# Edexcel Phy 2Watch

This discussion is closed.
13 years ago
#61
i dunno my text book said differently and i was going by exam, in all hoensty i think either will get u the mark, as the temp of the hot source is the initical temperature of the system, and the cold sink is the final temp of the system.
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13 years ago
#62
k then, don't know anymore, why are these questions like this, don't they want us to attain 100%. By the way, how was Physics 1 GUYS?
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13 years ago
#63
DAMN

Just remembered I forgor to change 4.2g into kg.

Sigh.

Hopefully I still did well.
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13 years ago
#64
We had to use that rate of temp in the next question. Since I used the wrong rate of temp will I still get marks on qs where you to find power(rate of energy)?
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13 years ago
#65
No...as long as you did everything right...it'll be error carried forward and you'll get all marks for that.
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13 years ago
#66
(Original post by tktaylor6)
i dunno my text book said differently and i was going by exam, in all hoensty i think either will get u the mark, as the temp of the hot source is the initical temperature of the system, and the cold sink is the final temp of the system.
HOW DID U DO THE THEREMAL, do u think one could say, Q/5 = M*4200? AND how did u find the Physics 1?
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13 years ago
#67
(Original post by fisfos815)
No...as long as you did everything right...it'll be error carried forward and you'll get all marks for that.
Ah I see.

I am now more worried that I forgot to change units or small things like that for other qs.
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13 years ago
#68
it was easy, for the specific heat capacity related question...

H/s = mcT/s

Notice per second under T, so put in the values = answer . That question always tricks so many people in using wrong formulas, the phy2 paper was so easy comparing to past papers. I thought the phy1 paper was a *******, phy3 was easy too.
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13 years ago
#69
What do you lot think you got, overall out of 3 papers today? I have a feeling I got a B.
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13 years ago
#70
80-093% in phy1
96-100% in phy2
80-097% in phy3

Now im gonna start revising phy4-phy6, and hopefully get 95-100% on phy4,phy5 to take strain off phy6.
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13 years ago
#71
(Original post by tktaylor6)
well heres how i did it

1. realises if a p.d of 3 volts is across the reistor, V2 combo, then 6V is across the voltmeter of resistance 10Mohm, use I = V/R to obtain the current of 6 x 10^-7 A
2. the current into the resitor V2 combo is 6 x 10^-7 and the p.d is 3V, so the total resistance of the combo is 5 x 10^6 ohms

1/ (5 x 10^6) = 1/R + 1/ (10 x 10^6) = 2 x 10^-7

1/ R = 1 x 10^-7

R = 10 Mohms
Hey Thats what I got aswell. I think what most people got was the resistance of components in parallel both RESITOR and a Voltmeter.

I just did the working again this morning turned out to be same answer.
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13 years ago
#72
page one very botton see my working,please somebody tell me if or where i have gone wrong!
No, you're definately right... it's all these other people that are wrong.

My way was:
3V across resistor means that there are 3V across the whole resistor/V2 combo

Therefore 6V across V1

Current through V1 = 6/10x10^6
Equals 6x10^-7 A

Current through V2 = 3/10x10^6
Equals 3x10^-7 A

Therefore current through R = 3x10^-7 (sum of currents entering a junction = that leaving)

Since V across R and I through R are the same as for V2, R must have same resistance as V2 = 10MOhms

Checking:

R = 3 / 3x10^-7
R= 10x10^6
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13 years ago
#73
[QUOTE=Narin]No, you're definately right... it's all these other people that are wrong.

hurrah well that means the very few of us who put 10Mohms can be looking and rather high marks, oh sum1 that repleid to me earlier, i only did phy2, it was a retake, i got 80/90 last time, but want to point incase i ruin phy6 lol
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13 years ago
#74
the resistor one i just redrew the circuit but did you guys leave the voltmeter in? and when you did this and u got the resistance for the v2 AND the resistor so i ended up doing rations until i got a resistance of like 4.82 m ohms? is this wrong the question really didnt make it clear. the heat engine wasnt too bad i just put in numbers til i gto a sensible answer like 42 degrees or sumthin.
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13 years ago
#75
heat engine =>

0.53 = [t1-t2]/t1

btw, taylor you are right for the resistor question...i didn't take in account for the voltmeter that was in parrallel to it, so i probs lossed 1 mark.
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