ALL CIE PHYSICS DOUBTS HERE !!!

[NidZ- Hero]

I know the answer But i want An explanation Can Some one explain

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w07_qp_1.pdf

Question 11, 16 , 17 , 26 , 37


http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w08_qp_1.pdf

Question 27 , 29

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s07_qp_1.pdf

Question  40 , 37


http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s08_qp_1.pdf

Question 14 , 16 , 31

THANNNNNKS

[EMO123]

I know the answer But i want An explanation Can Some one explain

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w07_qp_1.pdf

Question 11, 16 , 17 , 26

11.
Equal and Opposite Force Exerted on Road by the wheel so the answer is B

16.
The there is a elastic Potential Energy it is maximum at Position 3
and kinetic energy is minimum because when Potential Energy is Maximum then the K.E. is minimum.

17.
The density of the substance (x cm^3) = Density of the same substance (of unit volume)
       Keeping this in mind, density = mass/volume

        Mass of an atom = Mp, there Np atoms in all

        Thus, mass of Np atoms = Mp*Np

        Volume = unit volume
Density of P = (Mp*Np)/unit volume
Repeat the same with Q, density = (Mq*Nq)/unit volume
As density of P > density of Q
Mp*Np > Mq*Nq
(Unit volume gets cancelled)

26.
as Shown in the diagram that the field is moving from Positive to Negative
so if the electron is placed there then it will definitely the the right so
it will move towards the left

Hope it Helps

[Ghost Of Highbury]

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s09_qp_1.pdf
q 20

Stress = tension/area
Tension = stress*Area
stress = Y*strain
Tension = Y*Strain*Area
strain = e/l
Tension = YeA/l
Tension (P) = YeA/l
Tension (Q) = Ye0.5A/2l = YeA/4l
Tension(P)/Tension(Q) = 4/1

[Ghost Of Highbury]

I know the answer But i want An explanation Can Some one explain

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w07_qp_1.pdf

Question 11, 16 , 17 , 26 , 37


http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w08_qp_1.pdf

Question 27 , 29

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s07_qp_1.pdf

Question  40 , 37


http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_s08_qp_1.pdf

Question 14 , 16 , 31

THANNNNNKS

w08 - 27) Distance b/w 2 antinodes = 15mm
                 wavelength = 15mm*2 = 30mm = 0.03m
Frequency = c/wavelength = (3*10^8)/0.03 = 1*10^10
29) THe point charge is positively charged. SO inwards. Force on X > Y , because its closer

[EMO123]

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9702%20-%20Physics/9702_w08_qp_1.pdf

Question 27 , 29

27.
The Frequency of Microwave is 1*10¹⁰ if you have got the graph then see in that
so the answer is C

29.
as the field strength is moving from positive to negative that is from center to outwards
so the and X is more near to the Center so Magnitude of force on X is Greater Than on Y
and radially inwards as explained
so the answer is B

[NidZ- Hero]

11.
Equal and Opposite Force Exerted on Road by the wheel so the answer is B

Y not A

[Ghost Of Highbury]

I'm off, good luck for tomorrow guyz. Hope you all do well.

[EMO123]

I'm off, good luck for tomorrow guyz. Hope you all do well.

thanxx and good luck to all of us

[physichemaths]

Hai guys. How to get the answers for nov08 ques 1?

[SkyPilotage]

Hai guys. How to get the answers for nov08 ques 1?

Speed of light = 3x10^8 ms^-1
So Number of wavelength/second = frequency= speed/wavelength..

[EMO123]

May/June 2007
Paper 04

Q-2 c)
Q-3 b)
Q-4

more doubts coming from me soon

[$H00t!N& $t@r]

Please help me out with the following questions:

1) Two satellites orbit a planet of mass 'M' with Time periods of 3 days and 4.5 days, respectively. If the orbital radius of the first satellite is 2x10^6 m, determine the radius of orbit of the other. (Answer is 2.62x10^6)

2) A mass 'm' is moved from the North pole towards the equator. If the Earth is assumed to be a perfect sphere of radius 'R' , and if the mass moved is exactly 60kg, determine the difference in weight when measured at the North pole and at the equator.
R = 6.4x10^6 m
M of Earth = 6.0x10^24 kg
G = 6.67x10-11
m = 60 kg
    ( Answer is 2.0 N )

3) An astronaut travels from the Earth towards the moon where he experiences no net force. Determine the distance from the surface of the moon where this phenomenon occurs. Given that,

- the radius of the moon is 1.74x10^6 m
- mass of the moon is 0.0735x10^24 kg

(Answer is 3.71x10^7 m )

Please help me out.. i desperately need the explanation for these quesitons...

[tmisterr]

Please help me out with the following questions:

1) Two satellites orbit a planet of mass 'M' with Time periods of 3 days and 4.5 days, respectively. If the orbital radius of the first satellite is 2x10^6 m, determine the radius of orbit of the other. (Answer is 2.62x10^6)

2) A mass 'm' is moved from the North pole towards the equator. If the Earth is assumed to be a perfect sphere of radius 'R' , and if the mass moved is exactly 60kg, determine the difference in weight when measured at the North pole and at the equator. ( Answer is 2.0 N )

3) An astronaut travels from the Earth towards the moon where he experiences no net force. Determine the distance from the surface of the moon where this phenomenon occurs. Given that,

- the radius of the moon is 1.74x10^6 m
- mass of the moon is 0.0735x10^24 kg

(Answer is 3.71x10^7 m )

Please help me out.. i desperately need the explanation for these quesitons...

hello  

1. gravitational force = centripetal force. once we know that...we have solved the question.

   GMm/r²=mv²/r
where M is mass of planet and m is mass of satellite. since they orbit the same planet, M is equal in both cases.

now we know that v= 2*pi*r/T where 2*pi*r is the circumference of orbit and T is the periodic time (easy, velocity=distance/speed). so next we play around with the above equation... It is not difficult to see that

M= 4*pi²*r³/GT²

Voila...so now we plug in the data values of planet 1, converting days into seconds and the value of gravitational constant G (6.67x10^-11) to get the value of M as 7.04....x10¹⁹

Ok so now we play around with the equation for M to make r the subject of the formula...easy enough.

We know T for the second planet is 4.5 days (don't forget to convert to 4.5*24*3600 seconds) and we now know that M is 7.04...x10¹⁹. simple arithmetic............r is (to be exact lol) 2620741.394 m but due to rules of accuracy we quote it to 3 s.f hence 2.62x10^6[/sup

[tmisterr]

on so once you got the concept...we can save time doing all those arduous calculations.
all we need to see is this:

M=4*pi²*r3/GT²

and since M is the same for both planets, we can replace this equation directly into the equation of R (radius of second planet)

to get R=cube root( T²r³/t²)
try deriving for yourself

where T is the period of planet 2 (4.5 days converted to seconds), r is the radius of planet 1 (2x10⁶) and t is the period of planet 1 (3 days converted to seconds)

I'm sure you'll find this saves a lot of time

[$H00t!N& $t@r]

i love u mannn!  thanks alott