Physics: Post your doubts here!

[MYLORD]

Any more doubts ?

Not Now !!
But i am still doing so i ll disturb u again !!

 

[Thought blocker]

Not Now !!
But i am still doing so i ll disturb u again !!

Sure...

 

[Mohammed salik]

Already solved

ok

 

[MYLORD]

Nazr Aye?

Bhai ji me andha nahi houn so ji haan nazr a gaey

 

[Thought blocker]

Not Now !!
But i am still doing so i ll disturb u again !!

Well your one doubt was left I guess, that is w12_13.... Q11
so here you go
P initial = 2mu - mu =mu
And for momentum to be conserved, P ini = P final
So check it for all question..
Option A)
-((2mu)/3)+(5mu/3) = 3mu/3 = mu so momentum is conserved Pi = Pf
Option b)
-((2mu)/6)+(2mu/3) = 1/3mu = Momentum is NOT conserved Pi =/(not equal) pf
So answer is option B

 

[MYLORD]

Well your one doubt was left I guess, that is w12_13.... Q11
so here you go
P initial = 2mu - mu =mu
And for momentum to be conserved, P ini = P final
So check it for all question..
Option A)
-((2mu)/3)+(5mu/3) = 3mu/3 = mu so momentum is conserved Pi = Pf
Option b)
-((2mu)/6)+(2mu/3) = 1/3mu = Momentum is NOT conserved Pi =/(not equal) pf
So answer is option B

thnx !!

 

[Thought blocker]

MYLORD
Q23 w12_11
Total constant of the three forces would be:
For the two above, it would be 2k. Now, add the third one,
1/2k + 1/3k = 5/6k = 6/5k (Just add them together like you do in parallel circuits when they are connected end to end)

W = 6/5k * x
x = W *5/6k = A

 

[MYLORD]

MYLORD
Q23 w12_11
Total constant of the three forces would be:
For the two above, it would be 2k. Now, add the third one,
1/2k + 1/3k = 5/6k = 6/5k (Just add them together like you do in parallel circuits when they are connected end to end)

W = 6/5k * x
x = W *5/6k = A

Thnx again !!

 

[Thought blocker]

Thnx again !!

Why u unfollowed me bro ?

 

[MYLORD]

Why u unfollowed me bro ?

Sorry my account is opened on my computer maybe did by mistake my small bro is always on the computer so yeah!!

 

[Thought blocker]

Oki

Sorry my account is opened on my account maybe did my mistake my small bro is always on the computer so yeah!!

 

[MYLORD]

Oki

ok here's one more doubt !
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf Q 13, 15 ,17,

 

[Thought blocker]

ok here's one more doubt !
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf Q 13, 15 ,17,

we will solve it through two simultaneous equations
W-T=ma (for the 2.0 kg mass) As this mass is pulled down due to its weight. The resultant force acting on it will be W-T as W is greater. There is no component of weight acting in the direction of motion for the 8 kg mass, therefore it will move when the other mass moves down.
T-F=ma (for the 8 kg mass)
F is the frictional force. T is cancelled out. Solve for a which is 1.36. Closest answer is A.

 

[MYLORD]

we will solve it through two simultaneous equations
W-T=ma (for the 2.0 kg mass) As this mass is pulled down due to its weight. The resultant force acting on it will be W-T as W is greater. There is no component of weight acting in the direction of motion for the 8 kg mass, therefore it will move when the other mass moves down.
T-F=ma (for the 8 kg mass)
F is the frictional force. T is cancelled out. Solve for a which is 1.36. Closest answer is A.

THNX
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_11.pdf q22

 

[Thought blocker]

THNX
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf q22

Here you go bro!
use the formula for E
E = F L / Ax
rearrange to get the ratio x / L on one side (change in length / original length)
you'll get is x / L = F / E A : Where - (A = pi r ^2 )
= 20 / 2 * 10^(11) * pi x (2.5 * 10^(-4))^2
= 5.1 * 10^(-4)
multiply this by a 100 to get the percentage
5.1 x 10^-4 x 100 = 5.1 x 10^-2 %
So answer is B.

 

[ZaqZainab]

Take centre of moment 2 b on pivot n resolve
Clockwise moments = Anticlockwise moments
(10 × 100) + (50 × D) = (20 × 60)
1000 + 50D = 1200
50D = 1200 - 1000
D = 200/50
[ D = 4 cm ]

So distance from pivot = 4 cm
Distance from 0 mark = 40 + 4 = 44 cm

where did you get 10 from?
the 10*100??

 

[Thought blocker]

where did you get 10 from?
the 10*100??

LOL! he posted the answer again, I think he might not have got my alert.

 

[Hadi Murtaza]

where did you get 10 from?
the 10*100??

It is da distance of da pivot from da centre of mass

 

[MYLORD]

Here you go bro!
use the formula for E
E = F L / Ax
rearrange to get the ratio x / L on one side (change in length / original length)
you'll get is x / L = F / E A : Where - (A = pi r ^2 )
= 20 / 2 * 10^(11) * pi x (2.5 * 10^(-4))^2
= 5.1 * 10^(-4)
multiply this by a 100 to get the percentage
5.1 x 10^-4 x 100 = 5.1 x 10^-2 %
So answer is B.

one more !! Man i am sooo busted !!
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_11.pdf q 17

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_11.pdf

Q16,21,23,24 help!!!