help needed in physics paper 3

[Lyfroker]

cn any1 tel me hw 2 do Q 5 b in may/june/2009 paper variant 2

Thnx in advance

 

[HorsePower]

there arent any variants for may / june 2009, did u mean this paper http://www.xtremepapers.me/CIE/Cambridge IGCSE/0625 - Physics/0625_w09_qp_3.pdf
if so heres the answer -
a) use the formula P.E.=mgh
:. P.E. = 12x10x3 = 360 J
b) use the formula P= E/T
:. Useful power output = 360/60 = 6 W
( Hint : at the top it says 'Every second' , thats why we use 60 for 't' )

 

[respect1]

I suck at physics paper 3, i just barely pass the A mark... can any1 tell me how to do Q 4a (iii), (iv), whole of question 6, question 9c (i), (iii), question 10 and 11. sorry for all the questions but i have trouble understanding them. This is paper http://www.xtremepapers.me/CIE/Cambridg ... p_31.pdf...
I got 47/80 which is just below the A mark..

 

[Luzzo95]

occurding.to.the.threshold.you.got.an.A

 

[respect1]

49 is an A i think but ireally wanna clear that mark by a good amount of marks like 60 to 65
ny1 help?

 

[HorsePower]

49 is an A i think but ireally wanna clear that mark by a good amount of marks like 60 to 65
ny1 help?

which year is that in ? the link u entered is incorrect

 

[respect1]

woops ma bad , its oct/nov 2009 paper 31 thnx

 

[HorsePower]

woops ma bad , its oct/nov 2009 paper 31 thnx

No prob
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4 a) In direction of force.

b) It keeps the object moving in a circular path.

c) i) 1. The wieght should be given in newton so, 60 x 10 = 600 N (downwards)
2. Since lift is at rest, net force= 0 Therefore force exerted by lift is also 600 N (upwards)

ii) For this we'll use the equation F= ma , F = 60 x 2.5 = 150 N

iii) in this you have to read carefully, it says 'force exerted by the floor on the woman' , when it was at rest it was 600 N upward but when the lift accelerates up wards it means the total force exerted on the women will be 600 + 150 = 750 N

iv) its is moving at steady speed, so again netforce = 0 and the force will be equal to weight of women = 600 N
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6 a) i) Adding weights decreases the volume in cylinder, so pressure inside it increases.

ii) In this use the formula p1v1=p2v2
the final pressure = p2 = ( p1v1 )/ v2 = (1.05 x 10^5 x 860) / 645 = 1.4x10^5 Pa

iii) use the formula P=F/a rearrange it to F = Pa
you should get F = 1.4x10^5 x 5x10^-3 = 700 N ( total force after weights were added)
F= 1.05x10^5 x 5x10^-3 = 525 N ( total force before wieghts were added )
Force by additional weights = 700-525 = 175 N

b) i) Volume increases as V is proportional to T.

ii) Pressure needs to be constant , so 'No change'.

iii ) Increase the wights to increase pressure.

iv) Pressure increases as P is prportional to T.

 

[respect1]

Wow, that's simply amazing bro thanks a lot, it really helped
:friends: