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Physics: Post your doubts here!

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did u work out jun 2011? cud u help me out with q2? N i didn't get what is the difference between quantity and unit :(
Hey i am done with AS am in A2 now
but yea i sure can help:)
look....quantities are like mass, length, time
their units are kilograms, metres, seconds
so basically in the qn u asked, the qn wants quantities not units
so option A - ruled out coz it has 'metre' ( which is a unit)
option B - the statement itself is wrong, pd is not equal to energy per unit current
option C - the correct answer since it is the right definition and uses only quantities
option D - ruled out coz it has 'second' (which is a unit)

hope that helped....
 
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Hey i am done with AS am in A2 now
but yea i sure can help:)
look....quantities are like mass, length, time
their units are kilograms, metres, seconds
so basically in the qn u asked, the qn wants quantities not units
so option A - ruled out coz it has 'metre' ( which is a unit)
option B - the statement itself is wrong, pd is not equal to energy per unit current
option C - the correct answer since it is the right definition and uses only quantities
option D - ruled out coz it has 'second' (which is a unit)

hope that helped....
hmmm, for A, B and D it certainly helped :)
bt wat i still don't get is why area isn't a unit, bcoz its so close to cubic metre..
or do you mean they require only SI units? :O
 
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hmmm, for A, B and D it certainly helped :)
bt wat i still don't get is why area isn't a unit, bcoz its so close to cubic metre..
or do you mean they require only SI units? :O

okay look i am not sure about the SI unit thing, but area isn't a unit, a unit woud be someting yu write after a quantity, so for eg i can say 250kg, 300seconds, 500metres, but can i say 400areas? noo i can't......area is measured in metres.....not vice versa....hope you get it.....
 
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okay look i am not sure about the SI unit thing, but area isn't a unit, a unit woud be someting yu write after a quantity, so for eg i can say 250kg, 300seconds, 500metres, but can i say 400areas? noo i can't......area is measured in metres.....not vice versa....hope you get it.....
omg! :eek: thXx!! now i got it thnk u vry much :)
 
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what computer do you use? if it's normal pc or laptop - there's a printscreen button on the right of the keyboard
if it's a mac - just press command + shift + 3
that's as far as i know:)
ya, this i did but my pc stores the image in .bmp,, i think to upload an image here .jpg is required what u usually do after prnt screen?
 
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In the method in the marking scheme isn’t it they calculated a value of h using pressure as 1.01 x 105 and they used the formula Pressure = height x g x density. Then the h they calculated (which is 9.53) they subtracted that from 10 (which is the value of h the divers are taking in the question) to find delta h, which is 0.47 then they proceed to find the uncertainty and get the question right ;)
 
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Hey guys, I've got some mixed signals in doing Q 1 / part B of w02_qp_2.
I don't know why p/ro = m^-1 s^-2 / kg m^-3 which is equal to = kg^-1 m^-4 s^-2,
and then the marking scheme says square root of p/ro = ms^-1 <==== How?!?!?

QP: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w02_qp_2.pdf
MS: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w02_ms_2.pdf

Thanks in advance.

I have no idea what you are saying but ill still try to explain :p Unit of Pressure is N(m^-2) right, since its force over area. Now N (newton) is kgm(s^-2) so P's base units are kg(m^-1)(s^-2). Now P/ro will be (kg(m^-1)(s^-2))/(kg(m^-3)) right? So the kg's get canceled out and since the m^-3 unit is at the bottom, when we take it to the top the -3 becomes +3 so the units would be (m^(-1+3))(s^-2) = (m^2)(s^-2) but we arent done since theres still the square root (which also applies to units) so the unit would be m(s^-1) x (the constant) and m(s^-1) is the unit of velocity, therefore the constant has no units.
 
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Well, its relatively easy. Displacement is defined as the distance from a fixed point in a specified direction. We're all clear, right?
Let me give you an example:
I'm sitting in front of my computer and from this point, I will go out of the room and stop there. The distance between my initial position and the point I stopped at is my displacement. In reality, I'm still sitting in front of my computer typing this answer, which means I haven't moved, and my displacement is 0.
 
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Oct/Nov 2004, Paper2, Question 1, Part B) (ii)

I got the percentage uncertainty of diameter of wire as (0.02 / 0.50) * 100 = 4 %

Now when I try to get the uncertainty of the area of cross-section of the wire, the answer doesn't come up as 8%.

This is how I do it:
Cross-sectional area = pi * (r^2)
Diameter = 0.50 mm / 2 = 0.25 mm
pi (0.25^2) = 0.196 ~= 0.2 mm^2
% uncertainty of cross-sectional area = (uncertainty of area / area)* 100
% uncertainty of cross-sectional area = (0.02 / 0.2) * 100 = 10 % ????

Do I have to separately calculate the uncertainty of cross-sectional area? If yes, how?

QP: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_2.pdf
MS: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_ms_2.pdf

Thanks in advance.
 
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