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

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    (Original post by NEWT0N)
    You didn't include the speed but I'll assume it's pi/2 because I remember solving this. So in 6s it goes round (pi/2)*6=3pi=2pi+pi radians, that is a complete circle followed by a semicircle. The complete circle makes the car return to its original position and the semicircle makes it go to a point antipodal to its current position, i.e. a displacement equal to the diameter of the circle, i.e. 0.8*2=1.6m.
    Oh, yeah I didn't notice, it is pi/2
    It makes sense now, thank you
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    (Original post by fruity97)
    how many sig figs are we actually meant to give answers to? I remember in AS my teacher kept saying 2sf (unless stated otherwise), but after doing past papers it seems it is 3sf for A2?
    If it doesn't say to give your answer to an appropriate number of significant figures, then either two or three is expected, no more or less - although you won't be penalised for more (I was reading this somewhere on the aqa site the other day, possibly in the spec or an examiners report). If it does say to give them to a specific number of significant figures, it should match the piece of data with the least significant figures as newton explained above.
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    Given Newton has joined the conversation, I'm disappointed at the lack of Faraday, Coulomb, Lenz...


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    Can someone please explain Jan '11 Q5(c) from Section B.

    I do not understand the mark scheme or the examiner's report.
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    (Original post by CD223)
    Given Newton has joined the conversation, I'm disappointed at the lack of Faraday, Coulomb, Lenz...


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    Dirac was here before

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    (Original post by NEWT0N)
    Robert Hooke, bogeyman of science. He is much too short to be standing on the shoulders of giants like me, the exalted demigod.
    How is your buddy Huygen. Freinds forever

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    Light is a particle, whoever says otherwise will get executed. Shutup Huygens.
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    (Original post by CD223)
    

\dfrac{8}{2} = \dfrac{x^2}{(60-x)^2}

    Where x is the distance to the 8nC charge.

    

\therefore 4 = \dfrac{x^2}{(60-x)^2}

    

\therefore 2 = \dfrac{x}{(60-x)}

    

\therefore 2(60-x) = x

    

\therefore 120 - 2x = x

    

\therefore 120 = 3x

    

\therefore 40 = x

    Does that help?


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    Hello, great working out here, but was just wondering how did you derive the expression. It looks like you've done Q/q=R/(d-R)^2.

    What formula did you use, thanks!
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    (Original post by NEWT0N)
    Light is a particle, whoever says otherwise will get executed. Shutup Huygens.
    And it goes faster in denser medium right?

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    (Original post by Sonnyjimisgod)
    Hello, great working out here, but was just wondering how did you derive the expression. It looks like you've done Q/q=R/(d-R)^2.

    What formula did you use, thanks!
    It's not a formula. If it's x meters from one charge and the distance between the charges is 60, then clearly the distance from the other charge is 60-x. (Because x+(60-x)=60 if that makes sense.)
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    (Original post by a.a.k)
    And it goes faster in denser medium right?

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    Sir, I frame no hypothesis.
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    (Original post by Fvthoms)

    The mark scheme for 2 a) (ii). for this question, don't need numerical advice, just a description of the thought process and the method taken to calculate the answer.
    Attachment 421397
    In the earlier part of the question you know that the polar orbiting satellite is in an orbit of radius of 7370km and so the satellite requires a certain amount of force to stay in its orbit as its weight is pulling it towards the earth.

    As stated in the mark scheme there are more than one approaches to finding the centripetal force. You are already given the time period of the orbit of the satellite and that is extremely useful because it indicates the time taken by the satellite to orbit around the earth without falling towards the earth. If this time was greater it would definitely fall.

    The quickest method is the one involving less steps of calculation as it would be the first one in the mark scheme. Because you have the time period you don't need to take account of the Inverse square law of gravity as as this would be needed when you are given the radius to work out that time period or vice versa. You could still use it but then you would have to look at the formula sheet to get the mass of the earth. So because you have the time period you can simply consider the satellite as an object tied to a string and swung in a horizontal circle with the same time period. Then using F=mw2r would give you the centripetal force which will be equal to the weight of the satellite at that height above the earth.

    You might be thinking working with F=mv2/r but then you would have to know the linear velocity of the satellite, that is the velocity at which the satellite travels the circumstance of the orbit. But because the angular velocity is constant you just need to use the first equation as i said above or the first one in the mark scheme.
    Le me know if you still have any questions




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    (Original post by NEWT0N)
    Sir, I frame no hypothesis.
    Kudos to science super star

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    (Original post by a.a.k)
    Kudos to science super star

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    I be not understanding y^{\varepsilon} word Kudos that you use, but mere science super star I am not, for I am a science nebula cloud.
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    (Original post by NEWT0N)
    I be not understanding y^{\varepsilon} word Kudos that you use, but mere science super star I am not, for I am a science nebula cloud.
    For the sake of your magnificiant hair we must stop this conversation as it serves no purpose for the physicians on this forum.
    If u dont believe me then good luck proving scientifically:ahee:

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    (Original post by NEWT0N)
    Sir, I frame no hypothesis.
    Yes! But you have to admit that you have not been able to discover the cause of those properties of gravity from phenomena.


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    (Original post by a.a.k)
    For the sake of your magnificiant hair we must stop this conversation as it serves no purpose for the physicians on this forum.
    If u dont believe me then good luck proving scientifically:ahee:

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    I have a rigorous proof by Alchemy.

    But again you are mistaken, for contrary to popular belief, that is not my hair but bits of my large surface area brain flowing out of my head.

    Your most humble servant,

    Is. Newton
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    (Original post by NEWT0N)
    I have a rigorous proof by Alchemy.

    But again you are mistaken, for contrary to popular belief, that is not my hair but bits of my large surface area brain flowing out of my head.

    Your most humble servant,

    Is. Newton
    Oneday a tree grew freshly new apples weighing a ton.

    It was a newton of apples

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    (Original post by Mehrdad jafari)
    Yes! But you have to admit that you have not been able to discover the cause of those properties of gravity from phenomena.


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    God is the first and last cause, a most subtle gravitational spirit that pervades all and lies hid in all gross bodies.
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    (Original post by a.a.k)
    Oneday a tree grew freshly new apples weighing a ton.

    It was a newton of apples

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    Ha ha ha, I have not become this merry since the time when one of my students had the audacity to ask "what use is this book Euclid's Elements?" Funny lad.
 
 
 
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