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  1. A

    Paper 42 guyss !!!!!!!

    4i) Forceof 40 N makes an angle of 30 with the vertical. Block B mass=15kg Resolve Vertically=R+40cos30+Xsin30=150 Resolve horizontally: 40sin30=Xcos30 20=√3 x/2 40/√3=x ii) R=-40cos30-40/√3(sin30+150 R=103.8119 Ff=μR 10=103.8119μ μ=0.096 5ai R=0.3g=3N Ff=μR=0.7*3=2.1N Work done=Frictional...
  2. A

    Paper 42 guyss !!!!!!!

    Question 1i) you can use two methods for particle P u=11m/s a=-10m/s v=0m/s at maximum height. use v-u/t=a at=v-u t=v-u/a=-11/-10=1.1 seconds*2 to reach the ground. 2nd method set displacement to zero and use s=ut+1/2at^2=11t-5t^2=t(5-11t) t=11/5 seconds ii) Particle Q givens t=2.2 seconds...
  3. A

    Mathematics: Post your doubts here!

    Record in the givens: OA=1760 m=1100 kg u=0 m/s SinѲ=opposite/hypotenuse=160/1760=1/11 K.E=1/2mv^2=1/2(1100)(v^2-0^2)=550v^2 Gain in P.E=mgh*sinѲ=1100*10(x-0)*1/11=1000x Work done by driving force=gain in P.E + gain in K.E+Work done against resistance 1800x=1000x+550x^2+700x 100x=55ov^2 kv^2=x...
  4. A

    Mathematics: Post your doubts here!

    never heard of it, yah arab names are too long :D . it's nice to meet you as well. i am not taking the test tomorrow i took it last june, and i am retaking it in june.
  5. A

    Mathematics: Post your doubts here!

    Egypt :) and you?
  6. A

    Mathematics: Post your doubts here!

    mauritius O.o?
  7. A

    Mathematics: Post your doubts here!

    good to hear that :), your welcome :)
  8. A

    Mathematics: Post your doubts here!

    This was just expressing speed at time t seconds. i am not multiplying v by t. i am plugging whatever t is into equation of v
  9. A

    Mathematics: Post your doubts here!

    that was just a coincidence that at t=400 seconds speed was equal to k. in the equation equation for v, t lasts only for 400 seconds as it says in the question that both cyclists reach point B at 400 seconds. Equation is for part A to B only. now we can find the speed of cyclist Q at the 400th...
  10. A

    Mathematics: Post your doubts here!

    Cyclist Q starts from t=400 s and so on to move with constant acceleration. now, find the speed of Q at 400 seconds from the equation given v=0.04t-0.0001t^2+k. speed of Cyclist Q represent the initial velocity( from point B to C) which increases as time passes. v(t)=0.04t-0.0001t^2+k...
  11. A

    Mathematics: Post your doubts here!

    Cyclist P reached B at speed 5 m/s and travelled a distance from B to C of 1400 m Cyclist P moves with constant speed of 5 m/s t=1400/5=280 seconds to travel from B to C. at t=400 second the speed of cyclist Q is 4/3m/s this represents the intial velocity. plug that into the equation below use...
  12. A

    Mathematics: Post your doubts here!

    Integrate ∫ (0.04t-0.0001t^2+k)dt set your limits from 400 seconds to 0 seconds 0.04t^2/2-0.0001t^3/3+kx=1600 0.04(400)^2/2-0.0001(400)^3/3+400k=1600 3200/3+400k=1600 3200+1200k=4800 k=4/3 at Maximum speed a=0m/s^2 so differentiate with respect to t 0.04-(2*0.0001t)=0 t=200 seconds...
  13. A

    Mathematics: Post your doubts here!

    your welcome :)
  14. A

    Mathematics: Post your doubts here!

    we calculated potential energies at a distance x along the incline not when the particle reached the top at A
  15. A

    Mathematics: Post your doubts here!

    u=2.5m/s v=1.5m/s s=4m use v^2=u^2+2as v^2-u^2/2s=a=(1.5)^2-(2.5)^2/8=a=-1/2m/s^2 ii) apply newtons 2nd law sum of force in direction of motion-the sum of forces in opposite to motion=mass*acceleration -mgsinѲ=ma -m(10sinѲ)=-1/2m 10sinѲ=1/2 sinѲ=1/20
  16. A

    Mathematics: Post your doubts here!

    my pleasure :D
  17. A

    Mathematics: Post your doubts here!

    Record in the givens: OA=1760 m=1100 kg u=0 m/s SinѲ=opposite/hypotenuse=160/1760=1/11 K.E=1/2mv^2=1/2(1100)(v^2-0^2)=550v^2 Gain in P.E=mgh*sinѲ=1100*10(x-0)*1/11=1000x
  18. A

    Mathematics: Post your doubts here!

    http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_42.pdf Q7 i know that for the first part of the question we apply newton's 2nd law. sum of forces in the direction of motion-sum of forces in opposite to motion=mass*acceleration...
  19. A

    Chemistry: Post your doubts here!

    Thank you so much :D
  20. A

    Maths P12 !!!!!!!

    its okay, i hope you have done alright?
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