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Edexcel International A level (IAL): Physics Unit 6 (WPH06) - 19th May, 2016 - PM

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Reply 140
i got the time period as 3.20 seconds and the next question about the equilibrium as 0.80 seconds. Are any of these correct? for the uncertainty in the first question i got 2.2%
Original post by Owais.ik
For the uncertainity in question 1 I found the max possible value of W1 minus minimum possible value . and then divide by 2
I got something like 0.015N


I did the same!!
Did anybody get the acceleration at 0.4 s to be 1.96 ms^-2 ?
Original post by ab9826
i got the time period as 3.20 seconds and the next question about the equilibrium as 0.80 seconds. Are any of these correct? for the uncertainty in the first question i got 2.2%


All three correct.
Original post by ab9826
i got the time period as 3.20 seconds and the next question about the equilibrium as 0.80 seconds. Are any of these correct? for the uncertainty in the first question i got 2.2%

Same
Original post by Sandy_Vega30
Did anybody get the acceleration at 0.4 s to be 1.96 ms^-2 ?


I got 2 ms^-2
Reply 146
Will the p.d. across capacitor X drop to zero after it discharges through capacitor Y for some time? So what reading shoud we take as V2?
(edited 7 years ago)
Q4: I forgot to convert Celsius into Kelvinfffffffffuuuuuuuuu :frown:
(edited 7 years ago)
Reply 148
Original post by AvWOW
I wrote the same precaution (connecting the capacitor to the correct terminals) but I wrote repeat and find mean for accuracy... It might be wrong cause you don't repeat for circuits normally as it could heatup right? smh


I wrote repeat and average after cooling the setup!
Reply 149
Original post by Sandy_Vega30
Did anybody get the acceleration at 0.4 s to be 1.96 ms^-2 ?


I got 2m/s... btw what did you write for why was the line horizontal at 0.80 m/s in v-t graph?
If anyone manages to get hold of the exam paper, could you please link me to it?
How do you calculate the total percentage uncertainity in that density of clay question? Do you just add the percentage uncertainities?
(edited 7 years ago)
Original post by Aimen.
I got 2m/s... btw what did you write for why was the line horizontal at 0.80 m/s in v-t graph?


I wrote that at 0.8 s the mass experiences max. Velocity, therefore a horizontal tangent could be used to calculate max. acceleration. I don't think that's right though. :redface:
(edited 7 years ago)
Original post by ab9826
i got the time period as 3.20 seconds and the next question about the equilibrium as 0.80 seconds. Are any of these correct? for the uncertainty in the first question i got 2.2%

Yess I got the same 🎉🎊 but I dont remember % uncertainty value. 😞
Reply 154
Original post by Sandy_Vega30
I wrote that at 0.8 s the mass experiences max. Velocity, therefore a horizontal tangent could be used to calculate max. acceleration. I don't think that's right though. :redface:


Acceleration is always zero at max. velocity.
Original post by Aimen.
Acceleration is always zero at max. velocity.


What did you writet?
Reply 156
Original post by Sandy_Vega30
What did you writet?


I wrote velocity is maximum at 0.80 m/s^2 so the gradient is horizontal and acceleration=0 ms^-2
Original post by Aimen.
I wrote velocity is maximum at 0.80 m/s^2 so the gradient is horizontal and acceleration=0 ms^-2


That makes so much sense! :smile:
Reply 158
for question no. 4 my value of gradient was 1061 which is wrong but will i get carry forward marks for the next questions ??
Reply 159
Original post by Aimen.
I wrote velocity is maximum at 0.80 m/s^2 so the gradient is horizontal and acceleration=0 ms^-2


Correct. Acceleration is proportional to displacement: at t=0.80s, (wait not v=0.80ms^-1) its at center oscillation - displacement is zero so acceleration is zero and the tangent (dv/dt) = zero

I screwed up everything else tho. ggwp

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