Physics: Post your doubts here!

[gary221]

No i want to know how to make the graph. Lol

well, obviously... bt i meant tht since the ques is not asking 4 a graph, why do u want it??

 

[hellangel1]

well, obviously... bt i meant tht since the ques is not asking 4 a graph, why do u want it??

Lol.. i attached the wrong paper
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_22.pdf

 

[syed1995]

can you please help me in how to answer questions such as exam 9709/32/O/N10 the question 8 part b

its basically after making partial fractions...expanding it to ascending powers upto x^3 etc

how do you do these? plz help

Wrong thread

 

[sma786]

Hi
Does anyone have a good formulae sheet for AS Level Physics?

 

[1357913579]

can some one please help with this pleaseee
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_41.pdf
7c(ii)

 

[snowbrood]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_21.pdf
why Q5b(ii) which region is decreasing and which is constant and why. what if region is increasing than how would the diagram be

 

[Soldier313]

can some one please help with this pleaseee
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_41.pdf
7c(ii)

okay look, if you reduce the wavelength of the incident radiation, the energy of the photon increases (since energy of the photon is inversely proportional to the wavelength).
This will mean that each of the photons will now be carrying more energy. So, inorder to maintain a constant intensity, the rate of arrival of the photons on the metal surface must be reduced.
Therefore, the rate of emission of the photoelectrons will be reduced. (Photoelectric current is the rate of emission of photoelectrons).

Hope it helped.

 

[Soldier313]

Aoa wr wb

Can someone please explain 4b ii of this paper?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_ms_41.pdf

 

[daredevil]

I remember tht...kinda tough, hold on.

pressure = density * g * height
pressure in the hole = 17.5MPa = 17.5 * 10^6 Pa, which is a result of the pressures of both the liquids p1 n p2
So, P(hole) = P1 + P2
17.5 * 10^6 = (d1 * g * h1) + (d2 * g * h2) ------> d1 = density of water = 1000kgm^-3 h1 = 2000 - x m
d2 = density of oil = 830 kgm^-3 h2 = x m
17.5 * 10^6 = [10000 * 9.81 * (2000-x)] + [830 * 9.81 * x]
17.5 * 10^6 = 9810 x + 19620000 + 8142.3 x
x =1271 m

why is 9810 not -ve? i mean shouldn't it be -9810??

 

[cyclone94]

salamz plz help wid dis qs...frm m/j-10-43...qs 4(c)
gary221

 

[salvatore]

I always get confused in interpreting questions regarding a variable resistor or a length of a wire in a circuit.
Could anyone please explain when p.d across a variable resistor or a length of wire in a circuit is maximum/minimum using a jockey (contact) to vary the resistance?
Thanks

 

[selena]

A trolley runs from P to Q along a track. At Q its potential energy is 50kJ less than at P.
trolley
P
Q
At P, the kinetic energy of the trolley is 5kJ. Between P and Q, the work the trolley does against
friction is 10kJ.
What is the kinetic energy of the trolley at Q?
A 35kJ B 45kJ C 55kJ D 65kJ

 

[selena]

A metal cube of side l is placed in a vice and compressed elastically by two opposing forces F.
F F l
l
metal cube
How will ∆l, the amount of compression, relate to l?
A ∆l ∝ 2
1
l
B ∆l ∝
l
1
C ∆l ∝ l D ∆l ∝ l
2

 

[gary221]

salamz plz help wid dis qs...frm m/j-10-43...qs 4(c)
gary221

To calculate the work done, 1st find the change in potential...
For tht, find the potential of the electron at midpoint of line AB
This potential will be influenced by both the point charges at A n B -----> electric potential = kQ/r whr r = distance of separation.
So, potential at midpoint = potential due to charge A + potential due to charge B
= kQ(a)/r + kQ(b)/r
= [(6.4 × 10^–19) / (4πε0 × 6 × 10^–6)] + [(6.4 × 10^–19) / (4πε0 × 6 × 10^–6)]
= 1.92 * 10^-3 V
Note Q = charge at A/B

Potential at P = kQ(a)/ra + kQ(b)/rb
= [(6.4 × 10^–19) / (4πε0 × 3 × 10^–6)] + [(6.4 × 10^–19) / (4πε0 × 9 × 10^–6)]
= 2.56*10^-3 V
Now change in potential = potential at P - potential at midpoint
= (2.56 * 10^-3) - (1.92* 10^-3)
= 6.4 * 10^-4 V

since electric potential = work done in bringing UNIT charge
So, work done in moving an electron = potential * charge of electron
= (6.4 * 10^-4) * (1.6 * 10^-19)
= 1.0(24) * 10^-22 J
Hope u gt it!!

 

[1357913579]

okay look, if you reduce the wavelength of the incident radiation, the energy of the photon increases (since energy of the photon is inversely proportional to the wavelength).
This will mean that each of the photons will now be carrying more energy. So, inorder to maintain a constant intensity, the rate of arrival of the photons on the metal surface must be reduced.
Therefore, the rate of emission of the photoelectrons will be reduced. (Photoelectric current is the rate of emission of photoelectrons).

Hope it helped.

thanks alot for your reply
iam a bit still confused like isnt intersity the no. of photons travelling? so if thet are constant wouldn't the current remain constant?

 

[cyclone94]

gary221
Jazakallah bro....may Allah reward u fr d help...Aameen

 

[gary221]

A metal cube of side l is placed in a vice and compressed elastically by two opposing forces F.
F F l
l
metal cube
How will ∆l, the amount of compression, relate to l?
A ∆l ∝ 2
1
l
B ∆l ∝
l
1
C ∆l ∝ l D ∆l ∝ l
2

paper link??

 

[selena]

paper link??

winter 2011

 

[abruzzi]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w02_qp_2.pdf
Someone please help me with question 7 b(i), I don't understand the marking scheme.
Thanks

 

[gary221]

Aoa wr wb

Can someone please explain 4b ii of this paper?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_ms_41.pdf

Since both the nuclei have initial kinetic energy = Ek ---> total initial kinetic energy = 2Ek.
Now, as the nuclei(both positively charged) move closer together, they gain electric potential energy.
But, as the distance btw them is being reduced, they lose gravitational potential energy.
ie kinetic energy + gravitational potential energy is lost.
electric potential energy is gained.
So, 2Ek + ΔEgpe = ΔEepe ------> 2EK= ΔEepe – ΔEgpe
So, 2Ek = (6.06 × 10–14 ) - (1.93 × 10–49 )
Ek = 3.03 * 10^-14 J
Converting into MeV,
Ek = (3.03 * 10^-14)/(1.6 * 10^-13)
Ek = 0.19 MeV.
Hope u gt it!