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i wrote keep the conditions which were used previously same
keep voltage current same and keep wire of differnet lenght and take resistance (that's wrong)
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i wrote keep the conditions which were used previously same
ok will i get 1 markkeep voltage current same and keep wire of differnet lenght and take resistance (that's wrong)
I hope they be lenient and give you 2 marks. ISAok will i get 1 mark
what u wrote fr mcq of frequency of wave 2 x 60i think one mark each...
how many marks u expectingI hope they be lenient and give you 2 marks. ISA
the resistance of z was 20 and y 10, so it was 1/10 +1/20= 3/20 i.e. 0.15 total for the parallel. 10 + 0.15= 10.15...1/20 + 1/30 = 3/20 = 1/R
R= 20/3
Total resistance = 20/3 + 10 = 50/3 ohms
2 x 60what u wrote fr mcq of frequency of wave 2 x 60
we were given a battery so v was constant alreadykeep voltage current same and keep wire of differnet lenght and take resistance (that's wrong)
You get 2p/3, not 2 p.i
it was 2p
pressure remain same in all the vessel...... p1*v1=p2*v2
200*p=x*100 solve it we will get the 2p has a result
mine was 2p too... thank God! there's somebody who got the same answer... im feeling like a dumboi
it was 2p
pressure remain same in all the vessel...... p1*v1=p2*v2
200*p=x*100 solve it we will get the 2p has a result
You had to divide 1 by 3/20 to get 20/3 . This was for parallel and then add 10.the resistance of z was 20 and y 10, so it was 1/10 +1/20= 3/20 i.e. 0.15 total for the parallel. 10 + 0.15= 10.15...
No it is 200*p=300*p2i
it was 2p
pressure remain same in all the vessel...... p1*v1=p2*v2
200*p=x*100 solve it we will get the 2p has a result
I think it was downand where did we have to point the arrow at the water molecule X in paper 2? That wave question?
oops!You had to divide 1 by 3/20 to get 20/3 . This was for parallel and then add 10.
resistances in parallel = 1/R(1) + 1/R(2) = 1/R
so 3/20 wasn't R, it was 1/R.
The equation you get is 200*P=300*x because when the tap opened, the new pressure was 300 cm^3. The law is that for an equal mass of gas at constant temperature, p1*v1=p2*v2i
it was 2p
pressure remain same in all the vessel...... p1*v1=p2*v2
200*p=x*100 solve it we will get the 2p has a result
how many marks did it carry? and how many d'you think i'll get?You had to divide 1 by 3/20 to get 20/3 . This was for parallel and then add 10.
resistances in parallel = 1/R(1) + 1/R(2) = 1/R
so 3/20 wasn't R, it was 1/R.
wasn't the temp constant?The equation you get is 200*P=300*x because when the tap opened, the new pressure was 300 cm^3. The law is that for an equal mass of gas at constant temperature, p1*v1=p2*v2
It was supposed to be a double headed arrow, pointing upwards and downwards. The next one, when you had to mark D should have had 3 crests in it (2 wavelenghts)i did it upwards
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