Physics: Post your doubts here!

[seneria]

can anyone solve question 5 part b and c(11) of nov 12 paper 43 physics?

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_43.pdf​

 

[Vikings]

Need help with A2 Physics

Guyz i need help in the following question frm year 2003 Physics Paper 4 Q1 (c) :%)

C(i): Change in gravitational Potential = Potential at Earth surface(GM/R) - Potential at Altitude(GM/r) where (r=R + Altitude)
C(ii) Kinetic Energy = Change in Potential Energy
0.5m(v)2 = GMm/R
0.5m(v)2 = GM/R x m
o.5m(v)2 = Change in gravitational potential ( Ci) x m
"m" in Kinetic energy and the "m" in Potential energy will cancel out.
Therefore:
0.5(v)2 = Change in gravitational Potential.

Hope this was useful!!!!
Good luck for the exams.

 

[seneria]

vikings please reply to my question too... thanks

 

[Vikings]

can anyone solve question 5 part b and c(11) of nov 12 paper 43 physics?​

​

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_43.pdf​

5(b). Outside the area between the 2 magnets the magnetic field is zero. As the probe comes closer and move between the magnet
the magnetic field lines are cut, hence an e.m.f is induced. The induced e.m.f is maximum and constant inside the magnetic field. The e.m.f
decrease steeply to zero, once the probe moves out of the magnetic field, because no field lines are cut.
Hope this was useful.
Good luck for the exams!!!!

 

[Jiyad Ahsan]

right ok umm
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf

Q9 (b), can someone explain the working please.. in words.. if possible
Vikings ?

 

[cyclone94]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_42.pdf
qs 12 b(ii)
plz help!!

 

[VampBhums]

I had practical AS physics paper 33 n there was an electricity question that I mite hv done incorrectly. Pls c the thumbnail below to help me knw if i did correctly! I accidentally used the ammeter wire as a rheostat slider. The rheostat was not a real one but a one made up with wire coiled around a wood block.

 

[seneria]

5(b). Outside the area between the 2 magnets the magnetic field is zero. As the probe comes closer and move between the magnet
the magnetic field lines are cut, hence an e.m.f is induced. The induced e.m.f is maximum and constant inside the magnetic field. The e.m.f
decrease steeply to zero, once the probe moves out of the magnetic field, because no field lines are cut.
Hope this was useful.
Good luck for the exams!!!!

and c part (11)??

 

[Zari]

can anyone explain Q12 b) ii) ?
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_42.pdf

 

[gary221]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_42.pdf
qs 12 b(ii)
plz help!!

can anyone explain Q12 b) ii) ?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_42.pdf

Here, you just have to apply the formula signal to noise ratio = 10 * log(signal power/noise power) → [pg 522 in the coursebk]

Now, first of all v need to find the minimum signal power acceptable in the fibre
We know the minimum signal to noise ratio = 25 dB
Noise power in the fibre = 6.1 * 10^-19 W

So, signal to noise ratio → 25 = 10 * log(min. signal power/ 6.1 * 10^-19)
antilog 2.5 = min signal power/(6.1 * 10^-19)
So, minimum signal power = 1.93 * 10^-16 W

Now, we have to find the signal loss in the fibre,
Signal loss = 10 * log(power supplied/ min signal power)
Signal loss = 10 * log[(6.5 * 10^-3) / (1.93 * 10^-16)]
signal loss = 135 dB

Now, this is the total loss in signal → total attenuation
We know, attenuation per unit length = attenuation/ max uninterrupted length of cable → [pg 522 in the coursebk]
We need th find the max uninterrupted length of fibre,
Attenuation per unit length = 1.6 dB/km
Total attenuation = 135 dB
So, max fibre length = 135 / 1.6 → 85 km

Dist btw towns = 75 km

So, no amplifiers required.

Hope u gt it!!

 

[cyclone94]

gary221
jazakallah

 

[Zari]

Here, you just have to apply the formula signal to noise ratio = 10 * log(signal power/noise power) → [pg 522 in the coursebk]

Now, first of all v need to find the minimum signal power acceptable in the fibre
We know the minimum signal to noise ratio = 25 dB
Noise power in the fibre = 6.1 * 10^-19 W

So, signal to noise ratio → 25 = 10 * log(min. signal power/ 6.1 * 10^-19)
antilog 2.5 = min signal power/(6.1 * 10^-19)
So, minimum signal power = 1.93 * 10^-16 W

Now, we have to find the signal loss in the fibre,
Signal loss = 10 * log(power supplied/ min signal power)
Signal loss = 10 * log[(6.5 * 10^-3) / (1.93 * 10^-16)]
signal loss = 135 dB

Now, this is the total loss in signal → total attenuation
We know, attenuation per unit length = attenuation/ max uninterrupted length of cable → [pg 522 in the coursebk]
We need th find the max uninterrupted length of fibre,
Attenuation per unit length = 1.6 dB/km
Total attenuation = 135 dB
So, max fibre length = 135 / 1.6 → 85 km

Dist btw towns = 75 km

So, no amplifiers required.

Hope u gt it!!

JazakAllah May Allah Bless u

 

[ahmed abdulla]

mcqp ... mj 2002 q 15 ??

 

[Osaid Zahid]

In short, yes, the energy of all waves in the electromagnetic spectrum is carried by the photon; it is the quanta of electromagnetic radiation, and as such it carries the energy possessed in every wave, and it interacts with other particles by transferring this energy, etc.

In fact, electrons can be "made into" waves, since they exist in the form of waves; de Broglie, in his thesis, considered the possibility of matter having wave like properties, and he derived the following equation, that relates the momentum of a material object to its wavelength:

Wavelength = h/p
Where h = Planck's Constant
p = momentum of matter - based object concerned.

Usually, heavier objects have such small wavelengths that they can be ignored and assumed to not even exist; however, for matter, on the atomic level (e.g. electrons, etc), the values of momentum for the concerned particles are so small that the de Broglie wavelength is actually noticeable (by experiment) - for example, fast moving electrons can be diffracted; since only waves can be diffracted, the electrons can be proved to be possessing some sort of wave - like property.

Hope this helped!
Good Luck for all your exams!

Thank You very much .. That cleared a lot of my concepts... So then how align photons into a wave .. or are they disturbed from the surroundings like sound.

 

[ahmed abdulla]

go here youll fin your answer http://adf.ly/P9qyL

can u expain this question ie q15 and q 28 and 35 plz ?

 

[sagar65265]

Thank You very much .. That cleared a lot of my concepts... So then how align photons into a wave .. or are they disturbed from the surroundings like sound.

I'm sorry, I didn't really understand, but electromagnetic waves are not disturbances of air particles or any medium surrounding them; these waves are composed of oscillating electric and magnetic fields, that oscillate perpendicular to the direction of motion; they're not disturbances of the surrounding medium, just fluctuating/varying magnetic and electric fields.

Hope I understood right, sorry if I didn't!

Good Luck for all your exams!

 

[Zari]

can anyone explain me this as well M/J 2004 Q7 c) ?
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s04_qp_4.pdf

 

[Dug]

can anyone explain me this as well M/J 2004 Q7 c) ?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s04_qp_4.pdf

From (b), λ = h/√(2mE)
λ² = h²/2mE
E = h²/2mλ²

Also, E = qV
qV = h²/2mλ²

Insert values:
(1.6*10^-19)V = (6.63*10^-34)²/(2)(9.11*10^-31)(0.4*10^-9)²
V = 9.4 V

 

[Zari]

From (b), λ = h/√(2mE)
λ² = h²/2mE
E = h²/2mλ²

Also, E = qV
qV = h²/2mλ²

Insert values:
(1.6*10^-19)V = (6.63*10^-34)²/(2)(9.11*10^-31)(0.4*10^-9)²
V = 9.4 V

Thanks

 

[blueberryyums]

Hey, Anyone has notes on quantum physics and nuclear physics A2 please?