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Physics: Post your doubts here!

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hamidali391 said:
How did you get it completely? And I didn't get your question. Frequency is inversely proportional to wavelength. The greater the wavelength, the smaller the frequency. The speed remains same in the equation f=c/lamda and so you get the lowest F with greatest wavelength. You need to open the book and revise the concept of stationary waves before you ask anymore questions.
:oops: sorry 4 disturbing again nd again nd not reading the topic carefully. i read now and just understood that in a tube open from end and closed from the other has 1/4 lambda length. nd if u calculate the full wavelength by multiplying by 4 then u can get always the fundamental frequency. ''for a tube of length l in the closed tube the standing wave formed is one quarter of a wavelength, so the wavelength is 4l''
i asked something else that how is this the greatest wavelength i know the relation between v f nd lamda. thanks bro it is clear now
 
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Physics Columb's law doubt :(

Asallam o allaikum all :)
Guys I was just attempting some questions of columbs law (A2) and got stuck on a question:

Q) A metal sphere of radius of 20cm carries a positive charge of 2.0microC.
a. An identical metal sphere carrying a negative charge of 1.0microC is placed next to the first sphere. There is a gap if 10cm between them. Calculate the electric force that each sphere exerts on the other. (Ans is 0.072N)
b. Determine the electric field strength midway along a line joining the centres if the spheres. ( Ans is 4.32V/m)

My doubt is in the b. part, I cant even get close to 4.32V/m :( Hope some one can help me out :) Thank you

Regards,
Osama Anwer
 
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rviboy said:
Dayyanah said:
hey, can anyone help me with any sort of phy AS notes? i dont feel like reading the text book anymore
thnx
i have tried many notes, but the best one is the Cambridge international As and A level Physics coarse book by David Sang, Graham Jones etc. it is a great book and comprehensible.

still http://en.wikibooks.org/wiki/Advancing_Physics

http://miniphysics.blogspot.com/2010/11 ... h1-h2.html

Yeah agree with that I got that book too and itz Awesoooome!!! :D
 
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slayers said:
Is any Physics Books Available on Internet for FREE ? send the Link

Well dunno about any physics free books bt I do know about a website which will be very helpful :) it consist the whole syllabus A level (Including Physics) with detailed expiation and animations. Here is the link: http://www.s-cool.co.uk Hope it helps :)
 
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please if anyone of u know the answer to the question pls do help
i will also be thankful if you can give me a valid explanation for ur answer :D
thank u in advance
 

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Ok this is rather simple but i got a bit confused ...
A car of 5KN travels 1km along a sloping road ending up 100 m above the starting point.its engines provide forward motive force of 2KN. What is the work done against gravity?
 
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Arshiful said:
please if anyone of u know the answer to the question pls do help
i will also be thankful if you can give me a valid explanation for ur answer :D
thank u in advance

Hey mate :) I was also stuck once on this question but cleared my doubt :D so coming onto the question as you already know that 'E' is constant so 'F' is constant too (E=F/q) and as 'F' is constant 'a' will be constant too (F=ma). There fore velocity will increase at a constant rate. Hence you must have figured out that it must be a straight line graph from the origin :p but NO!!!! as this in ain't a Velocity/time graph but a velocity/displacement graph. Since, velocity increases at a constant rate, then the displacement will increase at an increasing rate, giving rise to a curve with a progressively decreasing gradient. but remember that the curve will not level out!!! it will end up at 9cm with slant curve. Hope it helps : ) you can ask as many questions as you want ;)
 
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Konfused said:
Ok this is rather simple but i got a bit confused ...
A car of 5KN travels 1km along a sloping road ending up 100 m above the starting point.its engines provide forward motive force of 2KN. What is the work done against gravity?

Well this is actually M1 Mathematics :p indeed:

Hope it helps :)
 

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In P2, electric field chapter, When two plates are horizontally aligned together, why don't we consider gravitational accln instead we just calculate out accln by calculating from electric force?
 
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Swoorup said:
In P2, electric field chapter, When two plates are horizontally aligned together, why don't we consider gravitational accln instead we just calculate out accln by calculating from electric force?

Well it is not consider because the mass of an electron/proton is very small, still for the sake of explanation if u find it out by using F=ma the force as a result of gravitation is negligible in front of electric force.

PS. F=ma cannot be used to find force experienced by an electron/proton due to gravity instead we have to use F=GMm/r^2 to precisely calculate the force. (A2 Part- Gravitational fields)

Summing up: Mass of an electron/proton is tooooooooooooooo!!!! small to be effected by gravitational acceleration, but the charge of an electron/proton experiences allot force by an electric field. Hope it helps :)
 
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I had practiced one question, dun remember which one,and mass do get considered. Don't remember which one. But majority of the questions have not involved any similar problems.
Never mind though, sticking to what you said! :)
 
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ahan I see but as far as my knowledge is concerned dat was my answer but if u get around a question in which mass is considered plzz let me knw ;) QUESTION ONLY IN P2 :p i MAY BE WRONG :(
 
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uable to solve a part. don't have a convincing answer for that

b) Electric field strength = Potential difference / distance between the plates. equate the values get the answer...

c) electric field strength = Force / charge so F here is equal to mg , equate both equations Eq = mg ..equate the values.

if u dnt get anything u can ask me !
 
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Hey.. Can you explain me the working of a current balance when it is being used to measure the magnetic flux density.?
 
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