IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: ruby92 on May 02, 2011, 10:45:36 pm
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o/n 2010 variant 41 question 1 c (i) and (ii)
Same paper variant 43 question 3 b (i) and (ii)
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hope im answering the ryt one :
paper 41 - 1ci) you have a gravitational field due to the earth, attracting any mass within its field to the left hand side of ur ppr , and you have another (smaller) gravitational field due to the moon attracting masses to your right hand side of th paper - (its smaller cuz the moon is smaller- simple) now at some point wont the resultant force exerted on unit mass, somewhere between the two planets, be equal and opposite? so thats why the grav field strength is zero
ii) grav. field strength = GM/r2 where M is the mass of the thing causing the field
so if the at a certain distance r from the centre of the earth - the grav field strengths of both planets are equal we form this eqn:
GMearth = GMmoon
_____ ____________
r2 (60RE-r)2
cancel the Gs and put in the masses - eventually you'll get ans answer in terms of RE
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paper 43 - 3bi) for cylinder X they gave you an eqn : x= -4.0Cos(220t)
now the phase diff for cylinder Y is 2pie/3
so the eqn for cylinder Y becomes : x= -4.0Cos(220t + 2pie/3) {everything else like freq and amplitude are same}
now the instant theyre talkin bout is wen x= -4 (you can see from th fig that cylinder X is in its lowest position where x=-4) so putting it in th eqn we get:
-4 = -4Cos(220t)
therefore 220t=0
put that in ur eqn for cylinder Y - you get its displacement at that instant :
x=-4.0Cos( 0 + 2pie/3 )
you get x=2
so draw it 2cm above AB
the next one is th same - th arrow direction is diff bcuz it has already done half a cycle (pie) and is continuing downwards ---> (as in 4pie/3 is larger than pie)
hope its clear - i got stuck here too a couple of days ago
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hello can u help me with physics paper 42 o/n 2009
1b why is the radius taken as 2.86x10^7 insted of 2.22x10^7
thank you
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hello can u help me with physics paper 42 o/n 2009
1b why is the radius taken as 2.86x10^7 insted of 2.22x10^7
thank you
In the equation F=GMm/r^2
r is taken from the center of mass. So r=Radius of earth+Distance of orbit from earths surface.
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hope im answering the ryt one :
paper 41 - 1ci) you have a gravitational field due to the earth, attracting any mass within its field to the left hand side of ur ppr , and you have another (smaller) gravitational field due to the moon attracting masses to your right hand side of th paper - (its smaller cuz the moon is smaller- simple) now at some point wont the resultant force exerted on unit mass, somewhere between the two planets, be equal and opposite? so thats why the grav field strength is zero
ii) grav. field strength = GM/r2 where M is the mass of the thing causing the field
so if the at a certain distance r from the centre of the earth - the grav field strengths of both planets are equal we form this eqn:
GMearth = GMmoon
_____ ____________
r2 (60RE-r)2
cancel the Gs and put in the masses - eventually you'll get ans answer in terms of RE
Thank you :D
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Thank you :D
can u plz have a look at the question. its actaully 2 equations used. can u plz see it
Thanks
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Yes GMm/r^2 =mw^r
but i dont understand what your doubt is then?
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ah ok stupid me din calcualte properly. THANK U SOO MUCH :D
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Your welcome and i'm sorry i'm very good at explaining.
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M/j 2005 question 7 c.
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lol how does the value become 1/10
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manganese decays to form iron ryt? so if
iron/manganese = 9
it means out of 10 molecules, 9 are iron and 1 is manganese
so its like 1/10 of manganese has decayed
so u use the eqn with 1/10 and the decay constant of manganese
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M/j 2004 7c,b , why is h squared?
M/j 2005 5(b)
o/n 2005 for question 3b (i) why is work done negative, isnt work dont on the system?
M/j 2006 4(c), Q7 (a) and (b)
O/n2006 7b
M/j 2007 Q3 d
o/n 2008 3c(ii)
M/j 2009 9b, 11 biii
Any help would be awesome!
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M/j 2004 7c,b , why is h squared?
so 2007 >> E=(1/2)mv2 and lambda=h/p where p=mv
so mv2=2E
v=sqroot(2E/m)
p=m x sqroot(2E/m) (muliply numerator and denominator by sqroot m) p=m x sqroot( 2E m) / (sqrootm)2
p= sqroot(2Em)
substitute p back into lambda=h/p
you get wat they're askin
part c) get E by substituting everything else in then use E=Ve where V is th pd and e is the charge on th electron
h is squared cuz amm if u want to make E the subject of the formula you have 2 square everything else if u still dont get it tell me ill attach a working
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M/j 2005 5(b)
ur plottin the gradient of th given graph against time it should look like this
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o/n 2005 for question 3b (i) why is work done negative, isnt work dont on the system?
ok so you calculated and its all fine - its not work done on the system - if you think about it the water or now steam would have 2 push against the atmospheric pressure - so its work done by the system - now:
delta U = delta W + delta Q
where U is internal energy, W is work done on the system, and Q is heat given to the system
notice how there aren't any negative signs in the equation
but when you come to calculate the last part, you cantttt take delta W as positive - since its by the system - therefore you put it in as : ''negative work done on the system'' = work done by system
get it?
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M/j 2006 4(c), Q7 (a) and (b)
use ur imagination (though its physix)
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Good job Sue T. Keep it up and thanks for helping out our members !
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Thank you :)
could you attached the working for the m/j 2004 question?
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thank you Ari Ben Canaan
M/j 2006 4(c), Q7 (a) and (b)
7 a nd b ) cant help with a sorry - wen i asked my teacher he simply said tht i shod no it...
so b) since electrons behave as waves, they get diffracted by the carbon which acts as a diffraction grating. the atoms in the carbons form rows and columns. The spaces between the columns are regularly spaced just like a diffraction grating - this causes the appearance of a central maxima and interference maximas on both sides of it. Then the atoms are also in rows with regular spaces. These give interference maxima above and below the previous central maxima - so you end up with shape 1 in the attachment (very rough sketch)
now there all several jumbled up layers of the atoms just described. so look at the second attachment - you end up with concentric circles as ur diffraction pattern
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Thank you :)
could you attached the working for the m/j 2004 question?
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can anyone explain question 5
November 02
parts a and b
how do we put the capacitors and why please :)
thx
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can anyone explain question 5
November 02
parts a and b
how do we put the capacitors and why please :)
thx
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i have come across certain places that 2*E(kinecticenergy) = change in potential energy?
howww?
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i have come across certain places that 2*E(kinecticenergy) = change in potential energy?
howww?
i can only help you if u specify where and what u saw ;) but in general energy is conserved so ek lost=ep gained
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october/n 2008 q5 a
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O/n2006 7b
M/j 2007 Q3 d
so if you increase the frequency, the wavelength decreases.Wavelength affects the number of photons per unit time so a dec in this means a dec in that. intensity remains same. dec in the number of photons per unit time means a derease in the number of photo-electrons per unit time which is the rate of emission.
keep in mind also an inc in frequency means an increase in th max kinetic energy of the photo-electrons but this is not required in this answer ;)
now mj2007 3d - see the more the electric feild strength, the larger th force ryt? and force is directly proportional to acceleration so for part (i) put it as d=o cuz thats the highest value for E. Then part (ii) put it as a description of E. as in go like >> it decreases and then increases -simply . Now wat would be wrong here is putting in a description of the gradient of the graph wen describing the acceleration - because the gradient of that graph is something else - not a quantity proportional 2 acceleration.
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October\November 2004 Q3 (c) (ii) How is the new amplitude of 4.5 cm calculated?
Thanx in advance
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October\November 2004 Q3 (c) (ii) How is the new amplitude of 4.5 cm calculated?
Thanx in advance
okkkkk sooooo E directly proportional to A2
E ---------> A2
15x10-3-------->(5x10-2)2
(80/100)15x10-3----------->a2
15x10-3 . a2 = (5x10-2)2 . (80/100)15x10-3
do the math im tired now ::)
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okkkkk sooooo E directly proportional to A2
E ---------> A2
15x10-3-------->(5x10-2)2
(80/100)15x10-3----------->a2
15x10-3 . a2 = (5x10-2)2 . (80/100)15x10-3
do the math im tired now ::)
Again thanx, that cleared it up, I have totally forgotten about that equation.
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o/n 2008 3c(ii)
M/j 2009 9b, 11 biii
on 2008 3cii) as it moves through the cloth, it is at a displacement of 3mm. there is a formula for the velocity at any point :
v = omega (sqroot(x2o- x2) where xo is ur amplitude
so just use this with a displacement of 3mm - ive attached smthin tht cod show why it is 3mm