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Physics: Post your doubts here!

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It varied depending on the values of t for example one set of values were 9.0 and 9.8s. Here the period would be 9.4/20 but the uncertainty would be ( (9.8/20)-9/20)/2 - basically the max and min value method of finding uncertainty: max value-min value/2
What variant is this?
 
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It varied depending on the values of t for example one set of values were 9.0 and 9.8s. Here the period would be 9.4/20 but the uncertainty would be ( (9.8/20)-9/20)/2 - basically the max and min value method of finding uncertainty: max value-min value/2
So if you do by this method, the mean time is 9.4±0.1s so isn't max value 9.41 and min 9.39? Error in T will then be 0.005
And if we apply the other rule that if a quantity is being multiplied or divided by a number, then the error in the calculated quantity is error in previous quantity times the number. In this case it would have been 0.1/20 = 0.005 for all values of T
 
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So if you do by this method, the mean time is 9.4±0.1s so isn't max value 9.41 and min 9.39? Error in T will then be 0.005
And if we apply the other rule that if a quantity is being multiplied or divided by a number, then the error in the calculated quantity is error in previous quantity times the number. In this case it would have been 0.1/20 = 0.005 for all values of T
What? Mean time would be 9.4s but with an uncertainty of 0.4s. Mean period would be 0.47 with an uncertainty of 0.02s.
And if we apply the other rule that if a quantity is being multiplied or divided by a number, then the error in the calculated quantity is error in previous quantity times the number. In this case it would have been 0.1/20 = 0.005 for all values of T
Oh we can't use the multiplied or divided rule in this because the calculations only involved one variable that had one uncertainty. So we had to use max-min values/2 - really the only method.
 
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Hi guys. For those who took Paper 21 in AS Level Physics today, which formula should have been used to calculate the charge of the oil drop in the electric field question?
 
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FREE HELP PROVIDED
Hi sorry to be a bother but if anybody needs help with any physics or maths topic im giving free help be it past paper questions or conceptual. Just email me on [email protected] and i'll reply asap good luck for exams yall
 
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Hi. Can someone help me with syllabus content of A Level Physics Chapter 2, Measurement Techniques.
Do we have to learn using calipers and micrometer?
And do we have to learn all techniques of measuring mass, temperature, time, etc? How much detail is required? And is the detail in Hodder book irrelevant?
 

PlanetMaster

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hi. um could i get some help with June 2011 Paper 23 Q 4c ?
Question Paper: https://papers.xtremepape.rs/CAIE/AS and A Level/Physics (9702)/9702_s11_qp_23.pdf

Answer 4c(i)
Since all the energy is converted to kinetic, we have
\(W=\frac{1}{2}mv^{2}\)
\(0.81=\frac{1}{2}mv^{2}\)
\(v=8.05ms^{-1}\)

Answer 4c(ii)
\(\text{speed }v_{2}=8.05\times2=16.1ms^{-1}\)
Therefore, \(W=3.24J\)
Now for compression, we can use
\(W=\frac{1}{2}kx^{2}\)
and with W=3.24J, we get \(x=72mm\)

Answer 4c(iii)
At maximum height. all KE would be converted to GPE, so
\(\frac{\frac{1}{2}k{x_{1}}^{2}}{\frac{1}{2}k{x_{2}}^{2}}=\frac{36^{2}}{72^{2}}=\frac{1}{4}\)

Hope this helps!
 
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State
(i) what is meant by the electric potential at a point,
(ii) the relationship between electric potential at a point and electric field strength at the
point.

(b) Two similar solid metal spheres A and B, each of radius R, are situated in a vacuum such that
the separation of their centres is D, as shown in Fig. 6.1.

q9uheW965.png

The charge +Q on sphere A is larger than the charge +q on sphere B.

A movable point P is located on the line joining the centres of the two spheres.
The point P is a distance x from the centre of sphere A.

On Fig. 6.2, sketch a graph to show the variation with x of the electric potential V between the
centres of the two spheres.

qw845hn87.png
 
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PlanetMaster

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Answer a(i)
The work per unit of charge required to move a charge from a reference point to a specified point, measured in joules per coulomb or volts.
Answer a(ii)
\(\text{The electric field}=-\text{ potential gradient}\)
It comes from expressing energy changes in two different ways.
You can say work done = force x distance = qE x distance
However you can also say the work done = change in potential x charge moved = qV

Answer (b)
g98eyuddncd.png
The region left of Sphere A and right of Sphere B is inside the sphere so there is some potential there i.e. straight lines. The rest should be self explanatory.

Hope this helps!
 
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