ALL CIE PHYSICS DOUBTS HERE !!!

[wstrawberries]

Question 9:-( I just need a confirmation on this )
Since the mass is going up and down..
At A its halfway going up..
At B its at its highest point...
At C its halfway going down..
And finally At D its at the bottom which is the lowest point of its motion..

Question 33:
Find the Pd of the resistors accross XQY or XPY..
The voltage across the 2 parallel wires is the same i.e 12 V...
To find the voltage of the 500 ohm resistor its 500/(500+1000) x12 = 4 V
ANd the voltage across the 2000 ohm resistor is 2000/(2000+1000) = 8 V
-->Therefore The potential DIFFERENCE is 8-4 volts = 4volts..
* you could also go using XQY and find the voltages accross the 1000 oh resistors and you will find the top one to be 4 and the bottom on eot be 8 so THe P.d between XandY is 4 volts...

@Ghost I forgot it too
Glad to help

Thanks a lot uhhmm for q 33 what formula are you using to calculate 4V and 8V?

[SkyPilotage]

Thanks a lot uhhmm for q 33 what formula are you using to calculate 4V and 8V?

I used it as A ratio..But If you are confused...
USE the formula R=V/I..
For example the current passing through X is v/R= 12/1500 = 0.008 A
To find the V in the 1000 ohm resistor : 0.008 x 1000 = 8 V...
Through Y : do the same thing..
I= 12/3000 = 0.004 A
V in 1000 ohm resistor= 0.004 x 1000=4 volts..

[wstrawberries]

I used it as A ratio..But If you are confused...
USE the formula R=V/I..
For example the current passing through X is v/R= 12/1500 = 0.008 A
To find the V in the 1000 ohm resistor : 0.008 x 1000 = 8 V...
Through Y : do the same thing..
I= 12/3000 = 0.004 A
V in 1000 ohm resistor= 0.004 x 1000=4 volts..

ohhhh okay thanks again!

[username]

qp11
7.  I have trouble in this too. How does horizontal component of velocity decrease to zero? Isn't it constant in a projectile?
8. It goes up 5m, comes down 3m. Displacement = 5-3 = 2m
9. v^2 = u^2 +2as
   s = v^2-u^2 / 2a
   v = 0, so 0 = u^2+2as -> 2a = -u^2/s or -u^2/x

Now, the new u = U+0.2u = 1.2u
V = 0
so s = 0-1.44u^2 /2a
s = -1.44u^2/(-u^2/x) = 1.44x
11. M1V1 -M2V2 = 0
    M1V1 = M2V2
V1/V2 = M2/M1
B
15. It can be either A or D (because they are closed vector triangles)
It's D, because the direction of the arrow representing the force by the cable is correct.
22. The length of the wire doesnt affect the extension. The load and cross-sectional area do. Here both are the same. So extensions of
     both the wires are same (1.5 and 2), that's y total e = 1.5 + 2 = 3.5mm

29. It's upward because electrons are attracted towards +ve plate. And ofcourse, it's not a part of the circle because the path is not exactly an arc.
40. Count th4e no. of protons in Argon, it's 18. Only option with 18 protons for argon is C. So C it is

qp12

9. There's a little bit of guess work involved here.
Given it's inelastic, the rebound velocity won't be v. Say, 40% of energy is lost, say around 0.6v
Change = (-0.6v*2m) - 2mv = -3.2mv
SO the change is almost 3mv.
(Even if u take 0.7v, u won't get 4mv+)

34. When x = 0, the voltage measured is across the whole of the wire, so it has to be V.
    when x decreases, and the pointer is moved at the right end of the part with resistivity p, the voltage is divided into 2 parts (p and 2p+3p)
   So the voltage across the length x is now, not V, but a lil less than V because some VOltage is lost in the remaining length of the wire.
Answer: B (note the graph is not a straight line because as x decreases, the resistance doesn't decrease proportionately)

13. Ofcourse it's D, gravitational pull is constant (mg). Mass is constant and a is a constant.
37. Check 1/R = 1/R1 + 1/R2 + 1/R3
And see when u get value of R as 50.

sry...
i got the links jumbled up:(

[EMO123]

sry...
i got the links jumbled up:(

so you got that or not
if not then tell i will post all in another method

[physichemaths]

I have trouble with question 24 june 2008? Anyone?

[username]

so you got that or not
if not then tell i will post all in another method

umm
i mean u answered the rong questions
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_11.pdf
questions: 9,13,34,37
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_12.pdf
questions: 7,8,9,11,15,22,26,27,29,40

got it mixed up:(

[physichemaths]

I have trouble with question 24 june 2008? Anyone?

please help me

[username]

please help me

answer is c
because
same steel so young modulus is the same

[physichemaths]

answer is c
because
same steel so young modulus is the same

thanks! How about ques 26?

[physichemaths]

answer is c
because
same steel so young modulus is the same

[physichemaths]

answer is c
because
same steel so young modulus is the same

[Ghost Of Highbury]

umm
i mean u answered the rong questions
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_11.pdf
questions: 9,13,34,37
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w10_qp_12.pdf
questions: 7,8,9,11,15,22,29,40

got it mixed up:(

Oh darn!

@physichemaths - 26) Ip = k1/r^2
                               Ip = k2*A^2
k1/r^2 = k2*(8*10^-6)^2

Iq = k1/4r^2

Iq = k1/4r^2 = k1/r^2 * 1/4 = k2*(8*10^-6)^2 *1/4 = 1.6*10^-11k2
Iq = k2A^2
A^2 = 1.6*10^-11
A = sqrt(1.6*10^-11) = 4*10^-6

[username]

Oh darn!

@physichemaths - 26) Ip = k1/r^2
                               Ip = k2*A^2
k1/r^2 = k2*(8*10^-6)^2

Iq = k1/4r^2

Iq = k1/4r^2 = k1/r^2 * 1/4 = k2*(8*10^-6)^2 *1/4 = 1.6*10^-11k2
Iq = k2A^2
A^2 = 1.6*10^-11
A = sqrt(1.6*10^-11) = 4*10^-6

sry bout dat...

[physichemaths]

sry bout dat...

how come the amplitude is times 2?