Physics: Post your doubts here!

[Thought blocker]

See you send me mine work, so see it and look where are you stopped at.

 

[SIstudy]

See you send me mine work, so see it and look where are you stopped at.

umm I don't get it :/
and neither the next part of this question
nvm i'll try later
thanks fr ur help tho

 

[Thought blocker]

umm I don't get it :/
and neither the next part of this question
nvm i'll try later
thanks fr ur help tho

Look you are asked to find the speed of the ball at a time of 0.40 s yes?
So here the graph is curved so we cant do the way you did.
So the other possible way is with the help of tangent.
So, make a line touching that 0.4 above and down, make sure no other points are touched in the graph.
So that you can now calculate it from gradient as s here = change in d/ change in time. Thats simply change in y/ change in x means (y2 - y1)/(x2 - x1)

That's it.

Be calm, do it after 30 minutes. I bet you will be able to do both. If not, do tell me. I will be here to help you Sls!

 

[SIstudy]

Look you are asked to find the speed of the ball at a time of 0.40 s yes?
So here the graph is curved so we cant do the way you did.
So the other possible way is with the help of tangent.
So, make a line touching that 0.4 above and down, make sure no other points are touched in the graph.
So that you can now calculate it from gradient as s here = change in d/ change in time. Thats simply change in y/ change in x means (y2 - y1)/(x2 - x1)

That's it.

Be calm, do it after 30 minutes. I bet you will be able to do both. If not, do tell me. I will be here to help you Sls!

ahan probably I was drawin the tangent wrong lol :') #non#math#student
thank yu

 

[Thought blocker]

ahan probably I was drawin the tangent wrong lol :') #non#math#student
thank yu

You're welcome. \(^o^)/

 

[SIstudy]

You're welcome. \(^o^)/

another favor
do yu hav an excess to the s14 p1 & p2
I fail to find em anywhere :')

 

[Thought blocker]

another favor
do yu hav an excess to the s14 p1 & p2
I fail to find em anywhere :')

Variant?

 

[SIstudy]

Variant?

any variant will be appreciated
i'm lookin fr all the variants tho :')

 

[Thought blocker]

any variant will be appreciated
i'm lookin fr all the variants tho :')

Variant 1 : Here.
Variant 2 : Here.
Variant 3 : Here.

 

[SIstudy]

Variant 1 : Here.
Variant 2 : Here.
Variant 3 : Here.

Thanks ALOT

 

[TimBluesWin]

Can anybody explain to me how to do no. 7 mayjune 2009 paper 04? Thanks a lot!!

 

[SIstudy]

Physicist can yu pls help me with this
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_21.pdf
Q2 B (ii) (iii) and Q3 (b) (iii) 2

 

[Bba321]

Please help me with qs 2 part a
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w08_qp_2.pdf
Why can't we take the distance as 29.3 to calculate the speed v of the car before the brakes are applied?

 

[SIstudy]

sure. but it won't right now.

I'll provide the explanations tomorrow insha Allaah

please also Q4 (B) and (C) and in Q7 c(ii) how to find volume?
(questions are of the same paper)

 

[Zepudee]

Hi anyone here, can help me to evaluate electric potential graph and electric field strength graph? I always find it confusing. Can anyone help me?

 

[Bba321]

Please explain me qs 3 part d
Thanks

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w06_qp_2.pdf

 

[manutd96]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_2.pdf

Paper 21 Q7(a)(ii), how do we know its 0.5R? how do we see that it is parallel? and lets say if a third resistor of same resistance is added in parallel, is the resistance AB still the same or different?thanks

 

[TimBluesWin]

look at qu 23 at
http://physics-ref.blogspot.com/2014/10/physics-9702-doubts-help-page-4.html

and SIstudy

solutions for the june 2014 papers are available at http://physics-ref.blogspot.com/2014/05/physics-9702-notes-worked-solutions-for.html

Heyy.. Sorry late reply.... But isn't the diagram shown at qu 23 is galvanometer? Also I don't understand the may/june 2009 no 9b... Any help would be appreciated. Thanks!!

 

[manutd96]

than

solutions have been posted at

http://physics-ref.blogspot.com/2014/11/9702-june-2009-paper-21-worked.html

thanks but i am still a bit unsure. does it mean that the resistance between BY BX CX are the same? meaning the total resistance between BY BX CX is the total resistance of the resistors in parallel. is this true?

 

[Bba321]

Hey can anyone please explain qs 5c part 2?

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w05_qp_2.pdf

Thanks