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---> Physics Paper 4 Past Paper Discussion <---

What do you find most difficult in physics P4?

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MaxStudentALevel

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sagar65265 thanks once again!!!
Could you please take a look at Q10 for ON 2009 41??

Im getting the opposite answer of the MS?
Wouldnt the red turn on for 18? And be green for 16? As V- would be greater an V+ here
Because the thermistor adds overall more resistance thus more voltage to that side?
 
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sagar65265

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1357913579 said:
thanks alot bro for your cooperation
this is whats written in the examiner report:
Very few candidates were able to cope with the two charges and determine the potential change
between the two points. A large number did not realise that the potential at each point was the
sum of the potentials for each charge. A number of candidates assumed the midpoint was at zero
potential or that there was only one charge. A significant number tried to determine the work done
using W = Fd or Vq. Very few were able to determine the change in potential between the two
points for both charges and then determine the work done on an electron.
Bro can you please solve this question to get the value of work done as i have tried for hours and hours still getting the wrong answer.
Click to expand...

Man, man , i'm sorry, there may not be any force on a stationary electron at that point and the net field strength may be zero, but the potential definitely isn't zero, man, i'm really sorry about that, Ashique was absolutely right - the potential energy of an electron in between the two positive charges will not be zero, it will be the sum of two systems, the electron and one of the charges and the electron and the other charge!

Just try your calculation by including the potential in the beginning, see if it works!

Good Luck for all your exams!
 
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sagar65265

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MaxStudentALevel said:
sagar65265 thanks once again!!!
Could you please take a look at Q10 for ON 2009 41??

Im getting the opposite answer of the MS?
Wouldnt the red turn on for 18? And be green for 16? As V- would be greater an V+ here
Because the thermistor adds overall more resistance thus more voltage to that side?
Click to expand...

The input potential to the inverting input of the op amp is the potential across resistor P, not across the thermistor T - the potential across T is the number of Volts lost when unit charge travels through T, and the potential into the inverting input is the remaining potential, which is the potential across the resistor R, not the thermistor T. So the calculation becomes:

16 Degrees Celcius:
Inverting Input: V- = (2000/4100) * 2 = 0.97 V
Non Inverting Input: V+ = (2000/4000) * 2 = 1 V

Since the potential applied at the non inverting input is greater than that applied at the inverting input, the output is positive (the op amp acts as a comparator) and the diode R is on while the diode G is off. The opposite process occurs for 18 degrees celcius, where the inverting input has a greater potential than the non inverting input and thus G is on while R is off.

Hope this helped!
Good Luck for all your exams!
 
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1357913579

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sagar65265 said:
The input potential to the inverting input of the op amp is the potential across resistor P, not across the thermistor T - the potential across T is the number of Volts lost when unit charge travels through T, and the potential into the inverting input is the remaining potential, which is the potential across the resistor R, not the thermistor T. So the calculation becomes:

16 Degrees Celcius:
Inverting Input: V- = (2000/4100) * 2 = 0.97 V
Non Inverting Input: V+ = (2000/4000) * 2 = 1 V

Since the potential applied at the non inverting input is greater than that applied at the inverting input, the output is positive (the op amp acts as a comparator) and the diode R is on while the diode G is off. The opposite process occurs for 18 degrees celcius, where the inverting input has a greater potential than the non inverting input and thus G is on while R is off.

Hope this helped!
Good Luck for all your exams!
Click to expand...
bro help required at this as wel please
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_41.pdf
number-4 please can u help iam totally confused about this question
thanks alot:)
iam only able to a(i) part nothing other than this iam totally 0 about how to solve the rest of the parts of this ques
 
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1357913579

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sagar65265 said:
Man, man , i'm sorry, there may not be any force on a stationary electron at that point and the net field strength may be zero, but the potential definitely isn't zero, man, i'm really sorry about that, Ashique was absolutely right - the potential energy of an electron in between the two positive charges will not be zero, it will be the sum of two systems, the electron and one of the charges and the electron and the other charge!

Just try your calculation by including the potential in the beginning, see if it works!

Good Luck for all your exams!
Click to expand...
thanks bro but iam still confused like if electric field is 0 isnt potential 0 can you please elaborate
 
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sagar65265

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MaxStudentALevel said:
Also another VERY small query, how to find Freq of information signal? On ON 2009 42 Q11 b (ii) ????
Click to expand...

Since the carrier wave is amplitude modulated with the information signal, the carrier wave will have an "envelope" that joins the crests of each wavelet on the carrier signal. The wavelength of the carrier wave is from one crest on this envelope to the next.
On the graph you can see that the first peak shown is the maximum height for that modulated wave, and that peak appears at t = 2.2 seconds or something (that isn't so important, you can take the part at time t = 0 and use the same zero displacement position for the next maximum). The next maximum amplitude of the carrier (and information) signal comes at t = 102.2 seconds or something of that sort. In other words, from one point on the envelope to the next point on the same envelope that is in phase with the first point is a time difference of 100 microseconds, i.e. 100 * 10^-6 seconds.

Since f = 1/T,

f = 1/(100 8 10^-6) = 10,000 Hz = 10 kHz.

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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sagar65265

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1357913579 said:
thanks bro but iam still confused like if electric field is 0 isnt potential 0 can you please elaborate
Click to expand...

Yahoo Answers is such a savior - I was doubting that statement right after I made it, but see if this explanation helps out:

http://answers.yahoo.com/question/index?qid=20120131161711AAekKKO

Think of it in this way - suppose the electron was being brought towards the two particles on the perpendicular bisector of the line joining them, i.e. along the line through the mid - point. There would be no net sideways pull, but there would still be a net forward pull, right? So this forward pull would do work, and to make sure the electron moves at a constant velocity, the work done on it by the external force (and thus the electric potential) would be negative. Meanwhile, if a proton was being brought into place there, it would be repelled along that same line and a positive amount of work would have to be done on it to get it there, so the potential would be positive!

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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1357913579 said:
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_41.pdf
number-4 please can u help iam totally confused about this question
thanks alot:)
iam only able to a(i) part nothing other than this iam totally 0 about how to solve the rest of the parts of this question
Click to expand...

Okay, so for a (ii), the basic statement is that both charges have the same sign, or the same polarity.
If both the charges have the same sign, they will both affect a charged particle in the same way, they will either both attract or both repel.
Taking a look at the graph, there is a point where the net field strength, and thus the net force (remember, E = Force per unit charge, so if the field strength at a point is zero, the force per unit charge is zero, and thus force on any charge at that point will be zero) will also be zero.

Imagine this - a positive charge is on the left, a negative charge is on the right. If you place a negative charge in the field, it will always be pushed away from the negative charge and towards the positive charge, so there will be no point where there is a zero net force! If you place a positive charge in the field at any point, the same conclusion can be arrived at - there is no point where the net force is zero!

So if the net force has to be zero at any poiny, the pull (or push) from the charge on the left has to be equal in magnitude and opposite in direction to the pull (or push) from the charge on the right, and in that case, the net force vectors add to zero. This can only happen if the two charges have the same polarity, or sign!

b) i) The relationship is that the Electric Field Strength at a point is the negative rate of change of the Electric Potential at a point with respect to time, i.e.

E = - Delta(Potential)/Delta(time)
The reason it is negative is that if the electric potential energy is increasing very fast, then the force on the particle will be a repulsive force, and since the field strength is a vector, it will be negative at this point, since the force is acting away from the charge causing the electric field.

ii) Since the rate of change of potential with respect to time is equal to -E (from the formula above), the rate of change of Electric Potential with respect to time is maximum when the field strength is minimum, which is at t = 11.4 cm on the graph, and the rate of change of potential is zero when the electric field strength is zero too, which is at around 8 cm or so on the graph (when the line touches the axis).

Hope this helped!
Good Luck for all your exams!
 
seneria

seneria

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blueberryyums said:
AL
ALRIGHT!!! I GOT THIS! (but not everything though)

2ai x x tick tick

x - because molecules are not moving therefore P.E is zero.

x - because K.E energy is increasing and P.E is decreasing

tick- because molecules are moving which means PE has increased / gained energy

tick - stretching involves moving the molecules therefore PE is increased.
Click to expand...

for the first option work is being done on the system , so there must be an increase in internal energy reight? this is confusing me.. as water is BEING freezed

however, for the third option the water is evaporating , meaning work is done BY the system... so internal energy must decrease... because increae in the internal energy is work done on the system and heat is supplied to the system....

can youclear my confusion?
 
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