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View attachment 65219 Can someone please help me solve this problem?
Explanation for X:View attachment 65229
Can someone please help me solve this problem?
Consider the weight of a column of air of cross-sectional area A and height h stretching from sea level to the edge of the atmosphere.Can anyone help me with this?? "B"View attachment 65228
Thankss! 😊😊go to
A positively charged oil droplet falls in air in a uniform electric field that is vertically upwards. The droplet has a constant terminal speed v0 and the electric field strength is E.
Physics Reference - A-Level 9702, Past Exam Paper Solutions ..physics-ref.blogspot.com
A small charge q is placed in the electric field of a large charge Q.
Both charges experience a force F.
What is the electric field strength of the charge Q at the position of the charge q?
A) F/(Qq)
B) F/Q
C) FqQ
D) F/q
Answer is D, why isn't it B? :O HELP
Hi everyone, AsSalamoAlaikum Wr Wb...
To get things organized in a better way, I am making this thread. As othewise, some queries remain unanswered!
So post your PHYSICS doubts in this thread. InshaAllah other people here will help me and you all.
NOTE: If any doubts in the pastpapers, please post the link! You can find links here!
Any Physics related notes and links will be added here in this post. Feel free to provide the links to your notes around the forum, or any other websites!
Thanks!
Jazak Allah Khair!
Physics Notes:
Some links & Notes - by destined007
As physics p1 MCQS YEARLY Solved [explaination]
Physics Practical Tips - by arlery
Notes for A2 Direct Sensing (Applications) - shared by sweetiepie
Physics Summarised Notes (Click to download)
AS and A-Level Physics Definitions
A2: Physics Revision notes - by smzimran
Paper:5 Finding uncerainty in log - by XPFMember
Physics Paper 5 tips - by arlery
Physics Compiled Pastpapers: <Credits to CaptainDanger for sharing this..>
Here are the compiled A level topical Physics questions in PDF form...
Paper 1 : http://www.mediafire.com/?tocg6ha6ihkwd
Paper 2 & Paper 4 : http://www.mediafire.com/?g65j51stacmy33c
(Source : http://www.alevelforum.com/viewtopic.php?f=15&t=14)
It is because the charge is constant but the potential is changing which means you've got to integrate the dE = QdV expression over the given limits of V1 and V2. Since the 1/2 factor is only introduced due to integration, therefore this gives the right answer. The formula (change in V) = E/Q doesn't have the 1/2 factor in it so you can't get the correct answer because of it.View attachment 65745
in the last part why cant we find the work done using change in potential=workdone/charge. and since workdone is the change in energy it should be valid right? it doesnt give the right answer.
To answer this question, you need to consider the free body diagrams (ignoring trolley weight of course) before and after the displacement of trolley. The right side can be considered as positive for force and displacement direction and vice versa. So initially the forces from both springs act in opposite directions in equal magnitude, thus keeping the trolley in equilibrium, i.e. F = ke.View attachment 65741can anyone explain Q11 (b) (i)? I read an explanation somehwere and it said that the two forces have to be subtracted because they are opposite in direction. Why subtract when both forces are in same direction? Shouldn't we add? Both springs will exert a force to the left to bring it to the equilibrium position right?
but the potential needs to be changing; we need to have two values of potential to get potential difference which is exactly what was given. What i have figured is that the charge is what's changing cuz in one of the earlier parts of question it says charge required to raise sphere to a potential, so when the potential drops, sphere loses some of its charge in the spark so we cant use potential difference=workdone/charge cuz its the charge that doesnt stay constant. for potential difference we need to two different values. so I found the energy using different potential and charge values at both points and substracted them to get the loss in the energy and it was correct.It is because the charge is constant but the potential is changing which means you've got to integrate the dE = QdV expression over the given limits of V1 and V2. Since the 1/2 factor is only introduced due to integration, therefore this gives the right answer. The formula (change in V) = E/Q doesn't have the 1/2 factor in it so you can't get the correct answer because of it.
yeah that is understandable...totally makes sense.but the potential needs to be changing; we need to have two values of potential to get potential difference which is exactly what was given. What i have figured is that the charge is what's changing cuz in one of the earlier parts of question it says charge required to raise sphere to a potential, so when the potential drops, sphere loses some of its charge in the spark so we cant use potential difference=workdone/charge cuz its the charge that doesnt stay constant. for potential difference we need to two different values. so I found the energy using different potential and charge values at both points and substracted them to get the loss in the energy and it was correct.
Re: Physics Help here! Stuck somewhere?? Ask here!
Problems in june 2011 paper 11
No 25, 27, 29, 32, 35,, 36,37..
And also can u send me some notes on potential divider, internal resistance along with some working examples which cover the whole chapter...
Thnks in adv..
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