Physics: Post your doubts here!

[Mathemagical]

http://papers.gceguide.com/A Levels/Physics (9702)/9702_s14_qp_41.pdf
qtn 10(a) & b(ii)
plzz help.....

3:

a) Binding energy is the energy required to separate the nucleons in a nucleus to infinity.

OR

Binding energy is the energy required to bring together nucleons in a nucleus from infinity. (first definition preferred)

b) ii)

Mass defect, ∆m = 211.70394 – 209.93722
= 1.76672u

binding energy = 1.76672 × 930

binding energy per nucleon = (1.76672 × 930)/210

= 7.82 MeV

Divide by 210 because there are 210 nucleons, and the questions asks for binding energy "per nucleon".

Hope this helps!

 

[selrey]

Q. When a potential difference V is applied between the ends of a wire of a diameter d and lenght l, the current in the wire is I. What is the current when a potential difference of 2V is applied between the ends of a wire of the same material of diameter 2d and the lenght 2I?
SOMEONE PLS HELP WITH THIS Q!

 

[Mathemagical]

Q. When a potential difference V is applied between the ends of a wire of a diameter d and lenght l, the current in the wire is I. What is the current when a potential difference of 2V is applied between the ends of a wire of the same material of diameter 2d and the lenght 2I?
SOMEONE PLS HELP WITH THIS Q!

Resistance is proportional to length, and inversely proportional to diameter/radius squared.

Let the resistance of the first wire be R.

Resistance of second wire = 2*(1/4) R = 0.5R

Current = 2V/0.5R = 4I

 

[selrey]

Resistance is proportional to length, and inversely proportional to diameter/radius squared.

Let the resistance of the first wire be R.

Resistance of second wire = 2*(1/4) R = 0.5R

Current = 2V/0.5R = 4I

THANYOUUU YOU MADE IT SO EASY! I was trying to solve it in twisted way

 

[Mathemagical]

THANYOUUU YOU MADE IT SO EASY! I was trying to solve it in twisted way

No problem!

 

[Thought blocker]

https://www.xtremepapers.com/community/threads/physics-component-marks-for-a-level-only.46161/

 

[kdawgtresbon]

This is paper 5 of 2015. I wanted to ask if the way of writing is okay ? Particularly if i should write in continuous prose or if points are fine

 

[Thought blocker]

This is paper 5 of 2015. I wanted to ask if the way of writing is okay ? Particularly if i should write in continuous prose or if points are fine

Hey there. I haven't read those points as I am sure you would have checked from marking scheme. Answering your question you can do whatever you want but I would strongly encourage you to write in paragraph format. I will tell you my personal experience about paper 5; I started it with writing in point format as that looked quick and easy method and as all the points if I cover in point format then no need of paragraph format, I use to get full marks in method but wasn't able to score in additional details. I thought lets try paragraph format may be that could help me. AND IT DID!!!!!!!!!!!!!!!!!! BINGO! In paragraph format there were many things that I could included which I was unable to add in point form, idk why, but the para foormat worked great with me I started getting 28 or 29 or 30 marks with that. So it all depends on you. If you can cover all points in point format then GO FOR THAT! but if u have problem like me try the way I did. xD

RIP my English. Ignore it.

 

[krishnapatelzz]

https://www.xtremepapers.com/community/threads/suggestion-for-9709-mechanics-p4-5.46187/
mechanics!!!!!!!!!!

 

[Mari_am]

Pls can I get electric fields A2 notes with explanations on how to plot the graphs in it

 

[Thought blocker]

Pls can I get electric fields A2 notes with explanations on how to plot the graphs in it

In this you can find it. Good luck.

 

[Mari_am]

Thank u

 

[Laveeza]

The question is 'describe the effect if any on the separation and on the maximum brightness of the fringes when the following changes are made.
The wavelength is increased by 1.5 factor keeping a and D constant
Effect on separation................. ....
Effect on maximum brightness of fringes ..

 

[nadeen64]

Hi, can someone please explain how to do the graph for part c? this is mj 2016 paper 42

 

[krishnapatelzz]

Hi, can someone please explain how to do the graph for part c? this is mj 2016 paper 42

 

[Thought blocker]

HAhahhaa.... MISS OUR TIME OF PHYSUCKS

 

[krishnapatelzz]

HAhahhaa.... MISS OUR TIME OF PHYSUCKS

yaaaaaaaaaa!!!!!!!!!!!!!!!!!!!!

 

[nadeen64]

mj 2016/42 q 9 bii
Help anyone??

 

[AnonymousX9]

Hey guys, I have this doubt that's been bugging me for so long now.

Why is the electric potential inside a charge conducting sphere constant?
Why is the electric field inside a charge conducting sphere equal to zero? (I know that since there is no change in potential, E = 0 but isn't there any explanation for that?)

It'd be really helpful if you guys could clear this for me and I'd love a simple/precise explanation that is A Level based.

Thank you.

 

[DragonCub]

Hey guys, I have this doubt that's been bugging me for so long now.

Why is the electric potential inside a charge conducting sphere constant?
Why is the electric field inside a charge conducting sphere equal to zero? (I know that since there is no change in potential, E = 0 but isn't there any explanation for that?)

It'd be really helpful if you guys could clear this for me and I'd love a simple/precise explanation that is A Level based.

Thank you.

Both questions can be answered by the same fact: electric charge is distributed uniformly inside the conducting sphere.

When outside the sphere, the electric field is not uniform - it is more concentrated nearer the sphere surface. This means there is a "concentration gradient" outside the sphere (with respect to distance).
Electric field is actually a measure of the magnitude of this gradient.
Electric potential is actually a measure of the integral of this gradient.

But when inside the conducting sphere, charge is uniform everywhere, meaning there is no such gradient, or gradient is zero.
Since gradient = 0 everywhere in the sphere:
- Electric field is zero.
- The integral is constant, which means electric potential is constant.