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The Ultimate AS Physics Thread

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guys,

can you help me with may/june 2008 ques 6b. i got everything alright its just the 2nd and the 3rd that i don't understand.

thanks
 
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Talha said:
libra94 said:
7bii
volatage in the thermistor 6-3.6= 2.4V
current in 2000Ω resistor = 3.6/2000 = 0.0018A
current in 5000Ω resistor = 2.4/5000 = 0.00048A
kirchoff's first law states that the current going into the point is equal to the curent leaving out of that point
so, the current in the thermistor is 0.00048 - 0.0018 = 0.00132A
R=V/I
V is 2.4V (as the voltage is the same in parellel connection)
I is 0.00132A
so R will be 1818.2Ω



even if you didn't use kirchoff's law which i happen to not understand here too, you will still get the answer.
- all you need to do is use the formula is (a) to find the total resistance of the Q and the thermistor.
- then use the normal parallel resistance formula and you will get the resistance of the thermistor.
 
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Please help in Nov07/02, q5 a) part. How we can show the phase difference between two waves that is 60 deg...??
 
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one complete oscillation is 360' = 1 wavelength

so half an oscillation is 180'.

then 180' divide by 3 = 60'

- the question made it really easy cause half an oscillation is 3 squares so 1 square = 60'.
 
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Hey guys, need help ASAP.
1st question: May 05, Q 5b(iii)- I dont understand the answer in ms.
2nd question: Oct 07, Q 4(c)-please explain the method, no clue about it.

THX
 
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I think someone already asked this question but no answer was posted:

M/J 2009, Q5, Variant is 21

Please explain in detail.

Secondly, I always get the wrong uncertainties so can someone please show me how to calculate them in questions such as:

M/J 2009, Q1, Variant is 21

Please solve the question and show full working.

Thanks.
 
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yubakkk said:
Please help in Nov07/02, q5 a) part. How we can show the phase difference between two waves that is 60 deg...??


the answer to this question is already possted on page 8 wd full details..u can have a look at it... :)
 
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o/n 05 q. 5

ci
just take the max amplitudes
I for A and Io for B. the corresponding amplitudes are 3 and 2
I=k x amplitide^2
so I/Io =3^2/2^2
Io=Ix4/3


c ii asume I. for new intensity
At the point of greatest amplitude
resultant amplitude is3-2=1
so, I./I=1^2/3^2
I.=I/9

d ii
just look at diplacements at 4 ms
corresponding displacements are (-2.6 and1.7) x 10^-4
so resultant displacement is (-2.6+1.7) x10^-4= 0.8x 10^-4 cm
 
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if its paper 2 then...here it goes.........

3b...i) upward force is equal to the downward force
570 + x = 950
so the reading on B will be 950-570 = 380 N

ii) it must be nearer to the Y end, cuz,again, upward forces are eual to the downward forcces, so to make the plank balanced, the weight must be somewhere near Y end!! make an arrow downwards labelling it 950N

iii) clockwise moments = anticlockwise moments
950 * x = (0.4*570) + (2.5-0.4 * 380)
so find the x
 
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7b....u use the equation d sinθ = λn
make sure all the units are in meters!!!
λ = 630 / 1000 000 000
n= 2
sinθ = (162 - 136) / 2
then find ur d, which is equal to 5.6*10^-6 m!!
 
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