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Recent content by shiningstar

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    IT Pending results

    I've just received an email from my school telling that my Nov 2012 IT result is frozen due to probabilty of being involved in a cheating case due to similar results with other students that I don't even know. They've asked me to write a report clarifying what had happened, and defending myself...
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    A2 Physics | Post your doubts here

    At the poles there is no centripetal force as it rotates about Earth's axis so the radius of rotation is zero. so ther is only gravitaional force: F = GMEm / R2 = mgo At the Equator, there are both Centripetal and Gravitational forces. But since the Gravitational force is greater the resulta nt...
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    A2 Physics | Post your doubts here

    This is the quantity that us proportional to the width of the resonance curve. As the damping inceases the frequency response also increases and the resonance curve becomes broader.
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    A2 Physics | Post your doubts here

    It said at a height o.5 above Earth's surface which means that r = R of Earth + 0.5 R(Height ABOVE Earth's Surface) = 1.5R Hope you get it :)
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    A2 Physics | Post your doubts here

    1. for an Ideal gas, PE = 0 and since constant temperature thus KE = o so internal energy U = 0. During compression Work is done ON the gas so W is +. Which leads to q is ( - ) as U= w + q 2. No Expansion thus w = 0. Heating means thermal energy is supplied to system so q is +, which leads to U...
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    A2 Physics | Post your doubts here

    No the TOTAL CHANGE in Intenal Energy of WHOLE Cycle PQRP = 0 So: 720-360+inc in internal energy of RP = o so increase in internal energy of RP = -360 J so work done = -360 - 480 = -840 J not -480 J Hope you get it :)
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    A2 Physics | Post your doubts here

    Total Capacitance in series = 67/2 C decreases to half its value and so will the charge Q as Q=CV and V is constant 12V So as Q is halved the Current will be halved Q=It and since P=VI then Power dissipated in the resistor is also halved.
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    A2 Physics | Post your doubts here

    Oh, I got it ! Thank you so much for correction :)
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    A2 Physics | Post your doubts here

    That is what I did, but it ended up as: 40x190 = 0.85 x 18 x c + Heat lost 7600 = 15.3 x c + Heat Lost (1) and 6840 = 15.3 x c + Heat Lost (2) But if I subtract the 2 equations both heat lost and 15.3c will be cancelled :/ What should I do next? Thank you :)
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    Dream, Believe, Achieve. ! Ya Rab :)

    Dream, Believe, Achieve. ! Ya Rab :)
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    A2 Physics | Post your doubts here

    Please I need your help in this question. A 850g of copper is being heated by a heater to determine its specific heat capacity. The block is initially at 12*C. The heater is switched on and the time taken for the temperature to rise to 30*C is recorded. The Block is cooled to its original...
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    Physics: Post your doubts here!

    Can anyone tell me from where can I get Physics AL Old Past papers before 2001?! Thanks in Advance :)
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    All May/June 2011 Papers + EXAMINER REPORTS here! :)

    I need the Examiner Reports for: IGCSE Accounting (0452) A Level Mathematics (9709) A Level Physics (9702) Thank You :)
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    Maths (Paper 3)

    which part do you want i) or ii) ??
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    Maths (Paper 3)

    For Q5: Solve (2x+7)/((2x+1)(x+2)) as partial fraction which will be 4/(2x+1) - 1/(x+2) so by Integrating this fractions with the limits 0 to 7 Which will give: (2 ln (15) - ln(9)) - (2ln(1) - ln(2)) = (ln (15^2/9) + ln(2)) = ln(25*2) = ln50 Hope I helped :)
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