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  1. D

    Mathematics: Post your doubts here!

    The tensions AP and AQ are equal, so the angles on both sides of 3sqrt3 are the same. The angles are (180-90-30)/2 = 30 Now take the direction of the 3sqrt3 force as your plane and resolve forces along it. Look at the horizontal part. Tcos30 + Tcos30 = 3sqrt3 T = 3
  2. D

    Mathematics: Post your doubts here!

    let p's time = t then q's time = t - 2 a = 1.75 and u = 0 for both Displacement of P = 0.875t^2 Displacement of Q = 0.875(t-2)^2 Displacement of P - Displacement of Q = 4.9 0.875t^2 - 0.875(t-2)^2 = 4.9 t = 2.4 but since it's talking about time in terms of Q in the question t = 2.4 - 2 = 0.4
  3. D

    Mathematics: Post your doubts here!

    If you are integrating an equation for speed, yes.
  4. D

    Mathematics: Post your doubts here!

    You may not know it but you ARE calculating the area. Area under v/t time graph is the displacement. Or in that that example, think of it this way, suppose you want to know the displacement between t = 0 and t = 10. So it will be displacement up till t = 10 - displacement at t = 0 so it will be...
  5. D

    Mathematics: Post your doubts here!

    Why are you putting them in the second derivative... Just take 1 derivative. That will give you acceleration. Velocity will be increasing when acceleration is positive and decreasing when acceleration is negative.
  6. D

    Mathematics: Post your doubts here!

  7. D

    Mathematics: Post your doubts here!

    suppose you integrate v and get this S = 2t^2 - 3t + c and you know s = 5 when t = 0 so c = 5 S = 2t^2 -3t + 5 now this is the term you will apply limits to so it becomes Upper limit - lower limit but whatever limits you use, the 5 will stay 5 because it has nothing to do with t so whatever...
  8. D

    Mathematics: Post your doubts here!

    It's not too bad as long as you understand that the v/t graph is the graph of the gradient of s/t and remember a few cases. constant velocity is a straight diagonal line for displacement straight diagonal line for velocity is increasing gradient graph for displacement curved graph for velocity...
  9. D

    Mathematics: Post your doubts here!

    It doesn't matter. You can take any direction as positive as long as the other direction is considered negative.
  10. D

    Mathematics: Post your doubts here!

    there are 2 cases of constant acceleration. Before reaching the liquid: a = 10 s = 5 u = o v = ? t = ? use constant acceleration formulas to find v and t v = 10, t = 1 Inside the liquid: a = 5.5 s = 4 u = 10 (v of previous) v = ? t = ? again use constant acceleration formulas v...
  11. D

    Physics: Post your doubts here!

    http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-13.html solution 73
  12. D

    Mathematics: Post your doubts here!

    Tension T is acting upwards on both sides of the pulley. So resultant is 2T.
  13. D

    Physics: Post your doubts here!

    Initial momentum = final momentum let the new speed = x Mv = 2/3(M)(x) + 1/3(M)(-6v) x = 4.5v
  14. D

    How did Maths 12 go? No discussion

    Tan theta was 2 and 3... I put the values of theta in the original equation and the equation was satisfied
  15. D

    Mathematics: Post your doubts here!

    No, the answer in the scheme doesn't work if x > .54 1/r(r+1)(r-1) ≡ A/r + B/(r+1) + C/(r-1) 1 = A(r+1)(r-1) + B(r)(r-1) + C(r)(r+1) at r = 1, C = 0.5 at r = -1, B = 0.5 at r = 0, A = -1 cosAOP = cosBOP (1,2,2).(1+2s, 2+2s, 2-2s)/3 = (3,4,0).(1+2s, 2+2s, 2-2s)/5 solve for s and you should get...
  16. D

    Mathematics: Post your doubts here!

    n of q = (1,b,c) Angle between 2 planes is 60 (1,b,c) . (1,1,0)/ sqrt(1 + b^2 + c^2) * sqrt2 = cos60 (1 + b)^2 = 1/4 * 2 * (1 + b^2 + c^2) 1 + 2b + b^2 = 0.5 + 0.5b^2 + 0.5c^2 0.5b^2 + 2b + 0.5 - 0.5c^2 = 0 --- eq 1 AB lies in q so its direction vector is normal to n (1,b,c)...
  17. D

    Mathematics: Post your doubts here!

    5)i) mgh = 1/2mv^2 10(0.45) = (0.5)(v^2) v = 3 m/s ii) 0.39 + PE = (.5)(.3)(3^3) PE = 0.96 J (0.3)(10)(h) = 0.96J h = 0.32m 4)i) motion of A: 3.6sin60 - T = (0.36)(0.25) T = 3.03N ii) motion of B: 3.03 - (Friction + 2.4sin60) = (0.24)(0.25)...
  18. D

    Mathematics: Post your doubts here!

    use 1/2 * BC * OA
  19. D

    Mathematics: Post your doubts here!

    w = cos2X + isin2X cos2X + isin2X - 1/cos2X + isin2X + 1 = cos^2(X) - sin^2(X) + i(2sinXcosX) - 1/ cos^2(X) - sin^2(X) + i(2sinXcosX) + 1 = -2sin^2(X) + i2sinXcosX/ 2cos^2(X) + i2sinXcosX = 2sinX(icosX - sinX)/2cosX( cosX + isinX) = tanX(icosX - sinX)/ (cosX + isinX) = tanX.i(cosX +...
  20. D

    Mathematics: Post your doubts here!

    Let the three times for the three stages be t1,t2 and t3 Area under the graph = 4000 (0.5)(20)(t1) + (240)(10) + (0.5)(20)(t3) = 4000 t1 + t3 = 160 since t1 + t2 + t3 = 240 t2 = 80 Now, the remaining time is 160s. This is divided between t1 and t3 in the ratio 1:3 (we know this from the...
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