ATAR Notes: Forum
Archived Discussion => 2010 => End-of-year exams => Exam Discussion => Victoria => Physics => Topic started by: DuBistKomisch on November 10, 2010, 02:09:24 pm
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Electric Power
1. Left through the inside of the solenoid and looping back around to the right on the outside.
2. Flux is zero, as the area is parallel to the magnetic field.
3. Downwards using the right hand rule (B left, I out of page).
4. F= BIln = 0.024 N
5. Force is zero as the current and magnetic field are parallel.
6. A; emf is negative time derivative of flux and slip rings collect AC.
7. Slip rings maintain contact between ends of loop and same terminal whereas a split-ring commutator alternates contact with each terminal each half-revolution. Slips rings for AC with generator/motor and split-ring for DC (expand though).
8. 0-1 below the axis, 1-2 on the axis, 2-4 above the axis but only half the magnitude of 0-1. Shouldn't matter whether you drew a line on the axis after 4 seconds or not.
9. Faraday's law (of induction).
10. 3.0V.
11. Q -> P using Lenz's law and right hand screw rule.
12. P = V^2/R = 120^2/48 = 1200W.
13. Connect in series to obtain 120^2/96 = 600W.
14. Power is lost but current must be the same throughout so voltage is less at the globe, hence less energy delivered and not as bright.
15. Current is 2.0A, voltage drop across lines is 2.0*4.0=8.0V, so 2.0+8.0=10V initially.
16. P_loss = I^2*R = 2.0^2*4.0=16W.
17. AC can be stepped up and down using transformers to reduce current and minimise power loss in lines.
18. D: 20.8*sqrt(2)*2 = 58.8V peak to peak.
19. 10:1 = 1460:x -> x=146 turns.
20. V2=2.0V and I2=2.0A, 10:1 leads to V1=20V and I2=0.2A, then P_loss=I^2*R=0.2^2*4.0=0.16W.
For questions 8 and 10 I am not sure how significant signs are. I am thinking they may accept any combination with each part opposite in sign since no reference polarity for the voltmeter was given.
For question 9, refer to the introduction of http://en.wikipedia.org/wiki/Lenz's_law and also http://en.wikipedia.org/wiki/Electromagnetic_induction#Technical_details.
Light and Matter
1. Young's double slit. Talk about diffraction and interference. Explain why only wave model explains these two phenomena.
2. Observation 2 is the only one that only the particle model supports. Explain how frequency = energy of packet and intensity = number of packets and the consequences.
3. E = hc/lambda = 2.14eV.
4. 2.5*580=1450nm.
5. D: intensity has no effect on threshold frequency or stopping voltage, only the current.
6. A: Magnesium has lower W, hence lower f_0 so its intercept on the horizontal axis will be to the left.
7. lambda = h/mv = 0.049nm.
8. A: greater velocity -> smaller wavelength -> smaller pattern.
9. Very similar wavelength -> very similar pattern.
10. Convering 600eV kinetic energy to joules and then finding the de Broglie wavelength, equating to the X-ray wavelength and finding E=hc/lambda=2.5*10^4eV.
11. E=hc/lambda=2.6eV which corresponds to n=4 (12.8) -> n=2 (10.2).
Photonics
1. C: Lasers are coherent so C was the only option with only incoherent sources.
2. C: Switch-on voltage of 2.0V indicates band gap of 2.0eV which corresponds to lambda=hc/E=621nm.
3. B: Voltage drop across LED is 2.0V so voltage drop across the 400 ohm resistor is 10V and current must be 10/400=25mA.
4. C: Blue light has a higher frequency and hence a blue LED must have a larger band gap energy. Therefore a larger voltage drop occurs over the LED and smaller over the resistor which means a smaller current through the circuit.
5. D: arcsin(1.38/1.44) = 73 degrees.
6. B: arcsin(sqrt(1.44^2-1.38^)) = 24 degrees.
7. B: When you bend it, the light is more likely to hit at an angle closer to the normal. Therefore angle of incidence becomes below the critical angle and is not reflected.
8. C: Water's refractive index is higher than the plastic so TIR cannot occur.
9. D: Absorption line is clearly higher on the graph at 2000nm.
10. C: Taking into account both lines (addition of ordinates), 1200nm had the lowest total attenuation. The only tricky question. :D
11. B: Graded index is an attempt to make the different modes arrive at the same time despite the different distances taken. Hence B modal dispersion.
12. C: Rayleigh scattering is caused by slight variations in the optical density in the core of an optical fibre caused by imperfections.
Anyone have reliable Sound or Synchrotron answers?
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FUCK FUCK FUCK i worked out the flux in question 2 by doing BA, i forgot it was parallel FUCK
... but that appears to be the only thing i've lost so far.
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I didn't halve the magnitude for question 8. I'd still get a mark though for putting 0-1 under and 2-4 over wouldn't I?
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Wait doesn't a smaller wavelength mean wider fringe spacing?
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1. Left through the inside of the solenoid and looping back around to the right on the outside.
Can you expand on this?
I had my lines going from the left of the page to the right of the page.
iTute says right to left =='
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I didn't halve the magnitude for question 8. I'd still get a mark though for putting 0-1 under and 2-4 over wouldn't I?
I don't remember how many marks it was worth, but I presume they'd look for the sign and the magnitudes to be correct, so you would get some marks but not full.
Wait doesn't a smaller wavelength mean wider fringe spacing?
Distance between fringes is proportional to wavelength.
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The south pole would have been on the right using the right hand grip rule, meaning the the lines through the centre would go from the right to the left.
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1. Left through the inside of the solenoid and looping back around to the right on the outside.
Can you expand on this?
I had my lines going from the left of the page to the right of the page.
iTute says right to left =='
The current comes up the front from the right side of the solenoid. Using the right hand screw rule, the magnetic field through the solenoid goes from the right end to the left end, then out and back around from left to right.
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I didn't halve the magnitude for question 8. I'd still get a mark though for putting 0-1 under and 2-4 over wouldn't I?
i uh, didn't do that either *cries*
this is getting worse and worse, i shouldn't have come here to look at the solutions, it's just making me depressed now.
That's at least 3 marks gone already.
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FUCK FUCK FUCK i worked out the flux in question 2 by doing BA, i forgot it was parallel FUCK
... but that appears to be the only thing i've lost so far.
dw i did this too, so far only 3 marks lost
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for 10 I pretty sure voltage was -3V (negative)
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I agree with everythign but light and matter Q2. I think its option 2, because I saw in a trial paper solution I did once that "intensity proportional to photoelectric current" is an explanation that is supported by both wave and particle model.
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For light and matter question 2, I thought that only 2 was acceptable - as the wave theory does stipulate that photo current depends on the intensity of incident light. . .
but it CANNOT explain why Ek is solely dependent on frequency. . .
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for 10 I pretty sure voltage was -3V (negative)
Hmm, good point. I'm not sure if the sign should really matter though since we're not told the polarity of the voltmeter. In fact I don't remember being told whether it is measured at the voltmeter at all...
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isn't question 2 of Light observation 2???
cause wave model suggests that higher intensity means higher energy, which isn't what the observation suggested, it suggested that intensity did not affect the energy emitted...only frequency.
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Question 10 in light was bloody hard. . . needless say, I figured out the answer in the last 2 minutes!!! phew, , ,
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I agree with everythign but light and matter Q2. I think its option 2, because I saw in a trial paper solution I did once that "intensity proportional to photoelectric current" is an explanation that is supported by both wave and particle model.
The question clearly stated that the wave model could account for 2 of the observations but not a third. Clearly then they only have one correct answer in mind.
I agree that it was 2.
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I agree with everythign but light and matter Q2. I think its option 2, because I saw in a trial paper solution I did once that "intensity proportional to photoelectric current" is an explanation that is supported by both wave and particle model.
For light and matter question 2, I thought that only 2 was acceptable - as the wave theory does stipulate that photo current depends on the intensity of incident light. . .
but it CANNOT explain why Ek is solely dependent on frequency. . .
I guess the "only" in the second observation makes a difference. The explanation you would use is the same for each though so I guess I'll lose a mark or two.
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for 10 I pretty sure voltage was -3V (negative)
Hmm, good point. I'm not sure if the sign should really matter though since we're not told the polarity of the voltmeter. In fact I don't remember being told whether it is measured at the voltmeter at all...
Voltage is a vector - because Flux is a vector (and Field is a vector)
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I think it may be Lenz's law rather than faradays. Faraday's would not imply any direction, Lenz's law takes into account the fact that voltage is above and below the axis.
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I think it may be Lenz's law rather than faradays. Faraday's would not imply any direction, Lenz's law takes into account the fact that voltage is above and below the axis.
Wrong. Faraday's Law includes the negative sign which gives the sign of the induced emf. Lenz's law is for determining the direction of the current.
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I think it may be Lenz's law rather than faradays. Faraday's would not imply any direction, Lenz's law takes into account the fact that voltage is above and below the axis.
According to Heineman Faraday's Law is the negative rate of change of flux, so it does predict the signs.
Lenz's law does not predict the magnitude of the voltage, which was a feature of the graph
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Only Faraday's law applied to that case. .
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for question 8 it did not state which polarity the voltmeter was in i drew above axis for 0-1 and half magnitude under for 2-4 i understand what you mean now partial marks u think??
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yeah you'll probably get partial marks.
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for question 8 it did not state which polarity the voltmeter was in i drew above axis for 0-1 and half magnitude under for 2-4 i understand what you mean now partial marks u think??
I'm starting to think it was pretty ambiguous as we were given no reference polarity for the voltmeter so they may accept either combination of signs as long as they are opposite. Don't take my word for it though.
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for question 8 it did not state which polarity the voltmeter was in i drew above axis for 0-1 and half magnitude under for 2-4 i understand what you mean now partial marks u think??
I'm starting to think it was pretty ambiguous as were given no reference polarity for the voltmeter so they may accept either combination of signs as long as they are opposite. Don't take my word for it though.
This is what I think too.
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for question 8 it did not state which polarity the voltmeter was in i drew above axis for 0-1 and half magnitude under for 2-4 i understand what you mean now partial marks u think??
I'm starting to think it was pretty ambiguous as were given no reference polarity for the voltmeter so they may accept either combination of signs as long as they are opposite. Don't take my word for it though.
Agreed, if they don't, they'll have to accept Lenz's Law for the following question as it is used to take into account polarity. Seems more correct imo to accept either for the graph than the law.
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i think u might be right but u never know i even stated at the top assuming positive voltage is p-q lol oh well we shall see
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=/ I was thinking about whehter you had to put the first section (t=0 to t=1) above or underneath the x-axis. I came to the conclusion that ti doesn't matter. You don't have a power source with a positive/ngative terminal in the diagram, therefore you can't tell which way si positive current and which way is negative.
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I think it may be Lenz's law rather than faradays. Faraday's would not imply any direction, Lenz's law takes into account the fact that voltage is above and below the axis.
I said Lenz's. I was under the impression that Faraday's was emf=d(phi)/dt, whereas Lenz's was the same except with the addition of a negative sign (emf=-d(phi)/dt). Can someone please clarify with an explanation?
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=/ I was thinking about whehter you had to put the first section (t=0 to t=1) above or underneath the x-axis. I came to the conclusion that ti doesn't matter. You don't have a power source with a positive/ngative terminal in the diagram, therefore you can't tell which way si positive current and which way is negative.
I ended up labeling the first emf as qV, where q is an arbitrary value(as they didn't actually give any values for flux) and the final section as -q/2 V.
Because they didn't give any indication of the voltmeters connection. So even though you know which direction the emf will be in you don't know what will come up on the meter.
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=/ I was thinking about whehter you had to put the first section (t=0 to t=1) above or underneath the x-axis. I came to the conclusion that ti doesn't matter. You don't have a power source with a positive/ngative terminal in the diagram, therefore you can't tell which way si positive current and which way is negative.
I wrote a note saying that I took X to Y (I drew it on the diagram) as positive.
I think it may be Lenz's law rather than faradays. Faraday's would not imply any direction, Lenz's law takes into account the fact that voltage is above and below the axis.
I said Lenz's. I was under the impression that Faraday's was emf=d(phi)/dt, whereas Lenz's was the same except with the addition of a negative sign (emf=-d(phi)/dt). Can someone please clarify with an explanation?
Lenz's law IS the negative sign in Faraday's equation. Lenz's law is about the polarity rather than the magnitude.
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=/ I was thinking about whehter you had to put the first section (t=0 to t=1) above or underneath the x-axis. I came to the conclusion that ti doesn't matter. You don't have a power source with a positive/ngative terminal in the diagram, therefore you can't tell which way si positive current and which way is negative.
I ended up labeling the first emf as qV, where q is an arbitrary value(as they didn't actually give any values for flux) and the final section as -q/2 V.
xD xD I just put 2k and -k
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I think it may be Lenz's law rather than faradays. Faraday's would not imply any direction, Lenz's law takes into account the fact that voltage is above and below the axis.
I said Lenz's. I was under the impression that Faraday's was emf=d(phi)/dt, whereas Lenz's was the same except with the addition of a negative sign (emf=-d(phi)/dt). Can someone please clarify with an explanation?
Faraday’s law of induction states that the magnitude of the induced emf is equal to the time derivative of flux. When direction is taken into consideration, the sign of the emf is negative:
E=-(∆Φ_B)/∆t
Lenz’s law states that the direction of the induced current is such that the magnetic field produced will be in the opposite direction to the change in flux that induced it.
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Also, http://en.wikipedia.org/wiki/Lenz's_law says:
Lenz's law is a common way of understanding how electromagnetic circuits must always obey Newton's third law. Lenz's law is named after Heinrich Lenz, and it says:
"An induced current is always in such a direction as to oppose the motion or change causing it"
Faraday's law of induction indicates that the induced electromotive force (emf) and the change in flux have opposite signs, and it also gives the direction of the induced (emf) and current resulting from electromagnetic induction.
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for 10 I pretty sure voltage was -3V (negative)
Hmm, good point. I'm not sure if the sign should really matter though since we're not told the polarity of the voltmeter. In fact I don't remember being told whether it is measured at the voltmeter at all...
Doesn't EMF have a modulus sign somewhere that makes it positive?
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for 10 I pretty sure voltage was -3V (negative)
Hmm, good point. I'm not sure if the sign should really matter though since we're not told the polarity of the voltmeter. In fact I don't remember being told whether it is measured at the voltmeter at all...
Doesn't EMF have a modulus sign somewhere that makes it positive?
EMF is a vector quantity - if they were to ask for the magnitude of voltage, then ye, you would mod it
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PEOPLE! did u have to label the voltage values? :O
dammit... i didn't do it, i just drew one voltage larger than the other ><
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Hey I heard someone really awesome sat afront of you in the exam Mr. DBK :smitten:
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Hey I heard someone really awesome sat afront of you in the exam Mr. DBK :smitten:
Haha, but he was too Iraqi for me and kept looking at my multichoice answers after the exam. :P
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PEOPLE! did u have to label the voltage values? :O
dammit... i didn't do it, i just drew one voltage larger than the other ><
There was no way to label voltage. They didn't give figures for the Area of the coil or field. . .
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PEOPLE! did u have to label the voltage values? :O
dammit... i didn't do it, i just drew one voltage larger than the other ><
There was no way to label voltage. They didn't give figures for the Area of the coil or field. . .
phewww lols
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Haha, I just gave arbitrary units to make it perfectly clear that the second emf = -0.5 the first. kyzoo did this too. But you didn't need it.
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Hey I heard someone really awesome sat afront of you in the exam Mr. DBK :smitten:
Haha, but he was too Iraqi for me and kept looking at my multichoice answers after the exam. :P
Ahaha Aye I was just checking to see if I got it all right! And i think I did woo! :D
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Hey I heard someone really awesome sat afront of you in the exam Mr. DBK :smitten:
Haha, but he was too Iraqi for me and kept looking at my multichoice answers after the exam. :P
Ahaha Aye I was just checking to see if I got it all right! And i think I did woo! :D
Good work <3
Haha, I just gave arbitrary units to make it perfectly clear that the second emf = -0.5 the first. kyzoo did this too. But you didn't need it.
Yeah, I didn't think it was overly necessary; I just made the distances as obviously 2:1 as I could.
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Jake, when is the new Pattomobile coming out?
Just kidding. <3