:

    Distance of Atlantis from Camera = 1/2.6
    Distance of Endeavor from Camera = 1/0.7

    Now put these figures into the ratio equation they give you in the question:

    (1/0.7) / (1/2.6)

    This will give you ~3.7(14...etc.), which is the right answer

    Hope this makes sense
    Thank You so much!!!!!!!! I forgot to read that the space shuttles are identical, so once i measured the shuttles i was a bit confused what to do next. As the question says ratio would 3.71 be the answer?
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    Out of all my exams, this is the exam I am most worried for. The contextual nature of G491 makes picking information out of the exam quite difficult.

    In regards to past papers, I've been getting ~45-54.
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    (Original post by ZahraP)
    Hey

    You're not the the only one
    How are you finding the electricity section i think its the part where i'm gna struggle in the exam

    (Original post by HappyCheesecake)
    Oooh, I'm sitting this exam too.

    How are you finding revision?

    (Original post by H0PEL3SS)
    Bits and pieces of it, definitely. But, as I'm also taking M1, G492 isn't as difficult.

    (Original post by Kuna9613)
    Help please! I'm stuck on question 5 of this paper:
    http://www.ocr.org.uk/Images/61672-q...-in-action.pdf

    The mark scheme answer is (at least 12 x 103 x 11 ) =1.3(2) x 105 (bit s-1) - But i don't know where to get the 103 from!

    Many thanks

    (Original post by ZahraP)
    Rate of transmission of a digital signal (bits per second)= sample per second x bits per sample

    Therefore it should be 12x10^3x11 which equals 132000, this is accepted by the mark scheme.

    I have no idea where the have got 103 from though

    (Original post by stupendousp)
    http://www.ocr.org.uk/Images/144795-...-in-action.pdf

    can someone please explain how to do 10. c) i), the mark scheme is useless

    (Original post by Knowing)
    I think the markscheme means 10^3 (10 cubed = 1000) - because the sampling rate is 12kHz (12,000 Hz) - so thats where the "103" or "10^3" came from
    Are you folks still at school? Could you ask your teachers if we need to know how to apply image processing techniques such as smoothing,edge detection and noise removal using the formulae in the book? I emailed my teacher but she's not in school until after the exam apparently and i can't find it in the specification(which we all know is rubbish)
 
 
 
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