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

1. (Original post by Aimen.)
oops yea it was cx/gradient , I did a mistake up there!
Alright. Well, i wrote V2 = 1/cy x Cx(v2-v1)

V2 as y -axis

Inverse of gradient is the value of Cy

and Cx(Net "V") as Y-axis.

The value of C intercept is zero because its straight line graph NOT LINEAR !

*It was mentioned in the question that how would a straight line graph would you form!
2. (Original post by WilsonDS)
Q4: I forgot to convert Celsius into Kelvinfffffffffuuuuuuuuu
same how many marks do u think we lose? + the curve was terrible haha
3. (Original post by abdullakhalaf)
same how many marks do u think we lose? + the curve was terrible haha
Maybe the entire marks from Q4 :'( NotLikeThis
4. (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%
I got the same answers for the period and same value. Dunno about the uncertainty thing.
5. (Original post by Aimen.)
I wrote repeat and average after cooling the setup!
That would have been a smarter move :/
6. (Original post by apocolyptic)
what do u guys think
Yep, It seems that this time the paper, had kinda a different format than most of the former papers T_T
7. (Original post by Aimen.)
oops yea it was cx/gradient , I did a mistake up there!
Hey I did the same thing..V1 on y axis and Vx on x axis..then Cy/Cx was gradient ..so Cy= gradient *Cx
Shouldnt we put V1 on y axis since we are gonna vary it..I think both the ways will be marked correct
8. (Original post by Choudhry Walid)
Alright. Well, i wrote V2 = 1/cy x Cx(v2-v1)

V2 as y -axis

Inverse of gradient is the value of Cy

and Cx(Net "V" as Y-axis.

The value of C intercept is zero because its straight line graph NOT LINEAR !

*It was mentioned in the question that how would a straight line graph would you form!
V1 against V2 is wrong? That gives the opposite gradient i.e. Cy/CxThen Cy =gradient*Cx
9. (Original post by Sandy_Vega30)
Did anybody get the acceleration at 0.4 s to be 1.96 ms^-2 ?
I got 1.94 ms^-2
10. (Original post by tgygt)
How about precaution for the capacitor question?
Check the capacitor's polarity. They must be connected in the right way to the battery.
or
The capacitors' must be rated at least to the maximum emf of the power supply. Otherwise they may get damaged.
11. (Original post by WilsonDS)
Maybe the entire marks from Q4 :'( NotLikeThis
wait the question to prove a straight line was worth 2 so we get that, and don't we get points for scales and plotting, and drawing a line of best fit? so that should be 2. also how many marks was the whole q worth?
12. is there no link to the paper?
13. (Original post by RizK)
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
Haha yea I meant t=0.80 s, v was sumfin like 1.40?
14. Hey guys, where do you do your rough work for physics? Is there any rough work sheet?
15. (Original post by andreashadj99)
Check the capacitor's polarity. They must be connected in the right way to the battery.
or
The capacitors' must be rated at least to the maximum emf of the power supply. Otherwise they may get damaged.

Well, I wrote that use a fixed resistor to protect the capacitor from damaging when the resistance of the whole circuit is zero.
16. (Original post by Choudhry Walid)
Well, I wrote that use a fixed resistor to protect the capacitor from damaging when the resistance of the whole circuit is zero.
Not sure if this is going to be acceptable. Also my second answer
"The capacitors' must be rated at least to the maximum emf of the power supply. Otherwise they may get damaged."
is incorrect since we do not know the capacitance of the capacitor Y.
17. (Original post by AvWOW)
I got 1.95x10^-19 for the charge.... everyone seems to have gotten something along the line of 1.55x10^-19
O_O pooping bricks here
i got something like 1.49
18. In no. 4 possibly everything was right except that i forgot to multiply 10^-3 with the temperature while calculating p and so the answer was 3.45 rather than 3450. The next two answers were right except that i took the wrong value. Does anyone have any idea whether i will get marks for the rest two i mean the vaue of e and comment about accuracy or not?
19. (Original post by Mahdia Akter)
In no. 4 possibly everything was right except that i forgot to multiply 10^-3 with the temperature while calculating p and so the answer was 3.45 rather than 3450. The next two answers were right except that i took the wrong value. Does anyone have any idea whether i will get marks for the rest two i mean the vaue of e and comment about accuracy or not?
You will.
20. (Original post by ali shahid1)
I got 2 ms^-2
same, i got 2

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