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link?w10_qp12 Q25. Help? :S
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link?w10_qp12 Q25. Help? :S
never mind i got itw10_qp12 Q25. Help? :S
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?
this ques. is nt clear to me...the formula for lambda@ is n@=dsintita....but here u r dividing sin35/(1*10^6)...how??
good jobnever 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
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
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.
ohhhhh.. i did what you said..i got 11 and answer is 10..are you supposed to get an exact answer or is this fine
and for question 30..they didnt tell us to find the avg though :/
It is 10 I just did that few days ago, ok I will give you my working and you try see your problem. square root of[(10cos30)^2 + (10-10sin30)^2)] check you get 10 N, if you don't understand my working then you don't know resolving forces or maybe you didn't recognize the angle just because 120 degrees is given. Sorry because it is hard to type out compared to writing down on paper
It is 10 I just did that few days ago, ok I will give you my working and you try see your problem. square root of[(10cos30)^2 + (10-10sin30)^2)] check you get 10 N, if you don't understand my working then you don't know resolving forces or maybe you didn't recognize the angle just because 120 degrees is given. Sorry because it is hard to type out compared to writing down on paper
ans D :/In this ques, we consider the moment of the forces.
http://www.s-cool.co.uk/gcse/physics/forces-moments-and-pressure/revise-it/moments
The moment of L is in the clockwise direction, whereas the moment of W = anticlockwise.
They balance each other → system is in equilibrium.
When, the dist of L from the pivot increases → the moment of L now increases.
[Since moment = force * dist from pivot]
To balance this, we need an increase in moment in the ANTICLOCKWISE direction.
Note → the force needs to be perpendicular to the pivot.
When R is shifted to the left, it produces a moment in the clockwise direction.
and towards the left, it produces a moment anticlockwise.
So, ans = B
Hope u get it
d is 1.0*10^-6 den y r u taking 1Did 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!
ans D :/
>>>>???
d is 1.0*10^-6 den y r u taking 1
and n is the no.of orders which is 2 here
can u help
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