Physics: Post your doubts here!

[Maiday Holsey]

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?

 

[Hamza Umair]

How do I study for physics practical

 

[hellodjfos;s'ff]

How do I study for physics practical

Look over mark schemes for question 2's last part and also just read the ECR for P3 - example candidate responses.

 

[FarGang]

Any idea regarding the grade boundary for A? I did AS and A level separately?

 

[FarGang]

Any idea regarding the grade boundary for A? I did AS and A level separately?

 

[Maiday Holsey]

Hi. Has anyone done Paper 11 today? How hard did you find it?

 

[Samatar011]

It was pretty good, just a couple questions I found hard.
What did you think of it?

 

[Maiday Holsey]

It was pretty good, just a couple questions I found hard.
What did you think of it?

Same

 

[ThePacifics]

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

 

[tabdeelisupporter]

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?

 

[fdshfdahda]

hi. um could i get some help with June 2011 Paper 23 Q 4c ?

 

[PlanetMaster]

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!

 

[Himani]

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.



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.

 

[PlanetMaster]

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)

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!

 

[fdshfdahda]

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!

thank you so much

 

[Rid@Ahmed]

how to solve this ??

 

[PlanetMaster]

View attachment 65032 how to solve this ??

From a velocity-time graph, the acceleration at time t is given by the gradient at time t. This gives the instantaneous acceleration at that time.
In this graph, the gradient and hence the acceleration is changing.

Note that the gradient at time t = 3s is obtained by drawing a tangent at that point and calculating its gradient. We cannot just take the coordinate at that point (3, 6) to obtain acceleration (as done for choice D).

A tangent at time t = 3s would pass almost at/close to points (0, 2) and (6, 10).
\(\text{Gradient}=\frac{(10–2)}{(6 – 0)}=\frac{8}{6}=1.33ms^{-1}\)

P.S. This is taken from an earlier post

 

[Rid@Ahmed]

From a velocity-time graph, the acceleration at time t is given by the gradient at time t. This gives the instantaneous acceleration at that time.
In this graph, the gradient and hence the acceleration is changing.

Note that the gradient at time t = 3s is obtained by drawing a tangent at that point and calculating its gradient. We cannot just take the coordinate at that point (3, 6) to obtain acceleration (as done for choice D).

A tangent at time t = 3s would pass almost at/close to points (0, 2) and (6, 10).
\(\text{Gradient}=\frac{(10–2)}{(6 – 0)}=\frac{8}{6}=1.33ms^{-1}\)

P.S. This is taken from an earlier post

Thank you

 

[fdshfdahda]

so i solved this and got B as the answer but ms says the right answer is 8cm/s. Can anyone please help me with this?

 

[Kar345]

How did Physics Paper 42 go???