• 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!

Messages
5
Reaction score
0
Points
1
You cannot use the actual capacitance of X for the first part because you don't know the don't know the potential difference across X. So it is easier to use the effective capacitance and the total potential difference. Otherwise you can also do it by the longer method if you want to use the actual capacitance.
First calculate the potential difference cross X.
Total resistance in series is 12 micro farad and 24 micro farad from the parallel combination. So, potential across X=
(capacitance of X/ total capacitance in series) * 9
(1/12)/(1/12+1/24) * 9 = 6v
Now use Q =CV for X
Q = 12 micro farad * 6 =7.2* 10^-5 =72 micro C
You get the same answer, just a longer calculation.

For Y you can use the actual capacitance because you already know the potential difference cross Y from your previous calculations.

Sorry, but farad is resistance? I thought it was the capacitance. So you can use it to calculate p.d. just like actual resistance?
 
Messages
887
Reaction score
466
Points
73
The Answer will be B that i think so,PLEASE do tell me if i am right?
A: In A weight is acting downward and the normal reaction force at the end of ladder is upward and the force P is the normal reaction force by the wall in straight direction.Which means the ladder is not sliding and on the equilibrium postion
B: In B weight is acting downward and the normal reaction force at the end of ladder is upward and the force P force P is acting slightly upward because the ladder is sliding .
C: In C weight is acting downward again and the normal reaction force at the end of ladder is now not acting against the direction of weight it is slighlt inward.
D : In D weight is acting downward again and the normal reaction force at the end of ladder is again not acting against the direction of weight it is slightly Outward
Thus the Answer will be B :)
look buddy this question need some focus the wall is rough but the floor is not
because of contact of ladder with both wall and floor there would be component of force perpendicular to the wall and the floor.
now since there is the floor has no friction so there would be no horizantal component and the result would be only vertical because of the contact force.
now the wall has both the frictional force and contact force. the resultant to vertical and horizantal force would be a hypotenuse
 
Messages
844
Reaction score
2,495
Points
253
Sorry, but farad is resistance? I thought it was the capacitance. So you can use it to calculate p.d. just like actual resistance?

I am so sorry, i wrote resistance instead of capacitance. :(
I meant the capacitance in series is 12 and 24 micro farad.
And you can use capacitance to calculate potential difference by using the ratio of capacitance * total potential.
 
Messages
1,442
Reaction score
7,812
Points
523
Can any solve my question?
A taut wire is clamped at two points 1.0m apart.It is plucked near one end.which are three longest wavelengths on the vibrating wire?
A: 1.0m, 0.50m, 0.25m
B: 1.0m, 0.67m, 0.50m
C; 2.0m, 0.67m, 0.40m
D: 2.0m, 1.0m, 0 .50m
E: 2.0m, 1.0m, 0.67m
 
Messages
43
Reaction score
19
Points
18
Can any solve my question?
A taut wire is clamped at two points 1.0m apart.It is plucked near one end.which are three longest wavelengths on the vibrating wire?
A: 1.0m, 0.50m, 0.25m
B: 1.0m, 0.67m, 0.50m
C; 2.0m, 0.67m, 0.40m
D: 2.0m, 1.0m, 0 .50m
E: 2.0m, 1.0m, 0.67m
Can u tell, which year it is from? and the answer?
 
Messages
887
Reaction score
466
Points
73
Can any solve my question?
A taut wire is clamped at two points 1.0m apart.It is plucked near one end.which are three longest wavelengths on the vibrating wire?
A: 1.0m, 0.50m, 0.25m
B: 1.0m, 0.67m, 0.50m
C; 2.0m, 0.67m, 0.40m
D: 2.0m, 1.0m, 0 .50m
E: 2.0m, 1.0m, 0.67m
FUNDAMENTAL SECOND HARMONIC AND THIRD HARMONIC ARE
WAVELENGTH=2L WAVELENGTH=L WAVELEGNTH 2/3L RESPECTIVELY NOW PLUG IN IN L AND U WILL GET UR ANS WHICH IS E
 
Messages
887
Reaction score
466
Points
73
Hi. I have a doubt in Photoelectric effect. I don't understand how the photoelectric current reduces when we reduce the wavelength.
Qs. 7C http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_41.pdf
Forget the photoelectric effect for a moment. Think about what is meant by the intensity of the radiation of the target surface.

Intensity = power / [target area]
Rearranges to:
power = intensity * area

If you are keeping the intensity constant, and the area of the photo-target is constant, then the power of the incident radiation must also be constant.

Each photon brings with it a certain amount of energy. If you reduce the wavelength, you increase the amount of energy carried by each photon.

The energy delivered = (energy of a photon) * (number of photons)

power = energy / time

So:
power = (energy of a photon) * (number of photons) / time

If the power is constant, and the energy of a photon is increased, then (number of photons / time) must decrease.

Now we are back to the photoelectric effect.
Fewer photons/s = fewer electrons/s = smaller current.

EDIT:

Nope!

The charge on an electron is fixed.
The total charge delivered to the photo-anode = (charge on electron ) * (number of electrons)

Electric current = [charge] / [time]

Current = (charge on electron ) * (number of electrons)/(time)

If you have fewer electrons/second emitted by the photo-surface, you have a smaller current.

How quickly the electrons travel does not change that. (this is what my friend said i got no idea about this)
 
Top