IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: sweetie on April 25, 2010, 07:33:38 pm
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Q10, 11 and 32
thanx 4 any help!!!!
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I'm on it :)
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Q10, 11 and 32
thanx 4 any help!!!!
PS; I haven't got the answers..soc heck'em pls :) :
Q.10: v1-v2/u2+u1=1
therefore u1+u2=v2-v1------> D
Q.11:
You must write down the equations for both objects:
1) box-----> T-6= 2a----> T=28+6
2) 2 KG---> 2g-T=2a----> T=2g-2a
now we got 2 Ts equations:
8a+6=2g-2a
10a= 2g-6
a= 1.4---->A
Hope you got that :D
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thanks + rep ;D
it goes up by 2 :D
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Thank you Nid ;D :-*
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PS; I haven't got the answers..soc heck'em pls :) :
Q.10: v1-v2/u2+u1=1
therefore u1+u2=v2-v1------> D
Q.11:
You must write down the equations for both objects:
1) box-----> T-6= 2a----> T=28+6
2) 2 KG---> 2g-T=2a----> T=2g-2a
now we got 2 Ts equations:
8a+6=2g-2a
10a= 2g-6
a= 1.4---->A
Hope you got that :D
for Q10 , arent v1 and v2 moving in same direction , so their signs should be the same , nd the signs of u1 and u2 should be diff.????
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lol no..
there's that thing called coefficient of restitution, e,=
e= v1+- v2/u1+-u2
sign is positive when directions are similar..e=1 when collision is PERFECTLY elastic :)
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For restitution
If moving in same direction closing speed or parting speed = bigger speed - smaller spedd
If movin towards each other clsong spped =sum of speeds
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And if they are moving apart separation speed = sum of speeds
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ok ppl,
cn sum1 xplain Q32 as well nd Q11 and 10 too ( am not gettin them :-[)
Thank You
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year? I suppose p1
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its nov08 ( i've attached the file) :)
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10 approach speed=u1+u2
separation speed=v2-v1
they are equal for elastic collisions D
11.res for both particles using F=ma
for 2g particle 2g-T=2a (1)
for 6g particle T-6=6a (2)
(1) +(2) 2g-6=8a so a/(2*9.8-6)/8=1.7 B
32)1/R=1/100+1/10+1/10+1/10+1/10+1/10+1/10=61/100
R=100/61 B
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hi would please help me in answering this question and plesa while answering explain how u progressed to this result
the paper is may/june CIE physics 2009
question 5 part b) :)
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hi would please help me in answering this question and plesa while answering explain how u progressed to this result
the paper is may/june CIE physics 2009
question 5 part b) :)
here u go
there is a formula u need to know
phase difference=( 2pi/ lamda) X path difference
phase difference = pi [ for destructive interference]
lambda(wavelength) = 2pi/ pi X path difference.
path difference is S2M - S1M
find S2M using Pythagoras = root of 802 + 1002
=128
path difference= 128-100=28
wavelength= 2pi/pi X 28
=56cm
At f= 1KHz wavelength= 0.33m/ 33cm
At f=4kHz wavelength= 0..0825/8.25cm
wavelength changes from 33cm to 8.25cm
for minima...phase difference should be either one of them pi, 3pi,5pi,7pi [sub them in the eqn]
minima is observed when lambda= 56cm, 18.7cm, 11.2cm, 8cm
which are the 2 values between 33 and 8.25?? 18.7 and 11.2...so only 2 minimas
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sorry what's the value for the pi plz
and what's pi????????????/
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pi radians = 180 degrees
one whole period corresponds to 2pi radians or 360 degrees...
For constructive interference they have to be in phase
For destructive interference they have to be out of phase...and for minima to have a value of zero...they should have a phase diff of pi or all odd multiples of pi
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10 approach speed=u1+u2
separation speed=v2-v1
they are equal for elastic collisions D
11.res for both particles using F=ma
for 2g particle 2g-T=2a (1)
for 6g particle T-6=6a (2)
(1) +(2) 2g-6=8a so a/(2*9.8-6)/8=1.7 B
32)1/R=1/100+1/10+1/10+1/10+1/10+1/10+1/10=61/100
R=100/61 B
Thank You :)
but i just wanted 2 know how do we know that the wires r connected in parallel in Q32????
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Thank You :)
but i just wanted 2 know how do we know that the wires r connected in parallel in Q32????
yes all of them parallel to each other
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hey agian am really sorry but how did u get 18.7and 11.2?????????????????? sorrrrrrrrrrrrrrrrrrrrrrrrrrrryyyyyyyyyyyyyyyyyyyyyyyy
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hey agian am really sorry but how did u get 18.7and 11.2?????????????????? sorrrrrrrrrrrrrrrrrrrrrrrrrrrryyyyyyyyyyyyyyyyyyyyyyyy
lol....it's ok
substitute for different phase difference...in the first case phase diff is pi, followed by 3pi,5pi,7pi ...like i said for destructive interference, it has to be an odd multiple of pi..
substitute these in the eqn to get wavelength.. phase diff= 2pi/ lambda X path diff
for phase diff wavelength/lambda
pi 56
3pi 18.7
5pi 11.2
7pi 8
hope ur getting it
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sorry i get the pi=56 but the 3pi is not coming out could u plz wirte it in numbers because is not coming
thx really ur so helpful :) :) :)
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sorry i get the pi=56 but the 3pi is not coming out could u plz wirte it in numbers because is not coming
thx really ur so helpful :) :) :)
no problem
for 3pi
wavelength= 2pi/3pi X 28
= 2/3 X 28
=18.7
got it?
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A-W-E-S-O-M-E Nid! You're awesome! :D
+ rep. :)
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CUD SUM1 PLZ XPLAIN THESE 2 QUE'S?????
may 06
Q.35
Q36
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A-W-E-S-O-M-E Nid! You're awesome! :D
+ rep. :)
yay!! :-* :-*
I finally got waves...thanks to you :-*
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CUD SUM1 PLZ XPLAIN THESE 2 QUE'S?????
may 06
Q.35
Q36
Q 35
one i've circled are in series...
10+10+10=30ohms
these are all parallel to the 10 ohm resistor
1/30 +1/10= 1/R
R approx 7.5 ohms
between 1 and 10 ...B
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Q36
It's the same diagram both of them..
in diagram 1
4 lamps...2 in series..parallel with 2 other is series on the same side of the source
diagram 2
4 lamps 2 in series parallel with 2 other on the other side of the source...
so they're just twisting the diagram...no difference on brightness...
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Thank You nid :)
+rep :D
cud u also xplain Q33 and 9 of the same year ??????
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9, B cos it is the biggest negative gradient
33. Across the top the voltage splits in ration 1:2 ie 4V:8V
but accross the bottom it is 8:V:4V
so Potential differenceD is 4V
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nid404 thx for the help and am really sorry for asking soo many times but the question so hard thxxxxxxxxxxxxxxxxxxxxxxxxxxx ;) ;) :) :)
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nid404 thx for the help and am really sorry for asking soo many times but the question so hard thxxxxxxxxxxxxxxxxxxxxxxxxxxx ;) ;) :) :)
no problem :)
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9, B cos it is the biggest negative gradient
33. Across the top the voltage splits in ration 1:2 ie 4V:8V
but accross the bottom it is 8:V:4V
so Potential differenceD is 4V
thanx :)
but for Q9 isnt the gradient at B zero??????
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gadient zero = max
at B graph is zero but gradient is maximum negative cos graph is going down
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i didnt get it, can u plz show me how?????
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At bottom of motion v=zero but acceleration is positive sp graident is positive. D
I had to read this question several time - I wasnt concetrating
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Thank You :D
i cudnt understand these Q's in may07 , help plz!!!!
Q40, 37,33,29,23,20 and 18
thanx in advance :-*
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Thank You :D
i cudnt understand these Q's in may07 , help plz!!!!
Q40, 37,33,29,23,20 and 18
thanx in advance :-*
I'm on it :)
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Thank You :D
i cudnt understand these Q's in may07 , help plz!!!!
Q40, 37,33,29,23,20 and 18
thanx in advance :-*
Q18: B---> the bottom curve is the one for unloading..and so area under curve equals energy recoverd..the lop's area equals to heat transformation
Q.20: B---> W acts downwards on vertical bar..and so tension at X..then just like the concrete and steel beam, when u put a load at the top, tension acts at the top, compression at the bottom
Q. 23: check this thread..5th page..my long post https://studentforums.biz/index.php/topic,6664.60.html
40: same as Q23
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hey you A.F, you rock 8) + rep
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Thank You soooo much A.F ;)
+rep
cud u also xplain the rest of the questions???????/
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my bad ;D
Q.33: u must find V1 and V2
V1: same resistence..therefore V1 across PS = 2/2= 1.0 V
V2: 3 ohms: 2 ohms: 5ohms
x voltage: : 2
do ratio..now x=3x2/5= 1.2
therefore V1-V2= -0.2 V----> answer is C
Q 37: Same resistors, same length, same material ---> C
Q29: Its exactly like a projectile..constant speed as its in vaccum..no rsistence, and acceleration upwards as its negatively charged..attracted upwaqrds towards the +ve plate
got that? ::)
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and Thanks 4 the + rep. ;D
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for Q33, why did u take the ratio's???
and Q37> in resistor Q arent the wires in parallel , so the total resistance should be x/7????
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Q33...current divides on the basis of the ratio of the resistance...
Q37 srry I can't explain :-[
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can sum1 please give me the paper, maybe i can help.
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R= pl/A
it says both wires have the exact volume..i.e the exact area and length x
These are the only factors on which the resistance depends on..
and so both have equal resistances
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here is the paper
thanx A.F
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I hate to disagree with you A.F, but even though the markscheme says C, I am not entirely convinced that that is the answer. I think we need astar to explain.
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Here's ur explanation guys:
Resistance of P= pl/A...agree?
Resistance of Q:
Area of each wire of the seveen wires= A/7...right?
now R=pl/A= pl/(A/7)= 7pl/A...so far so good..thats the are for each wire..
now total area of resistor Q:
RT = (1/R1 + 1/R2 ....)-1
=(1/(7pl/A) x7)-1 = pl/A
hope that's clear now
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Thanks A.F :) + rep
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I think I remember the question. Use 1/r=1/100+6x1/100/ on a phone now
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Thanks Nid :D
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I remember this question too, and I could answer it myself, but this is the problem, I dont think that would practically happen. Does this thing practically happen sir??
Its kinds hard for me to take in also.
I got it like this, reducing the diameter will reduce resistance, but since the wires are in paralell, that will even out. What I have a hard time swallowing is if this would really happen.
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Here's ur explanation guys:
Resistance of P= pl/A...agree?
Resistance of Q:
Area of each wire of the seveen wires= A/7...right?
now R=pl/A= pl/(A/7)= 7pl/A...so far so good..thats the are for each wire..
now total area of resistor Q:
RT = (1/R1 + 1/R2 ....)-1
=(1/(7pl/A) x7)-1 = pl/A
hope that's clear now
yup, now iam convinced :)
thanx A.F
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Q33 plz of the same year may07
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still not gotten it?
ok wait...i'll get back in a while
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in a parallel circuit voltage remains constant
In the first wire(resistors are in series), the 2 resistors have the same value, so voltage splits equally...V1= 2/2=1 V [voltage splits in series]
In the second wire, voltage splits according to ratio of resistance...V=IR (directly proportional) ...so in the 3 ohm resistor it would be 3/5 X2=1.2V (since total R=5...ratio is 3/5)
V1-V2= 1-1.2= -0.2
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Q10, 22,30 and 40
Thank You ;D
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Q 10) acceleration of the blocks is constant
the block has a Force= 4ma
F is directly proportional to m
X uses only 1/4 of the force to acc
3/4 of the force will be applies on Y...for both the blocks to move toether, hence D (acc is constant)
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22. each spring now support 1.5*2=3 times the weight
30charge=average current*time=(20mA+100mA)*8=480mC
40.2 ups and 1 down cos charge must add to1e
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okay...
if each spring now has extension 3x , so since they both r in parallel, isnt the total suppose to be 3x/2????????
and for Q30 why dont we consider the difference between the 2 values??
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The weight is doubled so the extension is doubled
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10 approach speed=u1+u2
separation speed=v2-v1
they are equal for elastic collisions D
11.res for both particles using F=ma
for 2g particle 2g-T=2a (1)
for 6g particle T-6=6a (2)
(1) +(2) 2g-6=8a so a/(2*9.8-6)/8=1.7 B
32)1/R=1/100+1/10+1/10+1/10+1/10+1/10+1/10=61/100
R=100/61 B
But , for Q11 how do we know that the acceleration of the 2 boxes r the same??????
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okay...
if each spring now has extension 3x , so since they both r in parallel, isnt the total suppose to be 3x/2????????
W=kx
k=W/x
now when one spring is removed, value of k reduced to 2/3 of it's previous...which is 2W/3x
2W=2w/3x x e
e=3x
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cos the boxes are always in contact
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W=kx
k=W/x
now when one spring is removed, value of k reduced to 2/3 of it's previous...which is 2W/3x
2W=2w/3x x e
e=3x
am sorry nid, but i didnt get how u got k as 2W/3x???????
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am sorry nid, but i didnt get how u got k as 2W/3x???????
See all the together had a k= W/x right?
when one spring is removed from the 3, 1/3rd of the k is eliminated...right? k reduce by 1/3 ...so remaining k= 2/3 X W/x
got it?
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yup..
Thank You ;)
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yup..
Thank You ;)
no problem :)
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nov07
Q35, 26, 25and 6
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26. the electric field points in the direction of the force on a positive charge. The charge on an electron is negative so the force points the other way, to the left.
25. 4th sinx=nw/d<1
If n=3 sinx=0.707 so w/d =0.707/3
nw/d<1 if n=4 but not n=5
6.A deflection +ve when current =0 so A or B
deflection increases more atr start so A
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Sir, can u xplain Q25 and 6 again plz???????
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as soon as you find w/d=0.707=0.236 use sinx =n*0.236
this has to be less than 1 to give a solution for x
If n=4 sin x =0.944 solutions exist
n=5 sin x =1.18 no solutions
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got it now,
cud u plz xplain Q35 too??????
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I couldnt find q 35 on the paper
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i've attached the que. paper...
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help plz......
may09> Q18
Nov08. Q4
May08. Q31,22,16,14,12
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When i get up
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ok....
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m08 q18 B - constant speed so kinetic energy constant
falling at constant rate so GPE decreasing at constant rate
B
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i guess thats the wrong Q. , Q18 is about the change in pressure........
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2008nov B
2+-1% is B
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Sorry not may 09 - is may 08
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May08. Q31,22,16,14,12
31 use E=QV Q=V/C=6000C D
22. B by definition
16.C by elimination
14.C use P=Fv and F=kv^2
12.ms says D but I understand by equilibrium B
My answer is
C
I
E
fullofshit
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Sir, cud u plz check the que's again , u xplained the ones that i understand ;D
*in may 08 i dont get Q14, 16 and 31
*and Q18 of may09
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When i get home