Physics: Post your doubts here!

[hela]

how to solve http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_41.pdf

question 4 b ii

 

[mshabir007]

i would like to get some help from my MCQ Paper which i did this year may june 9702/mj/13/q11
question paper 11
please help me to solve this questions
here are the questions 6,8,17,18,19,21,24,27,32,35 thank you

 

[mshabir007]

i would like to get some help from my MCQ Paper which i did this year may june 9702/mj/13/q11
question paper 11
please help me to solve this questions
here are the questions 6,8,17,18,19,21,24,27,32,35 thank you

 

[white ash]

any tips on acing your first physics AS EXAM at school?IM JUST TOO NERVOUS..

 

[hela]

HOW TO SOLVE http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_41.pdf
WINTER 2010 Q 2b iii

 

[Starlight97]

Could any of you plz provide me the name of phys book for As and A

 

[Suchal Riaz]

Could any of you plz provide me the name of phys book for As and A

Cambridge Internation A/As level Physics Coursebook by Cambridge University press

 

[Suchal Riaz]

Could any of you plz provide me the name of phys book for As and A

Cambridge Internation A/As level Physics Coursebook by Cambridge University press

 

[arsalanlaiqkhan]

i need compiled past papers of paper 5??

 

[Rrrrrr]

hello, is there any website like www.chemguideforcie.co.uk for physics ?

 

[jhon134]

How to solve this please help

 

[IGCSE O/L student]

How to solve this please help

Since the forces are all in equilibrium, the upward force must be equal to the downward force. So the vertical component of T must be equal to 5*9.81
So Tsin(60)=5*9.81
Find T by making it the subject.
Since the two forces acting horizontally are F and the horizontal component, and because they are both in different directions, they should be equal in magnitude. So Tcos(60)=F

For the next question, 2Tcos(30)=10*9.81

Find T

 

[richard_dell32]

Hi, can anyone please help me take a look at this Physics question from the picture attached:​

​

It is regarding AC transmission involving voltage supplied and power used.​

​

Any explanation is very appreciated, Thank you very much!​

 

[sj0007]

Can anybody plz help me wid QUESTION 12 of this paper the answer is A......and also wid QUeSTION 15 whose answer is D....

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_11.pdf

 

[sj0007]

Also question number 13 and 14 of http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf
whose answers are A and B respectively....

 

[IGCSE O/L student]

Can anybody plz help me wid QUESTION 12 of this paper the answer is A......and also wid QUeSTION 15 whose answer is D....

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf

Q12: we'll take this problem in terms of the forces acting on the barrel and the man
for the barrel: the force due to the barrel - the tension in the rope(T) = ma (because the barrel move's down, the force due to the barrel is greater than the tension)
(120*10) - T = 120a (we dont know the acceleration of the man and the barrel)
for the man: the tension (T) - the force due to the man = ma
T - (80*a) = 80a
find the value of a using simultaneous equations, which is 2 ms^-2
using v^2=u^2 + 2as
find the final velocity of the man at 9 m, taking u as 0 ms^-1

Q15 in order to keep the column from falling, the reaction force must move to the right.

 

[IGCSE O/L student]

Also question number 13 and 14 of http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf
whose answers are A and B respectively....

Q13: you use the same concept here too, just like the previous example
for the 2 kg mass: F - T = ma
2*9.81 - T = 2a
for the box: T - friction(Fr) = ma
T - 6 = 8a
solve the equations simultaneously, and get the value of a which is 1.362 ms^-2 = 1.4 ms^-2

Q14: the ladder is not stable, so the force P will not be exactly horizontal, but the other two forces will remain vertical

 

[sj0007]

Q12: we'll take this problem in terms of the forces acting on the barrel and the man
for the barrel: the force due to the barrel - the tension in the rope(T) = ma (because the barrel move's down, the force due to the barrel is greater than the tension)
(120*10) - T = 120a (we dont know the acceleration of the man and the barrel)
for the man: the tension (T) - the force due to the man = ma
T - (80*a) = 80a
find the value of a using simultaneous equations, which is 2 ms^-2
using v^2=u^2 + 2as
find the final velocity of the man at 9 m, taking u as 0 ms^-1

Q15 in order to keep the column from falling, the reaction force must move to the right.

Q13: you use the same concept here too, just like the previous example
for the 2 kg mass: F - T = ma
2*9.81 - T = 2a
for the box: T - friction(Fr) = ma
T - 6 = 8a
solve the equations simultaneously, and get the value of a which is 1.362 ms^-2 = 1.4 ms^-2

Q14: the ladder is not stable, so the force P will not be exactly horizontal, but the other two forces will remain vertical

OMG thanx alott........i don't know why i can't manange p1 of physics......just pray for me i have the school exam tommorrow!!

THANX AGAIN!!!!

 

[IGCSE O/L student]

OMG thanx alott........i don't know why i can't manange p1 of physics......just pray for me i have the school exam tommorrow!!

THANX AGAIN!!!!

it's a pleasure
All the best

 

[rockmoinul]

HELLO EVERYONE. HAPPY VICTORY DAY TO ALL.... ON THIS DAY IN 1971 PAKISTAN SURRENDERED THEMSELVES TO US N WE GOT OUR VICTORY AND OUR COUNTRY BANGLADESH....

 

[snowbrood]

HELLO EVERYONE. HAPPY VICTORY DAY TO ALL.... ON THIS DAY IN 1971 PAKISTAN SURRENDERED THEMSELVES TO US N WE GOT OUR VICTORY AND OUR COUNTRY BANGLADESH....

does this have to do something with physics i am afraid not please dont spam