Physics: Post your doubts here!

[The Sarcastic Retard]

Which question?

The same question u solved for me..
How iscalculations done??
How Fg = mg?

 

[ruby_fern]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_43.pdf
Q12 b(ii)
what do you mean by uninterrupted length and how do you solve this question ?

I think what they meant by 'minimum signal power' is the minimum power output from the cable and they've just asked for the length so you substitute all this in the formulae and find length.

 

[The Sarcastic Retard]

Physicist can u please answer all of my questions on electric field that I posted? with mentioning wht concepts u have used to answer the respective questions?
Thanks.

 

[The Sarcastic Retard]

i believe some of them are already solved at the blog. try to look for them there.
i'm a bit busy currently.

anyways, thanks.

 

[a_wiserME!!]

pls explain...

 

[ruby_fern]

if you could refer to the link instead of copy-paste, i would be good.


thanks

here is the link http://physics-ref.blogspot.com/2015/12/a-level-physics-9702-exam-tips.html

 

[a_wiserME!!]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_42.pdf

Q.4 (c) pls explain. TIA

 

[ruby_fern]

View attachment 59401

pls explain...

which year is this paper?

 

[The Sarcastic Retard]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_42.pdf

Q.4 (c) pls explain. TIA

same doubt.

 

[a_wiserME!!]

which year is this paper?

m/j 10 v.42

 

[nehaoscar]

View attachment 59378
View attachment 59379
How to calculate area under graph??

Divide the area into small sqaures
so say a 5*5 will be 1 sqaure
count approx the no. of squares under the graph
Count the pd in 1 sqaure and multiply this by the no. of squares
so say this red box is one sqaure

then count no. under graph

 

[nehaoscar]

View attachment 59380
c and d

part c
ke = work done
1/2mv^2 = VQ
And calculate

part d
F=ma
a is max when F is max
and F=EQ
so F is max when E is max so at d=0

ii - acceleration decreases and is min at 4cm and then increases

 

[nehaoscar]

View attachment 59381

b(ii) and c.

bii
V at A = V at B
QofA/4piEr = QofB/4piE(0.3-r)
And calculate

c
F is proportional to field strength.
Field strength is proportional to gradient
so f is prop to gradient
so F is largest at max gradient

 

[nehaoscar]

View attachment 59383
View attachment 59384
(ii) and (iii)

ii
intially ke + gpe = epe
so ke = epe - gpe

iii
fusion may occur

 

[nehaoscar]

View attachment 59394
View attachment 59395
So yes, electric field strength is a vector quantity ... But are the directions opposite coz the field lines will go INTO both the alpha particle and the proton?

And this ...
force per unit charge M1
on either a stationary charge
or a positive charge

Should we write either of the two .. or should we write 'either a stationary charge or a positive charge'??

Since it's a vector quantity and both charges are +ve the field at P due to the 2 particles are in opposite directions so the opp. signs

 

[The Sarcastic Retard]

Divide the area into small sqaures
so say a 5*5 will be 1 sqaure
count approx the no. of squares under the graph
Count the pd in 1 sqaure and multiply this by the no. of squares
so say this red box is one sqaure
View attachment 59419
then count no. under graph

I am so dumb :/
can u please show the steps

 

[a_wiserME!!]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w14_qp_22.pdf

Q.5 (a) (ii) pls explain this

 

[The Sarcastic Retard]

part c
ke = work done
1/2mv^2 = VQ
And calculate

part d
F=ma
a is max when F is max
and F=EQ
so F is max when E is max so at d=0

ii - acceleration decreases and is min at 4cm and then increases

I UNDERSTOOD!!!!!!!!!!!!!!!!!!!!!!! THANKS A TONNES!

 

[nehaoscar]

I am so dumb :/
can u please show the steps

So if u take that red box ie 5*5 sqaure the wd for that one box is E*d which is approx 2.5x10^3*1x10^-2 (we'll call this answer as Z ... I'm too lazy to work out the ans )
Now count approximately how many of these 5*5 boxes will fit under the curve. This will be say n boxes
So then multiply Z*n to get the total wd

 

[Catherine_1]

View attachment 59401

pls explain...

See Qs 500.
http://physics-ref.blogspot.in/2015/03/physics-9702-doubts-help-page-97.html