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

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    (Original post by Mehrdad jafari)
    but there wasn't anything on transformers and E.M though
    Yeah I know was just suggesting topics that could potentially come up in the paper this year, as it has not come up in a while. Do you have the ISA paper?


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    (Original post by Jimmy20002012)
    Oh really, could you upload it again or link me to the page your uploaded to. Thanks. They have not done circular motion, momentum, transformers and E.M Induction in a while so hoping these topics are kinda certain to come up in section B I think.


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    here it is
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  1. File Type: pdf AQA-PHY6T-P14-Test-Jun14.pdf (407.8 KB, 308 views)
  2. File Type: pdf AQA-PHY6T-P14-Final Marking Guidelines v2.pdf (318.7 KB, 117 views)
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    (Original post by Mehrdad jafari)
    here it is
    Thanks so much


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    (Original post by Jimmy20002012)
    Thanks so much


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    no worries
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    Has anyone ever heard of the formula:

    

B = \dfrac{{\mu_0}NI}{L}

    For magnetic field strength?

    Came across it in a really old paper yesterday.


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    Do you need an A* in the ISA for an overall A*?
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    (Original post by frankiejayx)
    Do you need an A* in the ISA for an overall A*?
    No, you just need 90% average UMS at A2 and 80% total UMS for both years. so 270 UMS this year, which could possibly be achieved with an E in the EMPA/ISA if you got full UMS in the written papers!
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    Please can somebody explain question 11 to me? Thanks. Also, for an A* in Physics you need to average 90% in the papers AND the coursework? That's a bit of a downer, thought it was just the papers.

    http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF
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    (Original post by Lau14)
    No, you just need 90% average UMS at A2 and 80% total UMS for both years. so 270 UMS this year, which could possibly be achieved with an E in the EMPA/ISA if you got full UMS in the written papers!
    I'd love to know if anyone has achieved that.


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    (Original post by CD223)
    I'd love to know if anyone has achieved that.


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    Lol, it is yet to be achieved


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    (Original post by JJBinn)
    Please can somebody explain question 11 to me? Thanks. Also, for an A* in Physics you need to average 90% in the papers AND the coursework? That's a bit of a downer, thought it was just the papers.

    http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF
    For an object to to appear weightless at the equator the speed of the rotation of the earth around itself must increase (therefore time period decrease), to such an extent that rotating faster would result in the object be lifted up the surface of the earth as the object is changing direction. At this time the centripetal acceleration is equal to the free fall acceleration of the earth.
    Name:  ImageUploadedByStudent Room1432299210.479456.jpg
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    (Original post by Mehrdad jafari)
    For an object to to appear weightless at the equator the speed of the rotation of the earth around itself must increase (therefore time period decrease), to such an extent that rotating faster would result in the object be lifted up the surface of the earth as the object is changing direction. At this time the centripetal acceleration is equal to the free fall acceleration of the earth.
    Name:  ImageUploadedByStudent Room1432299210.479456.jpg
Views: 59
Size:  122.3 KB


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    Thanks, I didn't know that. So at the point where objects appear weightless, the centripetal force is the same as the force of weight due to gravity on the object? However I thought both forces act in the same direction, which is confusing me a little.
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    (Original post by JJBinn)
    Thanks, I didn't know that. So at the point where objects appear weightless, the centripetal force is the same as the force of weight due to gravity on the object? However I thought both forces act in the same direction, which is confusing me a little.
    No worries. That's very good question. We've been told that always the centripetal force is acting towards the centre of the rotation, that's not the correct way of saying it. Rather, for an object to be in a circular motion the centripetal force "must act towards the centre". If not the the object will not remain in circular motion. So in the case of rotating an object with a string the tension in the string is certainly acting towards the centre of the rotation because you are holding the string. If the string snaps then the object will not keep rotating in a circle.
    In the case of an object rotating at the equator the centripetal force must act towards the centre which is essentially the weight of the object. But as the rotation gets faster the centripetal force , which would be the "inertia" of the object, overcomes its weight and this time there is no such a thing as the centripetal force


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    (Original post by CD223)
    Has anyone ever heard of the formula:

    

B = \dfrac{{\mu_0}NI}{L}

    For magnetic field strength?

    Came across it in a really old paper yesterday.


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    It's in the "advanced physics for you" book which covers the old a level syllabus. I think it's called Ampere's law, there's a nice experiment about it that I came across a while ago: http://www.cyberphysics.co.uk/topics...experiment.htm
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    Do we need to know the total energy formula for a system undergoing shm? E_T=E_K+E_P=\frac{1}{2}mA^2\omeg  a^2

    Edit: Also what about E_P for a mass on a spring? : E_P=\frac{1}{2}kx^2 (k=stiffness constant)
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    (Original post by JJBinn)
    Thanks, I didn't know that. So at the point where objects appear weightless, the centripetal force is the same as the force of weight due to gravity on the object? However I thought both forces act in the same direction, which is confusing me a little.
    Sorry, I've just seen an error in the last sentence in my explanation which i didn't mean to say.

    "But as the rotation gets faster the centripetal force , which would be the "inertia" of the object, overcomes its weight and this time there is no such a thing as the centripetal force"

    This should be something:
    But as the rotation gets faster the inertia of the body overcomes the weight. At this time there is no such a thing as the centripetal force. So the inertia is not the centripetal force


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    (Original post by Paul Dirac)
    Do we need to know the total energy formula for a system undergoing shm? E_T=E_K+E_P=\frac{1}{2}mA^2\omeg  a^2

    Edit: Also what about E_P for a mass on a spring? : E_P=\frac{1}{2}kx^2 (k=stiffness constant)
    We do need to know the first one but the second one is useful to know and it's a short cut


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    (Original post by Mehrdad jafari)
    No worries. That's very good question. We've been told that always the centripetal force is acting towards the centre of the rotation, that's not the correct way of saying it. Rather, for an object to be in a circular motion the centripetal force "must act towards the centre". If not the the object will not remain in circular motion. So in the case of rotating an object with a string the tension in the string is certainly acting towards the centre of the rotation because you are holding the string. If the string snaps then the object will not keep rotating in a circle.
    In the case of an object rotating at the equator the centripetal force must act towards the centre which is essentially the weight of the object. But as the rotation gets faster the centripetal force , which would be the "inertia" of the object, overcomes its weight and this time there is no such a thing as the centripetal force


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    For objects to appear weightless, R (the reaction force) always equals zero, right? Meaning only weight in this case provides the centripetal force?


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    (Original post by Paul Dirac)
    It's in the "advanced physics for you" book which covers the old a level syllabus. I think it's called Ampere's law, there's a nice experiment about it that I came across a while ago: http://www.cyberphysics.co.uk/topics...experiment.htm
    Thank you for this!


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    (Original post by Paul Dirac)
    Do we need to know the total energy formula for a system undergoing shm? E_T=E_K+E_P=\frac{1}{2}mA^2\omeg  a^2

    Edit: Also what about E_P for a mass on a spring? : E_P=\frac{1}{2}kx^2 (k=stiffness constant)
    In a past paper there was a question where you doubled f and m and were asked the effect on the total energy of the system. In that case it multiplied it by 8. That equation explains why.


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