Physics: Post your doubts here!

[shinnyyy]

5/3 nd 1/3 are their speeds after collision as given In opt A
Lol....my sir gav a long explanation abt whn to xhange signs nd when not to...ill try to snd u asap

Plx do send me that ..I'll be greatful to you.

 

[Wkhan860]

Plx do send me that ..I'll be greatful to you.

Ill try my best....actually im don wth my AS....so it will take me sm tym to fynd where i kept it.

 

[shinnyyy]

Ill try my best....actually im don wth my AS....so it will take me sm tym to fynd where i kept it.

Oh then you would have done chem and bio too ?

 

[Wkhan860]

Oh then you would have done chem and bio too ?

I ddnt take bio
Dn wth maths nd chem

 

[shinnyyy]

I ddnt take bio
Dn wth maths nd chem

So did you study hybridization in chem? That's also a new thing in syllabus and it's not in the book so I need someone to explain me that thing too

 

[Wkhan860]

So did you study hybridization in chem? That's also a new thing in syllabus and it's not in the book so I need someone to explain me that thing too

Yup I did.....thank god It ddnt cam In exams
ill send u my.notes nd explanation by tonight

 

[shinnyyy]

Yup I did.....thank god It ddnt cam In exams
ill send u my.notes nd explanation by tonight

Alhumdulilah lucky u
thankx

 

[shinnyyy]

Wkhan860 here comes another two questions for you.

 

[fantastic girl]

View attachment 55367
Wkhan860 here comes another two questions for you.

14-
Horizontal component of velocity = v cos(45)
Vertical component of velocity = v sin(45)

initial kinetic energy E = 1/2 m v^2
now see the question says KE at highest position which means there wont be vertical velocity as at the highest point vertical v is zero
we only hve horizontal so final KE = 1/2 m (cos 45)^2 v^2
final kinetic energy is 1/2 m 0.5 v^2......this is half of E
hope u got it!

 

[fantastic girl]

View attachment 55367
Wkhan860 here comes another two questions for you.

Is 13 B ?

 

[shinnyyy]

14-
Horizontal component of velocity = v cos(45)
Vertical component of velocity = v sin(45)

initial kinetic energy E = 1/2 m v^2
now see the question says KE at highest position which means there wont be vertical velocity as at the highest point vertical v is zero
we only hve horizontal so final KE = 1/2 m (cos 45)^2 v^2
final kinetic energy is 1/2 m 0.5 v^2......this is half of E
hope u got it!

Thankyouuuu

 

[Wkhan860]

View attachment 55367
Wkhan860 here comes another two questions for you.

srry....ddnt get the tag

 

[shinnyyy]

srry....ddnt get the tag

Ah no worries man :') you have already helped me alot
N fanta helped me so it's ok :')

 

[fantastic girl]

Thankyouuuu

most welcome

Is 13 B ?

????
got it?

 

[Boss201]

The problem is for c(ii)
i use Vout/V in=1 +Rf/R1
Replacing the values Vin=+0.4 , Rf =120 R1=5
i get answer :10 V
But the answer from marking scheme is 9 V . Can someone tell me the Reason ? is it linked to the diagram in part B ?

 

[Asfand Tayyab]

Can anyone tell me where to get Topical Mcq's P1 for As-Levels?????

 

[Syed Fakhar Hussain]

https://www.dropbox.com/s/kd8rxtx7fwig3wt/Final MCQ'S All Physics 9702.pdf?dl=0 bro visit this dropbox link.

Can anyone tell me where to get Topical Mcq's P1 for As-Levels?????

 

[Bharath 28 2 2000]

Can anyone please upload PHYSICS Notes for CIE AS level

 

[Robert Flockhart]

4/M/J/02 Q.4(b)
6/M/J/02 Q.5(a),Q.14(c),Q.15(a)

 

[The Sarcastic Retard]

4/M/J/02 Q.4(b)
6/M/J/02 Q.5(a),Q.14(c),Q.15(a)

s04
4bi)
change in momentum = [m(v - u)]/t
--------------------------> 0.035(4.5-(-3.5)) / 0.14 = 2
Direction will be upwards. (Hope u know this)

ii) K.E(l) = 0.5 * 0.035 * (4.5^2 - (-3.5)^2) = 0.14


s06
Please attach the link to the paper.