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    (Original post by Music With Rocks)
    It kind of makes sense? haha I can't argue with the fact you got the right value

    But then why was part (i) not 0.1/0.27?

    I feel like the +12mm should make no difference seen as the oscillation starts from the right doesn't it, but I have no idea as you can tell
    Oh yh good point, we just did part i normally. Im not sure
    And I only got 0.7 cos you showed the answer lol. Maybe someone else knows....
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    What do you guys predict the 6 marker to be on because it's what I struggle on tbh


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    (Original post by Cheesecake Ali)
    What do you guys predict the 6 marker to be on because it's what I struggle on tbh


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    My guess is it will be on transformers
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    Hey I am a bit confused by another question can anyone help me? for part (a) I calculate the potential gradient as 5.0jkg^-^1m(which is given) then for part (b) I know the gravitational field strength is the negative of the potential gradient so therefore equals  -5.0Nkg^-^1

    However the answer at the back gives it as just  5.0Nkg^-^1

    Why is it not negative?

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    (Original post by Music With Rocks)
    for part (b) I know the gravitational field strength is the negative of the potential gradient so therefore equals  -5.0Nkg^-^1

    However the answer at the back gives it as just  5.0Nkg^-^1

    Why is it not negative?

    Question: Name:  physics gravity question.jpg
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    How can a gravitational field strength be negative? If we're saying that higher field strength means higher density of field lines, how can you have a negative density of field lines?

    Does it say in the book that field strength is the negative value of potential gradient?
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    (Original post by MintyMilk)
    How can a gravitational field strength be negative? If we're saying that higher field strength means higher density of field lines, how can you have a negative density of field lines?

    Does it say in the book that field strength is the negative value of potential gradient?
    Yeah it does

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    It is written literally just above the exercise so I assumed it applied
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    (Original post by Music With Rocks)
    Yeah it does

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    It is written literally just above the exercise so I assumed it applied
    what book is that?
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    (Original post by jarjarmonkey)
    what book is that?
    AQA Physics A textbook, published by Oxford

    It is the standard one I think, this photo is from page 58

    EDIT: this one
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    (Original post by Music With Rocks)
    Hey I am a bit confused by another question can anyone help me? for part (a) I calculate the potential gradient as 5.0jkg^-^1m(which is given) then for part (b) I know the gravitational field strength is the negative of the potential gradient so therefore equals  -5.0Nkg^-^1

    However the answer at the back gives it as just  5.0Nkg^-^1

    Why is it not negative?

    Question: Name:  physics gravity question.jpg
Views: 321
Size:  49.7 KB
    I was about to help you with this, but I dont even get the first question :/
    How did you calculate the potential gradient? Could you also briefly explain what the potential gradient is please? Im sorry to be a pain, I didnt even know I didnt know this, thank you for putting it up on here
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    (Original post by boyyo)
    I was about to help you with this, but I dont even get the first question :/
    How did you calculate the potential gradient? Could you also briefly explain what the potential gradient is please? Im sorry to be a pain, I didnt even know I didnt know this, thank you for putting it up on here
    No problem at all!
    The potential gradient is the change of gravitational potential per metre at that point (the biggest changes are nearer to the surface of an object)
    The equation for potential gradient is  g=\Delta V/ \Delta r

    So the  \Delta r is the change in distance which we get from the fact the distance between equipotentials is 1.0km (so the closest equipotential either way is 1km away as shown by the rings on the diagram)

    then  \Delta V is the change in gravitational potential, the ring we are looking at has potential of 500kjkg-1 and the ones eitherside are 495k and 505k so either way the nearest equipotential is 5k

    So putting that together

    g= 5000 / 1000

g=5.0jkg^-^1m^-^1

    Hope that helped just say if you need anything clarified
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    (Original post by Music With Rocks)
    No problem at all!
    The potential gradient is the change of gravitational potential per metre at that point (the biggest changes are nearer to the surface of an object)
    The equation for potential gradient is  g=\Delta V/ \Delta r

    So the  \Delta r is the change in distance which we get from the fact the distance between equipotentials is 1.0km (so the closest equipotential either way is 1km away as shown by the rings on the diagram)

    then  \Delta V is the change in gravitational potential, the ring we are looking at has potential of 500kjkg-1 and the ones eitherside are 495k and 505k so either way the nearest equipotential is 5k

    So putting that together

    g= 5000 / 1000

g=5.0jkg^-^1m^-^1

    Hope that helped just say if you need anything clarified
    Ahh yhhh ok i see. Thank you lol. That makes sense, thanks
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    Can anyone please help me with this question?

    3. An insulated metal conductor is earthed before a negatively charged charged object is brought near to it.
    a) Explain why the free electrons in the conductor move as far away from the charged object as they can.
    b) The conductor is then briefly earthed. The charged object is then removed from the vicinity of the conductor. Explain why the conductor is left with an overall positive charge.

    so
    (a) the free electrons are negatively charged so as like charges repel they move away from the charged object
    (b) 1. conductor earthed
    2. Charged object discharged through object
    3. conductor no longer earthed, conductor and object both neutral
    4. object removed, conductor is neutral?

    What have I done wrong?
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    (Original post by Music With Rocks)
    Can anyone please help me with this question?

    3. An insulated metal conductor is earthed before a negatively charged charged object is brought near to it.
    a) Explain why the free electrons in the conductor move as far away from the charged object as they can.
    b) The conductor is then briefly earthed. The charged object is then removed from the vicinity of the conductor. Explain why the conductor is left with an overall positive charge.

    so
    (a) the free electrons are negatively charged so as like charges repel they move away from the charged object
    (b) 1. conductor earthed
    2. Charged object discharged through object
    3. conductor no longer earthed, conductor and object both neutral
    4. object removed, conductor is neutral?

    What have I done wrong?
    When the conductor is earthed, the conductor loses the negative electron charge through the ground. When it is removed from the earth again, it has a loss of negative charge so is positively charged
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    (Original post by Euclidean)
    When the conductor is earthed, the conductor loses the negative electron charge through the ground. When it is removed from the earth again, it has a loss of negative charge so is positively charged
    But does it not only lose electrons until it has reached a neutral charge? sorry I don't quite get it
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    (Original post by Music With Rocks)
    But does it not only lose electrons until it has reached a neutral charge? sorry I don't quite get it
    The conductor is actually neutrally charged to begin with (nuclei are positively charged remember), the electrons are just delocalised which means they can move somewhat freely.

    When the conductor loses electrons, it doesn't have those electrons to balance the positively charged nuclei anymore so the conductor becomes overall positively charged
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    (Original post by Euclidean)
    The conductor is actually neutrally charged to begin with (nuclei are positively charged remember), the electrons are just delocalised which means they can move somewhat freely.

    When the conductor loses electrons, it doesn't have those electrons to balance the positively charged nuclei anymore so the conductor becomes overall positively charged
    Oh, so is this how it works (sorry for the rubbish diagrams I did)

    EDIT: you have to click on the image again when it opens or else you can't see the text
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    (Original post by Music With Rocks)
    Can anyone please help me with this question?

    3. An insulated metal conductor is earthed before a negatively charged charged object is brought near to it.
    a) Explain why the free electrons in the conductor move as far away from the charged object as they can.
    b) The conductor is then briefly earthed. The charged object is then removed from the vicinity of the conductor. Explain why the conductor is left with an overall positive charge.

    so
    (a) the free electrons are negatively charged so as like charges repel they move away from the charged object
    (b) 1. conductor earthed
    2. Charged object discharged through object
    3. conductor no longer earthed, conductor and object both neutral
    4. object removed, conductor is neutral?

    What have I done wrong?
    https://www.khanacademy.org/science/...and-insulators

    6:58
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    Okay so I'm getting very confused with the mark schemes here, sometimes they give their answers to 2sf and sometimes they give them to 3. Also sometimes they use the unrounded values in their calculations and sometimes they use the rounded values. I know that when it specifies to use the correct number of significant figures you should use the same amount as the the most significant figures they give in the data, but what about any other question? and should we use rounded values or keep unrounded values in the calculation until the final answer?
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    Do we just circle the letters for the multiple choice, or is there an answer sheet that we write them on?


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    That explains it perfectly, thank you for the link!
 
 
 
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