IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: nid404 on July 12, 2010, 04:48:44 pm
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June 07 Q3
Can someone please explain part b and c
Thanks :)
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I have a test tom...please? :P
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I have a test tom...please? :P
OMG!! am reallllyyy sorry I can't help since I did only AS-level phy :-\
and u started A2 already?
walla am sorry..but I'll pray for u from my heart..insA u'll get it and do well :)
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lol thanks :)
If you know anyone who can help...please direct them here :-[
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oh! XENA needs some help?!!!!...the one who helps everyone..
hehe...Time for some action..i'll try to solve it.
it's paper 4 right.?
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yup yup paper 4 ;D
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3a) field strength is equal to the negative potential gradient.(its the concept u need to know)
b)find the area under graph by counting boxes.more than half counted as 1 box,less than half box not counted.calculate area of a box multiply with no of boxes.
c)use kinetic energy equals to electric potential energy.½mv2 = qV
½ × 9.1 × 10–31 × v2 = 1.6 × 10–19 × 530(u should get this value in b)
v = 1.37 × 107 ms–1
d)the gradient of graph indirectly shows the acceleration.d=0 ,it is steepest at that point.
from graph ,acceleration decreases(negative gradient) and increases(positive gradient).minimum at 4cm.
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3a) field strength is equal to the negative potential gradient.(its the concept u need to know)
b)find the area under graph by counting boxes.more than half counted as 1 box,less than half box not counted.calculate area of a box multiply with no of boxes.
c)use kinetic energy equals to electric potential energy.½mv2 = qV
½ × 9.1 × 10–31 × v2 = 1.6 × 10–19 × 530(u should get this value in b)
v = 1.37 × 107 ms–1
d)the gradient of graph indirectly shows the acceleration.d=0 ,it is steepest at that point.
from graph ,acceleration decreases(negative gradient) and increases(positive gradient).minimum at 4cm.
a little late...but thanks :) + rep
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A thermometer can be read to an accuracy of ±0.5 °C. This thermometer is used to measure a
temperature rise from 40 °C to 100 °C.
What is the percentage uncertainty in the measurement of the temperature rise?
A 0.5 % B 0.8 % C 1.3 % D 1.7%
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A thermometer can be read to an accuracy of ±0.5 °C. This thermometer is used to measure a
temperature rise from 40 °C to 100 °C.
What is the percentage uncertainty in the measurement of the temperature rise?
A 0.5 % B 0.8 % C 1.3 % D 1.7%
I'm not sure... Engraved or Astar can check. :-\
But still.
Rise = 60 °C
+ 0.5 or -0.5 => Interval = 1 °C
% uncertainty = 1/60 °C *100% = 1.7%
Have the MS?
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I think it should be B. 0.8% :-\
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should be 1.7%
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Should be B
0.5/60
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Should be B
0.5/60
It's measuring a change.
100 +- 0.5 - 40 +-0.5
the total uncertainty in the final ans will be +- 1 as far as my knowledge goes.
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It's measuring a change.
100 +- 0.5 - 40 +-0.5
the total uncertainty in the final ans will be +- 1 as far as my knowledge goes.
oooh yeah right, I get it now. Thanks. :)
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oooh yeah right, I get it now. Thanks. :)
That's what I think it is...anytime (:
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Adi has the MS?
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Adi has the MS?
1.7 is right
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1.7 is right
Sure? When I saw it was Physics, I became unsure... lol
By the way, thanks. ;)
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Anytime :)
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Q1,find the speed of the satellite which orbitz the moon near the moon's curface?
what is the kinetic energy per unit mass of the satellite?
(radius of moon: 1.74*10^6,mass of moon: 7.35*10^22)
Q2,astronomical observationz show the centre of mass of the earth-moon system is 4.7*10^6 m frm the centre of the earth.the distance b/w the centre of the earth n the moon is 384.4*10^6 m.find the mass of the moon M in terms of the mass of the earth E.
explain why both earth n moon must rotate about their common centre of mass,rather than the moon abot the centre of mass of the earth.
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Q1,find the speed of the satellite which orbitz the moon near the moon's curface?
what is the kinetic energy per unit mass of the satellite?
(radius of moon: 1.74*10^6,mass of moon: 7.35*10^22)
Q2,astronomical observationz show the centre of mass of the earth-moon system is 4.7*10^6 m frm the centre of the earth.the distance b/w the centre of the earth n the moon is 384.4*10^6 m.find the mass of the moon M in terms of the mass of the earth E.
explain why both earth n moon must rotate about their common centre of mass,rather than the moon abot the centre of mass of the earth.
Mass of the satellite? :\
Do you have the entire question? Is it from a past paper?
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this is the entire ques!!
itz frm unsolved topical pastpaperz,this one is frm year Nov88/p2/Q3.
do u have dese pastpprz??
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In the first question, the gravitation pull provides the necessary centripetal force. So you equate the two
GMm/ r2= mv2/r
You get v.
The second half, I don't really get it. The mass of the satellite ain't mentioned...I can't figure that.
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ok np!! :)
thanx for de help,you r the saver!! :)
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The second question. At 4.7*10^6 m frm the centre of the earth. the gravitational pull on any objects would be 0 apparently.
So.
GM/ (4.7*10^6)2= GM2(mass of the moon)/ (384.4*10^6 - 4.7*10^6)2
You get M2.
I don't get the second half of this one either :\
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hehe itz ok!! ;D
i ll let u knw de scnd part of both dese ques if i becum smart enough to get thm frm ma teacher'z xplanation!! :P
thanx alot once again!! :)