Post by: tmisterr on April 08, 2010, 08:52:11 am
Post by: nid404 on April 08, 2010, 10:32:10 am
a=v-u/t
V at M1= at1
V at M2= at2
v2=u2 +2as
v2-u2=2as
(at2)2-(at1)2=2as (s=h)
a=2h/ (t22-t12)
Post by: tmisterr on April 08, 2010, 10:40:19 am
Post by: nid404 on April 08, 2010, 10:45:35 am
ball A
mVA- mUA
ball B
mVB-(-mUB)(opp direction )
p of ball A=ball B (for elastic collision)
mVA- mUA=mVB-(-mUB)
VA-UA=VB+UB
VA-VB=UB+UA
Post by: nid404 on April 08, 2010, 10:48:12 am
the resultant of the horizontal and vertical components will be equal to the third component in the opp direction
If you notice the rest, they have no such corelation...
Post by: tmisterr on April 08, 2010, 12:45:07 pm
How bout if you use relative speed of approach is equal to relative of recoil?? this is how im doing it but getting a different answer
Post by: halosh92 on April 08, 2010, 02:37:09 pm
q2 d part ii
why dont we subtract the forces and then multiply the velocity? why do we have to add here exactly?
plz someone HURRY ??? ??? ??? ???
Post by: halosh92 on April 08, 2010, 03:48:05 pm
q5a
could someone plz sketch it :SSSS
Post by: astarmathsandphysics on April 09, 2010, 12:21:11 am
Post by: halosh92 on April 09, 2010, 07:32:51 am
paper 2 2007 may/june
Q1
Post by: halosh92 on April 09, 2010, 08:38:51 am
AS cie physics
paper 2 2007 may/june
Q1 and 5d i, ii
Q6 aii ................why isnt it 14+E...why do we subtract here?
Post by: astarmathsandphysics on April 09, 2010, 11:46:38 am
Post by: halosh92 on April 09, 2010, 02:28:11 pm
Which part? If you are asking which is better - that scale or a uniform linear scale I think it is the one cos it is easier to tell from zero so you are less likely to run out of petrolok and could you answer the rest of the questions plz..its rili rili URGENT
Post by: tmisterr on April 09, 2010, 02:36:38 pm
Post by: astarmathsandphysics on April 09, 2010, 05:21:11 pm
Post by: astarmathsandphysics on April 09, 2010, 08:32:05 pm
v_b-v_a =u+b+u_a
Post by: 3ishakay on April 10, 2010, 01:36:16 pm
in paper 2 .. may/june 09??
2nd variant ppr
Post by: halosh92 on April 10, 2010, 02:14:20 pm
cie AS physics 2006 nov
Q 3c ii
thx
Post by: astarmathsandphysics on April 10, 2010, 11:46:51 pm
Post by: astarmathsandphysics on April 11, 2010, 08:34:49 am
impulse =m(v-u)=1.2(-0.8-4))=-5.76
equal and opposite impulse is exerted on 3.6kg mass
so 5.76=m(v-u)=3.6v so v=3.76/3.6=1.6m/s
Post by: astarmathsandphysics on April 11, 2010, 08:40:58 am
cie AS physics 2006 nov
Q 3c ii
thx
conservation of momentum
before decay particle is stationary so momentum=0
hence 0=m_1 v_1 +m_2 v_2
0=4*1.8*10^7 +(208-4)*v
v=-7.2*10^7 /204 =-3.53*10^5m/s
Post by: tmisterr on April 11, 2010, 12:06:57 pm
Post by: astarmathsandphysics on April 11, 2010, 02:29:19 pm
If the are moving apart you subtract.
Iemphasise APPROACH SPEED AND SEPARATION SPEED
Post by: halosh92 on April 11, 2010, 02:38:42 pm
paper 2thankyou
cie AS physics 2006 nov
Q 3c ii
thx
conservation of momentum
before decay particle is stationary so momentum=0
hence 0=m_1 v_1 +m_2 v_2
0=4*1.8*10^7 +(208-4)*v
v=-7.2*10^7 /204 =-3.53*10^5m/s
but could u please solve the questions i posted on the first page its very important sir ;D
Post by: astarmathsandphysics on April 11, 2010, 04:18:22 pm
Post by: joel on April 11, 2010, 04:22:58 pm
ALL Collisions : Momentum of the system is conserved
Momentum = mass x Velocity
Elastic Collisions
Kinetic Energy of the system is conserved
Kinetic Energy = 1/2 x m x v^2
Post by: halosh92 on April 11, 2010, 05:03:40 pm
Q6 ...i dont get the whole thing
Post by: vanibharutham on April 11, 2010, 06:24:04 pm
Post by: astarmathsandphysics on April 11, 2010, 06:31:12 pm
Post by: astarmathsandphysics on April 11, 2010, 06:38:02 pm
q5a
could someone plz sketch it :SSSS
1/2 the intensity is sqrt(1/2)=0.707 times the amplitude
The sum of the two smaller waves is the bigger one
Post by: halosh92 on April 11, 2010, 07:02:20 pm
Paper 2?
oops i forgot to add that...yes paper 2
Post by: halosh92 on April 11, 2010, 07:13:36 pm
2007 oct/nov
q5a
could someone plz sketch it :SSSS
1/2 the intensity is sqrt(1/2)=0.707 times the amplitude
The sum of the two smaller waves is the bigger one
thankyou but when they say the wave has a phase difference of 60 degrees..wat do they mean exactly????
Post by: astarmathsandphysics on April 11, 2010, 08:31:34 pm
Post by: tmisterr on April 12, 2010, 10:52:32 am
Post by: astarmathsandphysics on April 12, 2010, 11:03:50 am
Post by: astarmathsandphysics on April 12, 2010, 11:13:27 pm
Post by: halosh92 on April 13, 2010, 07:24:12 am
C - for destructive interference the path difference must be a whole number of wavelengths plus an extra half wavelengthcould you explain this furthur please?
Post by: Amr Fouad on April 13, 2010, 07:44:36 am
could you explain this furthur please?
constructive interference- happens when waves are in phase...and so the path difference is a multiple of ?...
equation for path diff of constructive: n?, where n=1, 2,3....
destructive interference: happens when waves are out of phase, and so the path difference is a multiple of half ?
equation for path diff of destructive= (0.5+n)?
therefore the only possible answer is C...
got it?
Post by: sweetie on April 14, 2010, 05:51:18 pm
when a graph line is sloping downwards, it has a neg. gradient so if we had to cal. the value of the gradient will it be written as " -x" or " x"???????
Post by: halosh92 on April 14, 2010, 06:17:42 pm
constructive interference- happens when waves are in phase...and so the path difference is a multiple of ?...
equation for path diff of constructive: n?, where n=1, 2,3....
destructive interference: happens when waves are out of phase, and so the path difference is a multiple of half ?
equation for path diff of destructive= (0.5+n)?
therefore the only possible answer is C...
got it?
thx alot...but wats the "?" in ure explanation?
Post by: Amr Fouad on April 14, 2010, 06:52:13 pm
thx alot...but wats the "?" in ure explanation?
oh lol sorry SF cant comprehend that lol
i meant Londa..the wavelength symbol..substitute ? with Londa
Post by: *Hope* on April 18, 2010, 06:33:18 am
11) sum of momentum before collision=sum of momentum after collisionwhy did u take the momentum of A alone and B alone, shouldn't we take A and B before collision and A and B after considering the signs so like mUa-mUb=mVa+mVb??
ball A
mVA- mUA
ball B
mVB-(-mUB)(opp direction )
p of ball A=ball B (for elastic collision)
mVA- mUA=mVB-(-mUB)
VA-UA=VB+UB
VA-VB=UB+UA
plz explain nid
Post by: Saladin on April 18, 2010, 07:44:33 am
hey ppl,
when a graph line is sloping downwards, it has a neg. gradient so if we had to cal. the value of the gradient will it be written as " -x" or " x"???????
Post by: cashem'up on April 18, 2010, 11:41:01 am
Post by: astarmathsandphysics on April 18, 2010, 12:01:04 pm
Post by: *Hope* on April 18, 2010, 12:04:04 pm
hey i have doubt in this question its wo4 20 have uploadedin order for equillibrium to be reached, the pressure at both sides must be equal
so pressure at P= pressure atQ
since pressure= density*gravity*height
and density=mass/volume
pressure P= density*gravity*2h
pressure Q= density*gravity*h
so density of p must be half to get same pressure of Q
Post by: cashem'up on April 18, 2010, 12:11:45 pm
Post by: *Hope* on April 18, 2010, 02:19:28 pm
hey m still confused...................Ok no prob we will take it step by step:
1) what does equillibrium mean?
it means that they are not moving right?
This means there is equal forces on each side acting downwards
2)since pressure= force/area so equal pressure because equal forces acting on equal areas (vessel) and pressure is equal to density*gravity*height
3) however height of P is doubled than of Q so new pressure will be doubled of Q but
we must get EQUAL pressures and also gravity is not changed so density must be halved at P to get same pressure
so now we have half density of P since double height
Note:try with values if u still don't get it and see what happens to density
Post by: cashem'up on April 19, 2010, 02:23:43 pm
Post by: halosh92 on April 22, 2010, 06:39:56 pm
Q 6cii
Post by: astarmathsandphysics on April 22, 2010, 06:49:45 pm
Post by: Amr Fouad on April 22, 2010, 07:44:49 pm
paper 2 june 2004 cie physics AS
Q 6cii
Use the equation: londa = xa/D
therefore, :
(i) no change, as a, londa, and D are constant..also because x is proportional to a
(ii) Brighter, as now the gap is wider
(iii) again no change, as dark fringes remain dark!
got that? :)
Post by: halosh92 on April 22, 2010, 08:21:47 pm
Use the equation: londa = xa/Dyep thx alot! ;D
therefore, :
(i) no change, as a, londa, and D are constant..also because x is proportional to a
(ii) Brighter, as now the gap is wider
(iii) again no change, as dark fringes remain dark!
got that? :)
Post by: ~ A.F ~ on April 22, 2010, 08:27:55 pm
yep thx alot! ;D
haha any time ;)
Post by: halosh92 on April 22, 2010, 09:02:35 pm
Q6
Q12
Post by: ~ A.F ~ on April 22, 2010, 09:04:44 pm
2003 nov paper 1
Q6
Q11
in a couple of minutes
Post by: ~ A.F ~ on April 22, 2010, 09:11:17 pm
initial speed= u
final speed= -u
change in momentum= m(v-u)
m(-u-(u)_
=-2mu...got that?
Post by: halosh92 on April 22, 2010, 10:15:20 pm
Q. 11:could u do 12 also plz?
initial speed= u
final speed= -u
change in momentum= m(v-u)
m(-u-(u)_
=-2mu...got that?
Post by: ~ A.F ~ on April 22, 2010, 10:19:10 pm
could u do 12 also plz?
Alright..2 mins
Post by: astarmathsandphysics on April 22, 2010, 11:58:12 pm
separation speed =vy+vx
Since collsion is elastic these are equal A
answer =A
Post by: nid404 on April 23, 2010, 08:22:17 am
Paper 1
Q 20...I really don't understand :-X
Q23 how?
Q27 Explain again
Q40 Lowest speed would be one with greatest mass right...then how's it C...or is it cuz of the charge??
Thanks
Post by: halosh92 on April 23, 2010, 09:10:20 am
M/J 07Q27- A
Paper 1
Q 20...I really don't understand :-X
Q23 how?
Q27 Explain again
Q40 Lowest speed would be one with greatest mass right...then how's it C...or is it cuz of the charge??
Thanks
as u can c the arrows is first large..that means the amplitude is large thats the max point and then
its decreasing thus antinode then its decreasing a small amplitude there fore comes to minimum node similarly for the next 2..
continue the pattern ull find 2 antinodes and one antinode.
Post by: nid404 on April 23, 2010, 09:24:51 am
Q27- A
as u can c the arrows is first large..that means the amplitude is large thats the max point and then
its decreasing thus antinode then its decreasing a small amplitude there fore comes to minimum node similarly for the next 2..
continue the pattern ull find 2 antinodes and one antinode.
ok yep thanks a lot :)
Post by: halosh92 on April 23, 2010, 10:23:24 am
2003 nov
q2 b ,c
Post by: nid404 on April 23, 2010, 10:24:14 am
M/J 07
Paper 1
Q 20...I really don't understand :-X
Q23 how?
Q27 Explain again
Q40 Lowest speed would be one with greatest mass right...then how's it C...or is it cuz of the charge??
Thanks
A.F & astar, help? :-[
Post by: ~ A.F ~ on April 23, 2010, 11:34:41 am
A.F & astar, help? :-[
lol hey nid! :D
give me 5 mins :D
Post by: nid404 on April 23, 2010, 11:44:05 am
lol hey nid! :D
give me 5 mins :D
sure :)
Post by: ~ A.F ~ on April 23, 2010, 11:58:14 am
M/J 07
Paper 1
Q 20...I really don't understand :-X
Q23 how?
Q27 Explain again
Q40 Lowest speed would be one with greatest mass right...then how's it C...or is it cuz of the charge??
Thanks
Q20: W acts downward on X..therefore its tension..now its A or B
Also, any particle at point Y is pulled downwards, and therefore its under tension(this situation is exactly like the one where steel is beneath the concrete and weight is added..)
Q23: Actually this is A2 physics..lol but nvm :P..
You see..there are rules to memorize (and understand in A2):
* when there is maximum displacement, there is maximum acceleration and zero velocity..I magine a pendul swinging, at the far end there's maximum displacement, right? and also it momentarily stops, as it will move in the opposite direc. next---->therefore, max disp, max acc, zer velocity..
Also a pendulum swing exactly in the midpoint, will have its maximum velocity, where it has zero displacement (at the centre poiny of the system)
Therefore, in this case, point S has max displacement (on crest-amplitude) and so max acceleration..
Q27: You must exclude B and C at first..All arrows in each case point in the same direction, right? whislt a longitudinal wave is one in which compressions and rarefactions occur along the pathway of the wave...
So now its A and D...Again u must exclude D as arrows are pointing in the opposite direct, but this time AWAY from each other..There wouldnt be a wave then, as there wouldnt be a compression(no "squeazing" lol) ..and in D the particles would move AWAY from each other..
So the answer is A....
Q40:
You cudnt just depend on the mass alone..Its totally wrong..U must use the following equations:
E=qV...where q=charge, V= voltage
at the same time, E= 0.5 x m x v^2...right? Kinetic Energy..
therefore:
qV=0.5 m v^2
v^2= 2qV/m
however, Voltage V is constant..therefore:
v^2 is proportional 2 2q/m
Now u HAVE to substitute values in A, B, C, and D:
A---> m=1. q-1, therefore v= 2q/m
B---> m=4, q=2, therefore v= 2 x 2q/4m= q/m
C---> m=7, q=3, therefore v= 2qx3/7m= 6q/7m
D---> m=9, q=4, therefore v=2x4q/9= 8q/9m
C has the smallest ratio, and so C is the answer..
Tala! lol :P
Was I clear in everything? ::)
Post by: nid404 on April 23, 2010, 12:04:44 pm
Thank you so much :-*
Post by: ~ A.F ~ on April 23, 2010, 12:06:00 pm
ohhhhhh!
Thank you so much :-*
lol anytime dear :)
Post by: halosh92 on April 23, 2010, 03:30:23 pm
M1 edexcel mathscould someone answer this please????? ??? ??? ??? ???
2003 nov
q2 b ,c
Post by: astarmathsandphysics on April 23, 2010, 03:53:24 pm
Post by: astarmathsandphysics on April 23, 2010, 08:06:17 pm
v-12 =-28800/24000=-1.2 so v=10.8
c)impulse on T is 28800 from Newton's 3rd law
28800=m(v-u)=m(3.6--6)=9.6m so m=28800/9.6=3000Kg
Post by: halosh92 on April 23, 2010, 08:21:47 pm
b) -28800=m(v-u)=2000(v-12)
v-12 =-28800/24000=-1.2 so v=10.8
c)impulse on T is 28800 from Newton's 3rd law
28800=m(v-u)=m(3.6--6)=9.6m so m=28800/9.6=3000Kg
for c they say it doesnt change direction..so shouldnt 6 be positive?
Post by: astarmathsandphysics on April 23, 2010, 08:59:22 pm
Post by: halosh92 on April 23, 2010, 10:11:55 pm
from rest from a height of 1.6 m above the top of the post.
(a) Show that the speed of the pile-driver just before it hits the post is 5.6 m s–1. (2 marks)
The post has mass 6 kg and the pile-driver has mass 78 kg. When the pile-driver hits the top
of the post, it is assumed that the there is no rebound and that both then move together with
the same speed.
(b) Find the speed of the pile-driver and the post immediately after the pile-driver has hit the
post. (3 marks)
The post is brought to rest by the action of a resistive force from the ground acting for 0.06 s. By modelling this force as constant throughout this time,
(c) find the magnitude of the resistive force, (4 marks)
(d) find, to 2 significant figures, the distance travelled by the post and the pile-driver before
they come to rest. (4 marks)
how to do part c and d?
Post by: astarmathsandphysics on April 23, 2010, 10:57:02 pm
You need to find a, the deceleration
When the driver hits the post momentum is conserved
v(6+78)=6*5.6 so v=33.6/84=0.4
post comes to rest in 0.06s
so a=(v-u)/t=0.4/0.06=6.66m/s^2
F=ma=84*6.66667=560N
d)use v^2=u^2+2as
0=0.4^2+2*-6.66667*s
s=0.16/13.33333=0.012m
Post by: *Hope* on April 24, 2010, 04:25:38 pm
Q20: W acts downward on X..therefore its tension..now its A or Bfrom where did u get "E=qV...where q=charge, V= voltage"?
Also, any particle at point Y is pulled downwards, and therefore its under tension(this situation is exactly like the one where steel is beneath the concrete and weight is added..)
Q23: Actually this is A2 physics..lol but nvm :P..
You see..there are rules to memorize (and understand in A2):
* when there is maximum displacement, there is maximum acceleration and zero velocity..I magine a pendul swinging, at the far end there's maximum displacement, right? and also it momentarily stops, as it will move in the opposite direc. next---->therefore, max disp, max acc, zer velocity..
Also a pendulum swing exactly in the midpoint, will have its maximum velocity, where it has zero displacement (at the centre poiny of the system)
Therefore, in this case, point S has max displacement (on crest-amplitude) and so max acceleration..
Q27: You must exclude B and C at first..All arrows in each case point in the same direction, right? whislt a longitudinal wave is one in which compressions and rarefactions occur along the pathway of the wave...
So now its A and D...Again u must exclude D as arrows are pointing in the opposite direct, but this time AWAY from each other..There wouldnt be a wave then, as there wouldnt be a compression(no "squeazing" lol) ..and in D the particles would move AWAY from each other..
So the answer is A....
Q40:
You cudnt just depend on the mass alone..Its totally wrong..U must use the following equations:
E=qV...where q=charge, V= voltage
at the same time, E= 0.5 x m x v^2...right? Kinetic Energy..
therefore:
qV=0.5 m v^2
v^2= 2qV/m
however, Voltage V is constant..therefore:
v^2 is proportional 2 2q/m
Now u HAVE to substitute values in A, B, C, and D:
A---> m=1. q-1, therefore v= 2q/m
B---> m=4, q=2, therefore v= 2 x 2q/4m= q/m
C---> m=7, q=3, therefore v= 2qx3/7m= 6q/7m
D---> m=9, q=4, therefore v=2x4q/9= 8q/9m
C has the smallest ratio, and so C is the answer..
Tala! lol :P
Was I clear in everything? ::)
Post by: ~ A.F ~ on April 24, 2010, 04:27:48 pm
from where did u get "E=qV...where q=charge, V= voltage"?
from my physics book :P lol
its a general well-known rule
Post by: *Hope* on April 24, 2010, 04:35:04 pm
but is E energy?
and also for q27, isn't it A and not B cuz the waves interfere in opp direction?
Post by: nid404 on April 24, 2010, 05:31:07 pm
energy= force X displacement right?
E=Voltage / distance & also E= Force/charge (E=Electric field strength) [i hope ur aware of this one]
so Voltage / distance= Force/charge
Force= Voltage X Charge / distance
Energy= Force X displacement
= Vq/ d X d
= Vq
Got it?
Post by: tmisterr on April 24, 2010, 06:33:05 pm
Post by: student on April 24, 2010, 06:41:31 pm
T=W/P
W=F*d
T=F*d/P
T=180*4000/3000=240s
Post by: nid404 on April 24, 2010, 06:55:05 pm
Post by: tmisterr on April 24, 2010, 07:46:24 pm
Post by: *Hope* on April 25, 2010, 05:02:51 am
ohk...E=qv I'll tell u the derivationThanks but is electric field strength the same as energy??
energy= force X displacement right?
E=Voltage / distance & also E= Force/charge (E=Electric field strength) [i hope ur aware of this one]
so Voltage / distance= Force/charge
Force= Voltage X Charge / distance
Energy= Force X displacement
= Vq/ d X d
= Vq
Got it?
Post by: nid404 on April 25, 2010, 05:50:33 am
Thanks but is electric field strength the same as energy??
noooooo!
Post by: T.Q on April 25, 2010, 08:34:51 am
Thanks but is electric field strength the same as energy??
electric field: is a property that describes the space that surrounds electrically charged particles or that which is in the presence of a time-varying magnetic field. This electric field exerts a force on other electrically charged objects.
Post by: astarmathsandphysics on April 25, 2010, 09:50:57 am
but also W=Fd=Force *distance=Eqd
Post by: *Hope* on April 25, 2010, 02:26:03 pm
No electric field is more closely related to force via the equation F=Eq=Electric field*chargeok i get this one but for energy=VQ i am still confused since Energy is not = to electric field strength so we can not substitute there right?? ??? :-\
but also W=Fd=Force *distance=Eqd
Post by: astarmathsandphysics on April 25, 2010, 02:31:49 pm
Post by: sweetie on April 28, 2010, 10:08:33 pm
cud u help me with Q14, 19, 35 and 36 of nov02 ( AS cie)
Post by: astarmathsandphysics on April 28, 2010, 10:35:31 pm
v late here
Post by: astarmathsandphysics on April 29, 2010, 08:05:29 am
1000-1200+50x=0 so x=4
4cm from 40 point going clockwise ie 44cm mark
19.PE=mgh=m*9.8*200=1960m
0.6*1960m=1176m
1/2mv^2=1176m
c^2=2376
v=sqrt(2352)=48.5 C
35.D decreas since more of the voltage is across the variable resistor and nearer to Y cos resistances must remain in proportion
36.2/3V voltage at x is 2/3 and voltage at y is 4/3
Post by: sweetie on April 30, 2010, 12:44:48 am
14 moments about 40 mark 10*100-20*60+50*x=0
1000-1200+50x=0 so x=4
4cm from 40 point going clockwise ie 44cm mark
19.PE=mgh=m*9.8*200=1960m
0.6*1960m=1176m
1/2mv^2=1176m
c^2=2376
v=sqrt(2352)=48.5 C
35.D decreas since more of the voltage is across the variable resistor and nearer to Y cos resistances must remain in proportion
36.2/3V voltage at x is 2/3 and voltage at y is 4/3
thanx...
but cud u xplain 14 in more detail??
Post by: astarmathsandphysics on April 30, 2010, 06:26:56 am
Post by: sweetie on April 30, 2010, 10:11:06 am
Post by: astarmathsandphysics on April 30, 2010, 10:19:14 am
Post by: sweetie on April 30, 2010, 10:36:19 am
cud u help me in Q. 37,32,21,18,12 and 3 of may02..........
Post by: astarmathsandphysics on April 30, 2010, 11:20:30 am
37. A - V is constant E=V=d
32.P=V^2/R so R=V^2/P=240^2/100=576
originarl R=576/16=36 A
21.total mass is now 3pV in same volume so 3 times the mass occupies the same volume anddensity is 3 times bigger D
18. GPE+KE=work done against friction +KE
5+50=10+KE
KE=45 B
12.weight=upthrust+drag>upthrust D
3.only A m/s=sqrt(m/s^2 *m)=m/s
Post by: *Hope* on April 30, 2010, 03:25:31 pm
Post by: astarmathsandphysics on April 30, 2010, 10:13:20 pm
Post by: sweetie on May 01, 2010, 12:24:58 am
and can u xplain Q18 in more detail plz.......
Post by: nid404 on May 01, 2010, 08:10:29 am
k.e+p.e at Q = k.e + work done against friction
55=k.e+10
k.e=45kJ B
Post by: tmisterr on May 01, 2010, 05:23:05 pm
She estimates that her result is accurate to ±3 %.
Which of the following gives her result expressed to the appropriate number of significant figures?
A 327.7ms–1 B 328ms–1 C 330ms–1 D 300ms–1
Post by: Saladin on May 01, 2010, 05:32:58 pm
Post by: nid404 on May 01, 2010, 05:56:22 pm
A student makes measurements from which she calculates the speed of sound as 327.66ms–1.
She estimates that her result is accurate to ±3 %.
Which of the following gives her result expressed to the appropriate number of significant figures?
A 327.7ms–1 B 328ms–1 C 330ms–1 D 300ms–1
ok...3% of 227.66 is 9.8 approx 10...so the final answer is rounded off to the nearest 10 which is 330
Post by: tmisterr on May 01, 2010, 06:05:12 pm
how bout this one? I know the value but how do get the direction?
q36 ATTACHED PAPER
Post by: nid404 on May 01, 2010, 06:07:35 pm
Post by: tmisterr on May 01, 2010, 06:56:05 pm
Post by: nid404 on May 01, 2010, 06:58:55 pm
Post by: cashem'up on May 02, 2010, 10:55:51 am
Post by: astarmathsandphysics on May 02, 2010, 11:03:22 am
s_2=1/2at_2^2
subtrat
h=1/2a(t_2^2-t_1^2)
a=2h/(t_2^2-t_1^2)
Post by: tmisterr on May 02, 2010, 02:11:08 pm
field always from positive to negative...so from Q towards P
Thank you once again
Post by: *Hope* on May 03, 2010, 07:15:02 pm
Post by: T.Q on May 03, 2010, 07:46:25 pm
hey guys! Can u help me out in physics CIE paper1 nov03 Q31? am not able to solve it :-\
the total resistance of the circuit is = 4 ohms
find the current across the circuit using I=V/R I=3A
Find the voltage across 2 ohms resistor V=6v
SO , the voltage across 3 & 6 ohms resistors are V=6v
using I=V/R across 6 ohms resistor
I=6/6
I=1A
SO THE ANSWER IS A
Post by: *Hope* on May 04, 2010, 05:33:00 am
the total resistance of the circuit is = 4 ohmshow did u get 4 ohms in the 1st place?? and can u tell me which paths to take to get the total resistance??
find the current across the circuit using I=V/R I=3A
Find the voltage across 2 ohms resistor V=6v
SO , the voltage across 3 & 6 ohms resistors are V=6v
using I=V/R across 6 ohms resistor
I=6/6
I=1A
SO THE ANSWER IS A
Post by: T.Q on May 04, 2010, 10:19:36 am
how did u get 4 ohms in the 1st place?? and can u tell me which paths to take to get the total resistance??
for the parallel resistors 1/3 + 1/6 = 1/R , 1/R=1/2 , so R=2
now for the series 2+2=4 for the total circuit
Post by: *Hope* on May 05, 2010, 06:08:10 am
for the parallel resistors 1/3 + 1/6 = 1/R , 1/R=1/2 , so R=2Thanks!!TQ
now for the series 2+2=4 for the total circuit
+rep