Author Topic: phy p2 doubts =\  (Read 1481 times)

Offline Twinkle Charms

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phy p2 doubts =\
« on: June 07, 2010, 07:16:49 pm »
M/J07 Q3 c(ii)

paper is attached.
« Last Edit: June 07, 2010, 07:39:02 pm by Zara »
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Offline Twinkle Charms

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Re: phy p2 doubts =\
« Reply #1 on: June 07, 2010, 07:41:03 pm »
also in Q4 a(ii) [same paper attached] why do we add the P.E. and K.E here? they asked total kinetic energy so umm whats potential energy doing? =\
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Offline vserian

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Re: phy p2 doubts =\
« Reply #2 on: June 07, 2010, 08:29:26 pm »
potential energy is converted into KE when it falls down.
since there is a KE at the top ther will be anothe KE when it reach the sea and that KE is equal to the PE
so total is KE plus PE

Offline Twinkle Charms

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Re: phy p2 doubts =\
« Reply #3 on: June 07, 2010, 10:57:30 pm »
aite thx vserian

so Q3 still unanswered =\

in same paper M/J07 Q4d(i) and (ii)


O/N07 Q1 b , how to calculate the uncertainty, im gettin 6.9*10^-3 whereas in ms it says 0.009 ..??

this paper is attached to the post..
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Offline astarmathsandphysics

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Re: phy p2 doubts =\
« Reply #4 on: June 07, 2010, 11:15:11 pm »
3a)density=mass/volume
bi)mass=Ahp
bii)pressure=force/area=mg/A=Ahpg/A=hpg
ci)volume=mass/density so if masses are the same, colume of water vapour/volume of water=1/(1/1600)=1600
ii)cube root(1600)=11.7 remeber that volume =l^3 so l=cube root of volume
di)they have the same density
ii|)high melting point
iii)cannot deform easily, unlike liquids

Offline Twinkle Charms

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Re: phy p2 doubts =\
« Reply #5 on: June 07, 2010, 11:20:17 pm »
3a)density=mass/volume
bi)mass=Ahp
bii)pressure=force/area=mg/A=Ahpg/A=hpg
ci)volume=mass/density so if masses are the same, colume of water vapour/volume of water=1/(1/1600)=1600
ii)cube root(1600)=11.7 remeber that volume =l^3 so l=cube root of volume
di)they have the same density
ii|)high melting point
iii)cannot deform easily, unlike liquids
Thanks astar, explain other ques also please..
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Offline Twinkle Charms

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Re: phy p2 doubts =\
« Reply #6 on: June 07, 2010, 11:29:23 pm »
Q2 d (ii) in O/N07,  why do we add the forces 75 and 23 and why not subtract ???
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Offline astarmathsandphysics

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Re: phy p2 doubts =\
« Reply #7 on: June 07, 2010, 11:33:33 pm »
jun07 di) draw a vector 18 to the right and sqrt(25^2-18^2)=17.3 down
ii)tan^(-1(17,3/18))=44 degrees

Offline falafail

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Re: phy p2 doubts =\
« Reply #8 on: June 07, 2010, 11:34:24 pm »
aite thx vserian

so Q3 still unanswered =\

in same paper M/J07 Q4d(i) and (ii)


O/N07 Q1 b , how to calculate the uncertainty, im gettin 6.9*10^-3 whereas in ms it says 0.009 ..??

this paper is attached to the post..

i think this is how its done:

make R the subject of the formula, you get R=(V/pi X L)^1/2
to calculate the uncertainty here, you need to find the percentage uncertainty first then multiply by the value you get for R
in this case, % uncertainty in R = 1/2(% uncertainty in V + % uncertainty in L)
which is 0.019
the value you get for R is approximately 0.4886
so uncertainty = 0.4886 X 0.019 = 0.009

Offline astarmathsandphysics

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Re: phy p2 doubts =\
« Reply #9 on: June 07, 2010, 11:37:23 pm »
Q2 d (ii) in O/N07,  why do we add the forces 75 and 23 and why not subtract
I dont understand your question
use P=Fv=23*4.5=103.5W

Offline astarmathsandphysics

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Re: phy p2 doubts =\
« Reply #10 on: June 07, 2010, 11:38:39 pm »
thanks for doing the uncertainty question. Cat being overaffectionate at the mo

Offline Twinkle Charms

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Re: phy p2 doubts =\
« Reply #11 on: June 08, 2010, 12:05:30 am »
thanks to both =)
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Offline Twinkle Charms

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Re: phy p2 doubts =\
« Reply #12 on: June 08, 2010, 12:13:58 am »
Q2 d (ii) in O/N07,  why do we add the forces 75 and 23 and why not subtract
I dont understand your question
use P=Fv=23*4.5=103.5W
thats rong, ull be awarded only 1 mark out of 2.

in mark scheme they added both the forces and then used in this eqtn p=fv, why?
F=75N is taken from the previous sub question's answer.
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