Physics: Post your doubts here!

[TaffsAsLevel]

can some one explain october november 2009 questions 15,13,6
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf

For 15 ElasticPE is converted to KE and it moves both sides so more stretched it adds up. So add both KE that is the EPE stored then

For 13, T=F=3/0.05=60N so torque at P is 60x0.075 Nm I took the radius because wheel Q is driven from the centre so from that point to the top of wheel Q are the distances between the forces


for 6, If you see the pattern of the graph, it is very similar to the trajectory or path of the ball so y is displacement

 

[borabora]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_1.pdf
Please help with Q18!

 

[Sk. Shahriar Hossain]

View attachment 28780
HELP PEOPLE

Plz include the year of the question.....anyway i think the answer should be "C"...let me know if its correct then I'll post the working...

 

[aleezay]

Which is more plastic? A metal or its alloy? eg iron or steel? I say steel.. :/

 

[aleezay]

Plz include the year of the question.....anyway i think the answer should be "C"...let me know if its correct then I'll post the working...

Its D

 

[Z.S.M]

Which is more plastic? A metal or its alloy? eg iron or steel? I say steel.. :/

They alloy is more plastic.
You can bend iron easily. But not steel.

 

[Sara syed]

For 15 ElasticPE is converted to KE and it moves both sides so more stretched it adds up. So add both KE that is the EPE stored then

For 13, T=F=3/0.05=60N so torque at P is 60x0.075 Nm I took the radius because wheel Q is driven from the centre so from that point to the top of wheel Q are the distances between the forces


for 6, If you see the pattern of the graph, it is very similar to the trajectory or path of the ball so y is displacement

thx
for 15. when i add (0.5x2x2^2) and 0.5x1x2^2 i get 6 J but the answer is 12J

 

[Sara syed]

can some one explain winter 1o paper13 question2,31,17, aqnd why 13 is A not D

 

[TaffsAsLevel]

thx
for 15. when i add (0.5x2x2^2) and 0.5x1x2^2 i get 6 J but the answer is 12J

Speed is 2 ms^-1 is only for the 2 kg trolley

See, this is a momentum problem and we know the initial momentum is zero so the right equals the left (this is similar to explosion momentum) therefore m1v1=m2v2 if you see that m and v are inversely proportional so if the m2/m1=2 (you see its 2kg and 1kg) so v2/v1=1/2 (because the momentum is equal we have to make it equal get it?) and we got the other speed for the 2 kg so the other 1 kg will have double speed of the greater mass hence we can assume it is 4 ms^-1. Now get the KE for the individual and add up as I have said you will end up 8+4=12J so D

 

[zackle09]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_1.pdf
how do you do ques 3?
and for question 30..why do we have to find the average of the current...why not subtract the two values?

 

[daredevil]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_12.pdf

Q28 plzz

 

[taimoor.08]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_12.pdf

Q28 plzz

A that is

 

[omarjaved619]

w10_qp12 Q25. Help? :S

 

[daredevil]

A that is

yeah but how??

 

[daredevil]

w10_qp12 Q25. Help? :S

link?

 

[daredevil]

w10_qp12 Q25. Help? :S

never mind i got it
ok it is D bcz c in a closed pipe the smalest frequency possible is f. note that on an open end there is always an antinode and on a cosed one there is always a node.
so if there is an antinode on the open end of X and a subsequent node on the closed end the frequency is f
In Y however there should be an antinode on both the ends and for the smallest frequency only one node in between. so the pipe will have an antinode, a node and another antinode which will make the distance 2f. (antinode to node is f as in X)
i hope u got it

 

[TaffsAsLevel]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s03_qp_1.pdf
how do you do ques 3?
and for question 30..why do we have to find the average of the current...why not subtract the two values?

For q3. I'll tell you a method if they ask for resultant, choose an axis, let's say y vertical and x is horizontal. I hope till that part you know how to resolve the forces. Then if you know, get the resultant of the forces that is parallel to the y axis and then the same for the x axis. Now use pythagoras, so that Fresultant=Square root of [(resultant force parallel to y)^2 + (resultant force parallel to x)^2]

For q30.If you are asked to AVERAGE a value what you do? sum of all the values/number of values right? So stick to it.

 

[gary221]

this ques. is nt clear to me...the formula for lambda@ is n@=dsintita....but here u r dividing sin35/(1*10^6)...how??

Did u get how we calculated n?
Formula → nλ = dsinΘ
(1*10^6) * λ = 1 * sin 35
[d = 1, as it is a 1st order diffraction]
So, rearranging the eqn abv,
λ = (1*sin35)/(1*10^6)
λ = sin 35/(1*10^6)

Hope u get it!

 

[A star]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_11.pdf Q15 any one

 

[A star]

never mind i got it
ok it is D bcz c in a closed pipe the smalest frequency possible is f. note that on an open end there is always an antinode and on a cosed one there is always a node.
so if there is an antinode on the open end of X and a subsequent node on the closed end the frequency is f
In Y however there should be an antinode on both the ends and for the smallest frequency only one node in between. so the pipe will have an antinode, a node and another antinode which will make the distance 2f. (antinode to node is f as in X)
i hope u got it

good job