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can someone solve question 4 C(!!)
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_42.pdf
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_42.pdf
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can someone solve question 4 C(!!)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_42.pdf
seneriaThe graph starts off quite high on the y axis (but not at x = 0, at "surface of Earth") and decreases as a curve till a point closer to the "surface of Moon" than the "surface of Earth", at which point it goes to zero. Then it changes polarity (positive to negative of negative to positive, both are correct) and increases until it reaches a final value when the graph ends at "surface of Moon"; however, this maximum is not as high as that at the beginning of the curve at the "surface of Earth".
The explanation is this:
The gravitational field strength at the surface of the Earth is pretty high, and as you move further from Earth, this field strength decreases. At approximately the point you calculated in the previous part of the question, the net field strength (the sum of the gravitational pull per unit mass from both the Moon and the Earth) goes to zero and you can represent this on your graph by having the line decrease to zero.
If your initial potential is positive, the potential after that point then becomes negative; if the initial potential at Earth was negative, then the potential at the Moon is positive. The reason for this is that the force per unit mass is in opposite directions; the Moon pulls in one direction, the Earth pulls in the other direction; that is why the potential can be zero at one point; it is the point where the pull from the Moon is the same as the pull from the Earth.
However, since the Moon has a much lower mass than the Earth, this maximum is not as high as it is at the surface of the Earth.
Since the field strength decreases according to the inverse square law (Strength is proportional to 1/r^2), the gradient of the graph slowly decreases until the
zero-strength point and then the gradient increases(in the negative sense; literally, the gradient decreases). Hopefully the attachment will clear this up at least a little bit!
(The line is really scratchy, I hope it's not too bad!)
Hope this helped!
Good Luck for all your exams!
for the first option work is being done on the system , so there must be an increase in internal energy reight? this is confusing me.. as water is BEING freezed
however, for the third option the water is evaporating , meaning work is done BY the system... so internal energy must decrease... because increae in the internal energy is work done on the system and heat is supplied to the system....
can youclear my confusion?
Thank you very much!!For 3 b) 1, I guess you could just average out the maximum and minimum potential across the resistor, because the minimum is 4.8 V on the graph and the maximum is 6 V on the graph. Taking the average, it gives us 5.4 V, which should be the right way of getting the value (at least I hope so!).
Hope this helped!
Good Luck for all your exams!
can you draw it show please?seneria
sagar already answered one of your questions here
is it just me or the winter 2012 qp 42 is too easy? I have no idea how the grade boundary is still around 50 for A in this paper.
IKNOWWWW
I literally was like :O what is this???
Every single intuitive easy question! Whats gonna cme in our paper?? Lol!
The 2012 papers were kinda straight forward......INSHALLAH IT STAYS LIKE THAT FOR US THIS YEAR.IKNOWWWW
I literally was like :O what is this???
Every single intuitive easy question! Whats gonna cme in our paper?? Lol!
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