• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Physics: Post your doubts here!

gary221

gary221

Messages
6,440
Reaction score
31,077
Points
698
zain786 said:
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 :(
Click to expand...
paper link??
 
KurayamiKimmi

KurayamiKimmi

Messages
191
Reaction score
80
Points
38
sufyanalizaib said:
What happens when one of the slit is closed in YOung's double slit experiment.. ??????????? kindly reply
Click to expand...
when one slit is closed light still passes from the other slit
therefore the light passes through that slit and you get light on the screen BUT no fringe pattern , just a completely lit surface
 
syed1995

syed1995

Messages
1,824
Reaction score
949
Points
123
zain786 said:
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 :(
Click to expand...

Wrong thread :p
 
Soldier313

Soldier313

Messages
958
Reaction score
3,499
Points
253
1357913579 said:
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)
Click to expand...

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.
 
daredevil

daredevil

Messages
1,394
Reaction score
1,377
Points
173
gary221 said:
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
Click to expand...
why is 9810 not -ve? i mean shouldn't it be -9810?? :confused:
 
salvatore

salvatore

Messages
382
Reaction score
315
Points
73
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
 
S

selena

Messages
41
Reaction score
21
Points
18
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
 
S

selena

Messages
41
Reaction score
21
Points
18
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

gary221

Messages
6,440
Reaction score
31,077
Points
698
cyclone94 said:
salamz plz help wid dis qs...frm m/j-10-43...qs 4(c)
gary221
Click to expand...

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!!
:p
 
1

1357913579

Messages
466
Reaction score
101
Points
53
Soldier313 said:
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.
Click to expand...
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?
 
You must log in or register to reply here.
Top