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  1. A

    June 03 P3 Maths

    8ii x(n+1) = 2 / 3-ln x(n) just substitute alpha a into x a(n+1) = 2 / 3-ln a(n) then, remove the n terms a = 2 / 3-ln a rearrange it and you will get 3 = (2/a) + ln a
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    June 03 P3 Maths

    q3, | x - 2 | < 3 - 2x -(3 - 2x) < x - 2 < 3 - 2x solve one by one -(3 - 2x) < x - 2 x < 1 x - 2 < 3 - 2x x < 5/3 do line bar with x < 1 and x < 5/3 then you will see that it intersect when x < 1. that is the answer.
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    help needed in maths p3

    for ON03 q7, to answer the part iii, you need to have a sketch on part ii first. | z - u | = 2 | z - 1 - 2i | = 2 it means that you have to sketch a circle at center (1,2) with radius 2. then, make a tangent line from origin to the circle. the tangent line will pass through point around...
  4. A

    chem p4 help plzzz ;''(

    zain786, it is a-level. chem p4 paper 43 ON10 i also don't know to answer that. :p
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    Can i complete A-level course in a month

    it's hard actually to cover all topics in a month. but, maybe you can look through the syllabus and from there, you can know what you should study. other, maybe you can ask any of the hardworking student to borrow their notes for a while and photocopy them.
  6. A

    Do you get your percentage marks with your grades???

    whoaaaa. how can we know the percentage? i just get the grades.
  7. A

    Chemistry P1 Help

    i need help MJ08 Q25 please. :)
  8. A

    Chemistry P1 Help

    yeah, need help too MJ06 2
  9. A

    PHY P2 HELPPP PLZZZZ!!!!!!!!!

    MJ 05 4b (iii) max strain energy = area under graph. SE = 1/2 Fx = (1/2) (60N) (3X10^-4)m = 9 X 10^-3
  10. A

    Physics P2 Help Required

    can anyone gives some ideas on how to outline an experiment to demonstrate diffraction of transverse and longitudinal wave like in ON 08 Q6 b (i) and (ii). do i have to draw 3D diagram? any diagram attached is highly appreciated. :)
  11. A

    Phy P2 urgent help please!

    @ azfar kashif, we know that the wavelength is 64.8cm. so, from a node to an antinode, it cover only 1/4 the wavelength. means 1/4 of 64.8cm = 16.2 cm te antinode should be 16.2cm above the water surface or u can also answer (16.2-15.7)=0.5cm above the top of tube. :)
  12. A

    Physics P2 Help Required

    hey guys. can anyone help me with this question. MJ06 Q7 b. why is the answer state that 'shorted lamp A'. i don't understand.
  13. A

    Phy P2 urgent help please!

    separation between plates is 1.5 right? distance between electron to plate is half of it. which is 0.75
  14. A

    Phy P2 urgent help please!

    use equation s=ut+1/2at^2 to answer this question. s= 1/2 (2.46X10^15) ((2.4X10^-9)^2) s = 7.08 X 10^-3m = 0.708cm this mean that at the end of the plate, electron do not hit the plate as it traveled 0.708cm vertically. which is less that (1.5/2)=0.75cm if the electron travel >0.75, for sure...
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    Physics P2 Help Required

    MAVtKnmJ, yes it's correct. :)
  16. A

    Physics P2 Help Required

    May/June 2006, Q5 (c) both springs lets say A and B have initial length 4.5cm. after moving the trolley, spring A will extent 1.5cm while the spring B will compress 1.5cm. so, final length spring A is 4.5 + 1.5 = 6 cm spring B 4.5 - 1.5 = 3 cm use expression in b convert every cm to meter...
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    Physics P2 Help Required

    @ MAVtKnmJ maybe u can just directly find uncertainty for radius from this equation R(square) = V/(L X pi) then it will be like this 2(uncertainty in R) /R = (uncertainty in V) / V + (uncertainty in L)/L 2 (uncertainty in R) /0.489 = 0.5/15 + 0.1/20 2 (uncertainty in R) = 0.0383 X 0.489...
  18. A

    paper 1 Chemistry 2006

    ON 06 Q1 CaSO4 + 2KOH ---> K2SO4 + CaO + H2O no of mole of KOH = (25/1000) X 1.0 X 10^-2 = 2.5 X 10^-4 no of mole of CaSO4 is half KOH = 1.25 X 10^-4 mol concentration of CaSO4 = 1.25 X 10^-4 / (50/1000) = 2.5 X 10^-3
  19. A

    PAPer 1 chemistry 2004

    for Q7, PH3 consists of 3bond pair and 1 pair of lone pair. so, it is trigonal pyramidal and the angle is about 107degree. answer B (104degree) is the nearest answer since C (109degree) is for tetrahedral (4bond pair)
  20. A

    PAPer 1 chemistry 2004

    for Q3, 1mol of CH3SH react with 3 mol of O2, so this means that 10cm3 of CH3SH react 30cm3 of O2. since 60cm3 of O2 is used, there is 30cm3 of O2 left. after reaction, produce gases (1 mol of CO2 and 1mol of SO2) means 10cm3 of CO2 and 10cm3 of SO2 resultant mixture = 30cm3 of O2 + 10cm3 of...
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