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Physics: Post your doubts here!

rz123

rz123

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Doctor Nemo said:
Why not at right angles to the field? Imagine the proton traveling between the two parallel plates in the figure above this question. There will be a downward force on it and you calculate it from the electric field strength as F/Q so answer is charge on proton X 2.40 × 10–6 V m–1.
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1st of all thanks 4 the response. ok if we take the figure on the previous question and imagine the proton to be coming initially bf4 entering the field at right angle to the field means coming straight nd the field is from positive to negative plate downward. now in the c part he asks us to calculate the acceleration of the proton at right angles to the field which i m not getting dat how it is zero. the proton will be coming like a mass in a projectile motion so it must have some acceleration at right angles to the field as well? :/ the marking scheme says its zero nd the force is at right angles to the field ? :|
 
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Rvel Zahid said:
1st of all thanks 4 the response. ok if we take the figure on the previous question and imagine the proton to be coming initially bf4 entering the field at right angle to the field means coming straight nd the field is from positive to negative plate downward. now in the c part he asks us to calculate the acceleration of the proton at right angles to the field which i m not getting dat how it is zero. the proton will be coming like a mass in a projectile motion so it must have some acceleration at right angles to the field as well? :/ the marking scheme says its zero nd the force is at right angles to the field ? :|
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I agree that makes no sense. Maybe the mark scheme is wrong.
 
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Silent Hunter said:
http://www.xtremepapers.com/CIE/index.php?dir=International A And AS Level/9702 - Physics/&file=9702_s11_qp_21.pdf

aslamoalikum...... wanted help in this paper question 5 part (b) ..?? :)

Thank You
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When the arrow of the potential divider is on the far left the 6 ohm resistor is in parallel with just the wire connecting the potential divider to the circuit. The voltage across this is 0 so the minimum current is 0.

When the arrow is on the far right the arrow is now in parallel with all the resistance of the variable resistor and in parallel with the battery. Now the full 12 volts is across the 6 ohm resistor and the maximum current is two amps.
 
Ashique

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Silent Hunter said:
http://www.xtremepapers.com/CIE/index.php?dir=International A And AS Level/9702 - Physics/&file=9702_w10_qp_12.pdf

Q 34 ...?
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Now, we know from the question that the second portion of the wire has twice the resistivity of the first portion, and the third portion has thrice the resistivity of the first portion. This will alter the gradient, and not be straight line. So this eliminates A from the options, Now we have B and C to work with. Notice how the second portion of B is two times more steeper, and the third portion is three times more steeper than the first portion. The higher the resistivity, the higher will be the steepness (or the numerical value of the gradient) So the answer will be B.
You may have doubt to why C is not the answer, it confused me too. It appears to me that the first portion of the graph of C appears to have be the steepest of the three lines, so a higher negative gradient.
 
Ashique

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Doctor Nemo said:
When the arrow of the potential divider is on the far left the 6 ohm resistor is in parallel with just the wire connecting the potential divider to the circuit. The voltage across this is 0 so the minimum current is 0.

When the arrow is on the far right the arrow is now in parallel with all the resistance of the variable resistor and in parallel with the battery. Now the full 12 volts is across the 6 ohm resistor and the maximum current is two amps.
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When the arrow is towards the right, the the resistance of the variable resistance is 12 ohms, as it will provide resistance in the circuit. The two resistances are now in parallel, and the resistance is
1/12+1/6= 1/4.
So the reciprocal, gives us 4 ohms in the circuit, since I=V/R, shouldn't the it be 3 A? The maximum current? I know it's wrong, but I don't really get what you did.

And when the arrow is pointing towards the right, the resistance of the variable resistor is 0, so the resistance in the entire circuit is 6 ohms, and hence 2 A.

I know, I make no sense at all, but can you please elaborate what you did?
 
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unique840

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MEGUSTA_xD said:
Help pls?
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the question has the table of metals along with their work function energies. we hav 2 lights. one of wavelength 350 and one of 700. we will find the energies of both the lights and compare them with the work function energies. only potassium and light of 350 nm wavelength gives energy more than work function so it will give rise to photoelectric emission
 
rz123

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worksheet 12, Q. 6 c part] (why is the ms saying dat current in lamp increases proportionally with resistance?)
R= V/I so current decreases with increase in resistance...:/
 

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DANGERBP

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Rvel Zahid said:
worksheet 12, Q. 6 c part] (why is the ms saying dat current in lamp increases proportionally with resistance?)
R= V/I so current decreases with increase in resistance...:/
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mm man is there by any chance ms would b wrong ?? coz if that was r8 ide kill myself =/ ! =@
 
rz123

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DANGERBP said:
mm man is there by any chance ms would b wrong ?? coz if that was r8 ide kill myself =/ ! =@
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yo man remain easy! :p me not getting this 1. i guess the ms is faulty. ur life is much expensive then da filament lamp in the question causing the prob so chill man. :D
 
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Ashique said:
When the arrow is towards the right, the the resistance of the variable resistance is 12 ohms, as it will provide resistance in the circuit. The two resistances are now in parallel, and the resistance is
1/12+1/6= 1/4.
So the reciprocal, gives us 4 ohms in the circuit, since I=V/R, shouldn't the it be 3 A? The maximum current? I know it's wrong, but I don't really get what you did.

And when the arrow is pointing towards the right, the resistance of the variable resistor is 0, so the resistance in the entire circuit is 6 ohms, and hence 2 A.

I know, I make no sense at all, but can you please elaborate what you did?
Click to expand...


You are only interested in the current going through the 6 ohm resistor not the total current leaving the battery. The answer does not depend on the resistance of the variable resistor only the amount of the variable resistor that is in parallel with the 6 ohm resistor.

If the arrow was in the middle of the variable resistor then the 6 ohm resistor would be in parallel with 1/2 of the variable resistor. There is a 12 volt drop across the variable resistor and one half or 6 volts would also be across the 6 ohm resistor and there would be 1 amp through the 6 ohm resis
 
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