Author Topic: Physics help A2  (Read 2766 times)

nid404

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Physics help A2
« on: July 12, 2010, 04:48:44 pm »
 June 07 Q3

Can someone please explain part b and c

Thanks  :)

nid404

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Re: Physics help A2
« Reply #1 on: July 12, 2010, 04:55:28 pm »
I have a test tom...please? :P

Offline *Hope*

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Re: Physics help A2
« Reply #2 on: July 12, 2010, 05:31:43 pm »
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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nid404

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Re: Physics help A2
« Reply #3 on: July 12, 2010, 05:32:22 pm »
lol thanks :)

If you know anyone who can help...please direct them here  :-[

Offline Nobody

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Re: Physics help A2
« Reply #4 on: July 12, 2010, 05:58:32 pm »
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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nid404

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Re: Physics help A2
« Reply #5 on: July 12, 2010, 05:59:32 pm »
yup yup paper 4  ;D

Offline Light

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Re: Physics help A2
« Reply #6 on: July 14, 2010, 12:59:38 pm »
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.

nid404

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Re: Physics help A2
« Reply #7 on: July 14, 2010, 03:18:55 pm »
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

Offline Ghost Of Highbury

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Re: Physics help A2
« Reply #8 on: July 28, 2010, 02:44:40 pm »
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%
divine intervention!

Alpha

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Re: Physics help A2
« Reply #9 on: July 28, 2010, 02:59:03 pm »
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?
« Last Edit: July 29, 2010, 08:33:27 am by ~Alpha »

Offline Vin

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Re: Physics help A2
« Reply #10 on: July 28, 2010, 03:11:14 pm »
I think it should be B. 0.8% :-\

nid404

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Re: Physics help A2
« Reply #11 on: July 28, 2010, 03:27:38 pm »
should be 1.7%

Freaked12

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Re: Physics help A2
« Reply #12 on: July 28, 2010, 03:31:29 pm »
Should be B

0.5/60

nid404

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Re: Physics help A2
« Reply #13 on: July 28, 2010, 03:33:24 pm »
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.


Offline Vin

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Re: Physics help A2
« Reply #14 on: July 28, 2010, 03:41:54 pm »
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. :)