Post by: Bozo on May 25, 2011, 09:40:05 pm
Would anybody have a list or summary of what features a graph must have to suit a specific characteristic of a material?
Cheers 8-)
Post by: Plan-B on May 25, 2011, 10:57:24 pm
Toughness is how much energy a material can absorb before fracture. Therefore if comparing materials, the material that possesses the largest area under it is the toughest. As the energy it can absorb is equal to the area under the graph multiplied by volume.
Brittleness: How much strain a material can withstand before fracture. They are considered brittle when they fracture shortly after reaching it's elastic limit. Therefore brittle materials generally have a linear line.
Ductility: How much plastic deformation before fracture. Materials are considered ductile if after it's elastic limit (linear region), it continues without fracture in non-linear ways for extensive values of strain. Sooo the line looks long :P
Stiffness: How much force it takes to change the shape of an object. This is the gradient of the line (a.k.a Young's Modulus). Soo the steeper the gradient, the stiffer the material is.
Strength: The amount of stress it can withstand before failure. Therefore a strong material in comparison would be the line that reaches the highest point on the graph.
Ehh, that's how i understand it :)
Post by: Bozo on May 26, 2011, 06:00:39 pm
Another question. Has anyone here done a neap diagnostic test for electronics and photonics?
Post by: Shark 774 on May 26, 2011, 07:48:31 pm
Thanks man.
Another question. Has anyone here done a neap diagnostic test for electronics and photonics?
No I haven't, do you have a copy of one?
Post by: cranberry on May 26, 2011, 08:01:19 pm
is this correct?
Post by: Bozo on May 26, 2011, 09:01:23 pm
Post by: soopertaco on May 27, 2011, 07:01:27 pm
On a company exam i did, an answer was tensile strength = stress at elastic limit on stress-strain graph, not @ the max stress...
is this correct?
I'm pretty sure tensile strength of a material should be equivalent to its 'ultimate strength' which refers to the max value of stress regardless of whether it is in the elastic region or plastic region unless of course the material fractured at the elastic limit??
Post by: cranberry on May 27, 2011, 07:27:41 pm
On a company exam i did, an answer was tensile strength = stress at elastic limit on stress-strain graph, not @ the max stress...
is this correct?
I'm pretty sure tensile strength of a material should be equivalent to its 'ultimate strength' which refers to the max value of stress regardless of whether it is in the elastic region or plastic region unless of course the material fractured at the elastic limit??
thats what i thought too....probs a mistake, but can anyone confirm?
Post by: Bozo on June 03, 2011, 10:35:51 pm
What would this Vout wave look like?
Thanks!
Post by: b^3 on June 03, 2011, 10:57:30 pm
Post by: Bozo on June 03, 2011, 11:35:00 pm
Post by: b^3 on June 03, 2011, 11:44:49 pm
Post by: Bozo on June 04, 2011, 04:06:52 pm
if so what would it look like.
Post by: b^3 on June 04, 2011, 05:07:47 pm
Edit: Sorry I forgot to shrink the image file down, so save the image if you cannot see it properly.
Post by: Bozo on June 04, 2011, 05:59:56 pm
Post by: Bozo on June 04, 2011, 09:12:47 pm
Post by: cohen on June 04, 2011, 09:16:30 pm
My teacher said it'll be ok, but If i get a chance tomorrow at lunch, or during the breaks, I'll ask the lecturers :)
Post by: onur369 on June 04, 2011, 09:18:17 pm
Post by: Whatlol on June 04, 2011, 09:45:13 pm
Does anyone know anything about using pre transposed formulae in the exam. If anyone is going to the TSFX lecture tomorrow could they find out. Because for e.g i came across a 3 mark question the other day for banked corners. And i had a pre transposed formula for it which was a simple sub in and bang the answer. I just want to make sure i don't get penalised for not putting enough working out.
yes that should be allowed ,and as quoted from the vcaa examiners report :
"The required angle was 5.70.
There were a number of correct ways students answered this question. Some students had a standard formula involving
tanθ on their A4 sheets. Others did vector diagrams with weight, normal and resultant centripetal force to form a rightangled
triangle "
So i think that implies it is acceptable. But if your really concerned just do the vector diagram its not too hard and you can even just write it on your cheat sheet if your worried.
Post by: Vincezor on June 04, 2011, 09:59:22 pm
Does anyone know anything about using pre transposed formulae in the exam. If anyone is going to the TSFX lecture tomorrow could they find out. Because for e.g i came across a 3 mark question the other day for banked corners. And i had a pre transposed formula for it which was a simple sub in and bang the answer. I just want to make sure i don't get penalised for not putting enough working out.
I personally would only use a pre-transposed formula to check my final answer (not including gravity :P). I'm not sure what would happen if you used a pre-transposed formula and got the answer wrong though... 0/3? :S
Post by: Bozo on June 04, 2011, 10:43:07 pm
Post by: cohen on June 04, 2011, 10:52:31 pm
Post by: Bozo on June 04, 2011, 11:18:06 pm
Post by: Bozo on June 05, 2011, 12:29:32 am
Post by: examsman on June 05, 2011, 10:42:16 am
The answer is C?
You have to use torques about ankle. Thus finding the force that is perpendicular to the foot from the normal reaction force of the toe. then, you find the perpendicular force acting at the heel, and use trig to find tension in the achilles.
If you want further explanation ring me or come my house later today.
-t.ztila
Post by: Bozo on June 05, 2011, 12:52:44 pm
Post by: cohen on June 05, 2011, 08:03:24 pm
1) You are able to use pre-transposed formulae (He even gave us a list of some projectile motion formulae, and energy per unit volume equations)
2) Significant figures don't matter that much in Physics, but try and stay within the least amount of numbers that you use in your calculations.
Do you want me to make a thread with all of the pre-transposed formulae that we were given, and the ones i got from my teacher? (My teacher has given me quite a few of these formulas ^.^)
Post by: jinny1 on June 05, 2011, 08:31:06 pm
Ok, i got back an hour ago or so, and here's what geoff said at the start
1) You are able to use pre-transposed formulae (He even gave us a list of some projectile motion formulae, and energy per unit volume equations)
2) Significant figures don't matter that much in Physics, but try and stay within the least amount of numbers that you use in your calculations.
Do you want me to make a thread with all of the pre-transposed formulae that we were given, and the ones i got from my teacher? (My teacher has given me quite a few of these formulas ^.^)
yes please! :)
Post by: xZero on June 05, 2011, 08:46:05 pm
Post by: Bozo on June 06, 2011, 11:55:38 am
Post by: Bozo on September 04, 2011, 10:14:55 pm
Understanding that with transmission lines, to reduce power loss the voltage is "stepped up" henceforth decreasing the current and decreasing the power loss as modeled by P=I^2*R. But why is it that according to ohms law (V=IR) that when voltage increases current also increases with it. Does anyone have an explanation to this contradiction?
Post by: b^3 on September 04, 2011, 10:21:07 pm
When it is being stepped up, the resistance of the wire has nothing to do with it as the power has to stay the same, hence P1=P2, V1I1=V2I2
With the power loss, we cannot use the voltage drop in the wire unless it is stated, so we must use I and R, so Ploss=I2R
Post by: Bozo on September 04, 2011, 10:34:25 pm
Post by: b^3 on September 04, 2011, 10:41:02 pm
That still doesn't answer the question, cause Voltage step up reduces current, but ohms law says the opposite. And your assuming resistance is fixed in both situations?Ok think of it this way, when dealing with stepping up and stepping down, we can throw the resistance of the wire out of the window, since it won't make a difference, we have to keep power of the transformer the same. We can't apply ohms law to this here because we can't apply the resistance of the transformer. The transformer (ideal) won't have a voltage drop and so we can't use ohm's law. Someone check this please.
Post by: Bozo on September 04, 2011, 10:41:46 pm
Post by: b^3 on September 04, 2011, 10:45:20 pm
Post by: xZero on September 04, 2011, 10:47:04 pm
Post by: Bozo on September 04, 2011, 10:48:10 pm
Post by: Bozo on September 04, 2011, 10:49:41 pm
Post by: b^3 on September 04, 2011, 10:51:27 pm
in this case we assume P is constant, so P=I^2R and P=VI are both constant right? we have the equation V=IR, lets sub I=V/R into the power equation so P=V^2/R. What this equation is saying that as we increase voltage, resistance of the transmitting line will also increase(the rate of increase is higher than voltage). Now back to V=IR, if v increase and r increase at a faster rate, I must decrease to balance things out. what does this mean? it means that the resistance of the transmitted line must be very large if we want to step the voltage up by heapsThat makes it a lot clearer.
Post by: xZero on September 04, 2011, 10:55:18 pm
Post by: Bozo on September 08, 2011, 08:48:24 pm
Post by: cranberry on September 09, 2011, 10:25:00 am
Post by: Bozo on September 12, 2011, 07:00:23 pm
Post by: Bozo on September 16, 2011, 07:51:48 am
Post by: b^3 on September 16, 2011, 08:16:27 am
1. Download http://www.cutepdf.com/ and install
2. Open the pdf you.
3. Click print and change the printer to cutePDF writer and put in the pages you want, click print
4. When the dialog box comes up select where you want to save it.
Post by: Lasercookie on September 17, 2011, 01:45:12 pm
Does anyone here, know how to edit pdfs to get rid of the detailed studies your not doing. Cause I want to go to officeworks to mass print all my exams, but like 1/3 of the stuff is redundant paper that will be charging me more.Since I print at home or school, I can print using Adobe Reader. I just take a look at the pages that I need and just print off those ones (you can restrict pages in the print dialog).
I don't know what officeworks is like.
You could use something like CutePDF and just print out the pages you need into a new PDF file. Or you could use Adobe Acrobat and actually delete the pages, but that would depend on if the file is unsecured. the CutePDF method is probably easier (CutePDF is a virtual printer that allows you to save files as PDF).
edit: Didn't notice b^3's post. I described the same thing that he did.
Post by: Bozo on September 17, 2011, 03:39:00 pm
Post by: Bozo on September 19, 2011, 11:38:58 am
Post by: Bozo on September 27, 2011, 11:15:21 am
Post by: cranberry on September 27, 2011, 01:08:19 pm
Post by: cranberry on September 27, 2011, 01:12:53 pm
Look at their response, shouldn't red light be more spread out? Or am I missing something?...
Post by: HarveyD on September 27, 2011, 04:47:41 pm
Post by: cranberry on September 27, 2011, 05:37:46 pm
Post by: Lasercookie on September 27, 2011, 05:52:17 pm
Post by: HarveyD on September 27, 2011, 06:16:12 pm
how did you do this one
Post by: Lasercookie on September 27, 2011, 06:22:17 pm
Follow the current along and pick one of the lines. Right hand grip rule.
I picked the right-most line, so current (thumb was pointing up).
Fingers give the direction of the field, and they are pointing left.
So answer is Z.
Post by: HarveyD on September 27, 2011, 06:46:20 pm
Post by: Lasercookie on September 27, 2011, 07:02:49 pm
Post by: cranberry on September 28, 2011, 10:20:10 am
The heineman diagram actually shows the "S" and "N" with the cool arrows which tell you if its clockwise or anticlockwise.
Post by: Lasercookie on September 28, 2011, 10:59:47 am
The diagram is right for both. Current is going in the opposite direction though. You can use the grip rule, but it's much easier to determine which pole is south and north (south - clockwise, north - anticlockwise). Then you know that field goes from south to north inside the magent and north to south outside.thanks for that clarification.
The heineman diagram actually shows the "S" and "N" with the cool arrows which tell you if its clockwise or anticlockwise.
Post by: HarveyD on September 28, 2011, 02:21:55 pm
Post by: Lasercookie on September 28, 2011, 02:31:47 pm
wait I'm confused, how would you do the question then? :SIf you want to do it the method that cranberry described:
Picture it the solenoid side on.
e.g. The X side, the coil starts to coil in clockwise. Therefore that is the South pole on the outside. Therefore, from the Z side is the north pole on the outside.
On the inside it would be opposite:
X is the north pole. Z is the south pole.
The direction of the field is described to be from the North pole to South pole. So the field goes X to Z, or in other words leftwards.
This is the Z arrow.
Personally, I find the right hand grip rule to be easier. You should get the same answer using both methods of working out.
Post by: HarveyD on September 28, 2011, 02:34:12 pm
how come the right hand rule doesnt work for the heinemann diagram?
Post by: Lasercookie on September 28, 2011, 03:09:50 pm
ahh okayI just tried it with right hand rule. It does work. I think when I tried it earlier, I forgot that inside the solenoid: South to North and outside the solenoid: North to South. This makes sense if you look at the field lines.
how come the right hand rule doesnt work for the heinemann diagram?
I assumed that it would be north to south, which is what gave me the incorrect answer. But that assumption is what I made when trying that Lisachem question using the "pole method". Unless I determined the poles incorrectly. My mistake may have been in what I initially determined (X is clockwise, therefore south) and how I assumed that would be the outside pole.
I have no idea what I meant by outside or inside pole.
Let me restate it:
X is the south pole. Therefore Z is the north pole.
We are looking for the direction of the field inside. The direction inside will be south to north.
Therefore X to Z.
That makes sense now (I think).
So it seems by using the right hand rule, you figure out the middle field line i.e. the inside field. I looked over any questions I did similar to this one, and that seems to be what I've done - I just never thought about it in this way explicitly. That's where I was going wrong with the Heinemann diagram (assuming that what I've said in this post is all correct), thinking about this stuff in the wrong way (but coincidentally ending up with the correct answers when doing questions).
Post by: Bozo on September 28, 2011, 08:46:24 pm
Post by: Lasercookie on September 28, 2011, 10:44:15 pm
How many prac exams you guys done?A grand total of two so far.
Post by: Bozo on September 28, 2011, 11:12:05 pm
Post by: Bozo on September 28, 2011, 11:13:46 pm
I've probably opened my heinemann text book, once this year.
Post by: Lasercookie on September 28, 2011, 11:18:21 pm
Do you guys actually use the text book?Yeah, I've used the textbooks pretty heavily this year.
I've probably opened my heinemann text book, once this year.
My teacher set the Heinemann exercises as a work requirement, so I had to do them to pass.
I've read over the Heinemann and Nelson book a fair bit.
I finished the questions from Nelson a while back as well.
Now that I'm finished with that, I really only read over iTute summaries, Wikipedia, Feynman (most of it is irrelevant and beyond the course, but it's damn interesting reading and well explained) and the Nelson textbook.
Post by: Bozo on September 28, 2011, 11:19:50 pm
Post by: Lasercookie on September 28, 2011, 11:22:10 pm
I've basically been living off the A+ notes the whole year. As for work requirements, we get set the checkpoint chapters. So all that needs to be completed.I borrowed the checkpoints book from the school library earlier this term. I did some of them, but I hated it. The questions are so dull, repetitive and boring.
I've never seen the A+ notes, but they must be good if they're detailed enough for you to learn the course off them.
Post by: HarveyD on October 01, 2011, 01:05:55 pm
Post by: Bozo on October 01, 2011, 07:48:08 pm
Post by: HarveyD on October 01, 2011, 07:52:20 pm
16) 3.9 × 10^–5 V
17) Anti-clockwise
Post by: Bozo on October 01, 2011, 07:57:59 pm
As for 17, imagine the ring dropping veritcally infront of her cutting the magnetic field. Since the magnetic field lines are running from the magnet i.e towards her (head on), use your right hand grip rule and pretend your the teacher standing there. The field lines are going to be running into you so your thumb will be facing the right direction ( i.e fist upside down). Hence the current will flow anti clockwise. If that doesn't make sense i'll draw a diagram.
Post by: HarveyD on October 01, 2011, 07:59:08 pm
Post by: Bozo on October 01, 2011, 08:06:04 pm
I can't draw 3D, so yeah hopefully this will make you understand.
Imagine as if the hand is horizontal matching the field lines exactly.
Post by: HarveyD on October 01, 2011, 08:08:28 pm
Post by: Bozo on October 01, 2011, 08:09:12 pm
Post by: HarveyD on October 03, 2011, 12:32:44 am
with this one:
wouldnt the brightness increase since theres more current running through it ---> More power?
Post by: Bozo on October 03, 2011, 10:33:05 am
Post by: HarveyD on October 03, 2011, 12:37:36 pm
Post by: Bozo on October 05, 2011, 05:02:34 pm
Am I missing something?
They say the answer is B =S
Post by: Bozo on October 05, 2011, 05:22:34 pm
Because for on question i was given the energy and momentum of X-ray photons. Then i was asked what the wavelength of the X-rays was, so i went straight to lamda=h/p (using the momentum), but in the solutions they used the energy and it gives an answer that varies by .37???
Post by: Lasercookie on October 05, 2011, 05:32:33 pm
Another question, is there a difference in the actual value of a de Broglie wavelength and a normal wavelength?No there isn't any difference. What I think is happening is that trial exams give us rounded off values sometimes which leads to the small differences. That's my theory anyway.
Because for on question i was given the energy and momentum of X-ray photons. Then i was asked what the wavelength of the X-rays was, so i went straight to lamda=h/p (using the momentum), but in the solutions they used the energy and it gives an answer that varies by .37???
Post by: Bozo on October 05, 2011, 05:41:34 pm
But .35 is a huge difference.
Post by: Lasercookie on October 05, 2011, 06:11:09 pm
Just if you didn't see, look at 2 posts up, I had another question.Yeah, but if you calculate stuff yourself (and not round off), you'll find that both methods give the same answer.
But .35 is a huge difference.
The two equations can actually be derived from one another:
E=pc is just e=mc^2=mc*c, mc is momentum so e=pc.(since we're talking about photons)
If you did relativity then you would be familiar with
Which gets us:
Compton's equation:
de Broglie:
I've got further proof that the momentum value supplied by STAV is incorrect.
So the energy is
Now plug this into the debroglie equation. What do you get? The same answer for both methods.
Just speculating here: but in STAV's defence, they might have used a more accurate value for the speed of light. Perhaps that's what gave us the different answers.
Post by: Lasercookie on October 05, 2011, 06:17:50 pm
A compass needle points towards the south pole. (the north end of a compass needle is north).
So that's what it would point up, towards the southerly field lines.
More on the momentum/energy, it could be the energy calculation that's wrong (you can go figure it out based on the momentum) - either way I'm sure that one of the values provided in the question is wrong.
Post by: Bozo on October 05, 2011, 07:02:31 pm
BTW what did you get for this exam, i thought it was fairly easy (STAV 08)
Post by: Lasercookie on October 05, 2011, 07:08:40 pm
BTW what did you get for this exam, i thought it was fairly easy (STAV 08)I haven't done it :P - I've only done a few VCAA, a few iTute and Neap 2010. Oh and STAV 2011 as well.
I found those all fairly easy, though I think it's because I've enjoyed Unit 4 more than Unit 3 (which was probably partly because I'm not pressuring myself in the back of my mind of getting a 45+ score anymore).
Post by: Bozo on October 06, 2011, 01:05:50 am
Post by: paulsterio on October 06, 2011, 01:09:44 am
Slip Rings - are used to maintain contact with the coils in an AC Generator. They are used instead of connecting the wires directly to the coils to prevent tangling as the coil spins. Thus, the slip rings act merely as connections to the coils, connecting the coils to the circuit like a simple wire would.
Split Ring Commutators - change the current direction every half cycle, hence they are used in DC motors to change the direction of the force couples every half cycle, causing the motor to rotate in one direction. In electricity generation, they are used to reverse half of an AC signal, hence, acting like a bridge rectifier and producing unsmooth DC Voltage.
Post by: HarveyD on October 06, 2011, 03:39:31 pm
Post by: Bozo on October 06, 2011, 04:21:50 pm
Post by: Bozo on October 06, 2011, 06:19:23 pm
Post by: HarveyD on October 07, 2011, 01:05:06 pm
so yeah, found it quite good as well :P
Post by: Lasercookie on October 07, 2011, 09:09:55 pm
Looks like it's time for me to revise sound
Post by: Bozo on October 08, 2011, 01:00:58 am
Post by: paulsterio on October 09, 2011, 09:55:07 pm
Post by: Bozo on October 10, 2011, 12:44:31 am
Post by: Bozo on October 12, 2011, 09:30:47 pm
Post by: Bozo on October 22, 2011, 08:28:34 pm
Post by: Lasercookie on October 22, 2011, 08:41:13 pm
So how's everyone going thus far?Finished with all the VCAA ones now, did pretty damn well on most of those. Focusing on non-VCAA exams now. I'm making more silly errors on these than I'd prefer :'(
I'm going to have to sit down and figure out why I keep making these mistakes.
Something like 3 weeks and five days left until the exam :|
You, Bozo? (no pun intended)
Post by: Bozo on October 23, 2011, 07:31:56 pm
Post by: cranberry on October 24, 2011, 07:41:07 pm
Im 99.99967% sure that the left side is the south pole as current flows clockwise (from the positive terminal)!?
Any one else agree? 8)
Post by: Prizate on October 24, 2011, 07:58:31 pm
Post by: b^3 on October 24, 2011, 08:17:57 pm
I think I may be wrong, but it kinda makes sense. Anyone got anything to add (laserred)?
Post by: xZero on October 24, 2011, 08:32:09 pm
I've got a feeling what I'm about to say is wrong, but it's something to think about. The magnetic field that is produced by the current carrying wire will be north Q, to P south. But wouldn't this field cause a change in flux in the electromagnet, using lenz's law the electromagnet will try to oppose this change in flux producing a magnetic field in the opposite direction? So the field that the elctromagnet is expirencing is from north P to Q south?
I think I may be wrong, but it kinda makes sense. Anyone got anything to add (laserred)?
lets suppose that you're right, but after like 0.1 sec of switching the current on, there won't be any change in flux so the 'magnetic field' thats trying to oppose the change will disappear and the only magnetic field left would be from the current
Post by: b^3 on October 24, 2011, 08:35:16 pm
Yeh I thought of that about 5 mins after I posted but was trying to make the stupid question work. How would you go about ansering it xZero? Do you think they are wrong or how do you explain electromagnets (this is an area I'm hazy on and need to work on)I've got a feeling what I'm about to say is wrong, but it's something to think about. The magnetic field that is produced by the current carrying wire will be north Q, to P south. But wouldn't this field cause a change in flux in the electromagnet, using lenz's law the electromagnet will try to oppose this change in flux producing a magnetic field in the opposite direction? So the field that the elctromagnet is expirencing is from north P to Q south?
I think I may be wrong, but it kinda makes sense. Anyone got anything to add (laserred)?
lets suppose that you're right, but after like 0.1 sec of switching the current on, there won't be any change in flux so the 'magnetic field' thats trying to oppose the change will disappear and the only magnetic field left would be from the current
Post by: Bozo on October 24, 2011, 08:41:50 pm
Post by: paulsterio on October 24, 2011, 08:43:09 pm
Post by: b^3 on October 24, 2011, 08:45:47 pm
Post by: xZero on October 24, 2011, 08:48:59 pm
Post by: paulsterio on October 24, 2011, 08:51:48 pm
Yeh now that I realise it's insight, they are probably wrong. I hate the way the whole year 12 physics is set up, it's like nothing is set in stone and changes depending on the textbook you read or the exam you pick up. Can't wait to finish VCE, Only one day to go!
Off topic, but yeah, I agree with you, plus none of the VCE Physics textbooks seem to contain all the material you actually need in order to do exams! D:
Most of them don't even have all the formulas and somehow you're given these massive paragraphs of text that are completely irrelevant to the course and are just beating around the bush
Makes studying for Physics hard, I just use iTute and my teacher's notes now
Post by: Lasercookie on October 24, 2011, 08:54:29 pm
They've gone the fingers is the direction of the wires (clockwise/anticlockwise) and then the thumb indicates north. I learnt this one at the Connect lecture - personally it confuses me because I keep thinking that the thumb becomes the direction of the field (I prefer to stick with current = thumb, fingers=field). I think Insight made this mistake - they forgot that the field direction on the inside is North to South.
That's my guess anyway.
Using the way I would do the question - I'd pick one line of wire. I then figure out the current direction - "down" in this case. Then I use the RH Grip rule (thumb = current) and use that to figure out the field. Which is P to Q.
Post by: b^3 on October 24, 2011, 08:55:11 pm
yeah imma say that its wrong, and b^3, just wait till uni physics, you'll be crying so hard that you wish you're still doing VCE physics :P (although i dont miss vce maths at all)I'm hoping you're not saying that uni physics is more ambiguous than VCE physics :(. I'll have engineering maths to keep me sane then :).
Post by: xZero on October 24, 2011, 09:12:37 pm
well its not ambiguous at all, infact the exam is so god damn hard that they had to scale everyone up, almost everyone failed mid sem test, 8am start for first years and the fact that you have to learn 4 different topics in a semister from shitty lecturers (susan... pray that you dont have her) makes VCE physics much more appealing.yeah imma say that its wrong, and b^3, just wait till uni physics, you'll be crying so hard that you wish you're still doing VCE physics :P (although i dont miss vce maths at all)I'm hoping you're not saying that uni physics is more ambiguous than VCE physics :(. I'll have engineering maths to keep me sane then :).
Post by: Bozo on October 24, 2011, 11:04:32 pm
Post by: xZero on October 24, 2011, 11:18:21 pm
Post by: cranberry on October 25, 2011, 07:12:45 pm
Post by: Bozo on October 27, 2011, 07:22:04 am
Post by: HarveyD on October 28, 2011, 04:45:43 pm
do you guys think that the graph given for the induced emf with the frequency halved is wrong?
doesnt really look like the period has doubled...
Post by: paulsterio on October 28, 2011, 04:48:44 pm
Post by: HarveyD on October 28, 2011, 05:04:16 pm
typical
Post by: HarveyD on October 29, 2011, 01:07:02 am
but any ideas on how to do this one?
Post by: Lasercookie on October 29, 2011, 10:40:23 am
(sorry to hijack thread)Well the electrons in the atom are being bombarded with energy. I presume it's not that different from being bombarded with photons.
but any ideas on how to do this one?
This means we follow "Bohr's Postulates" and that any of the possible energies must be discrete. My approach to finding out the discrete energies is inefficient, it involves guessing at what a number will be and then punching it into the calculator (and then repeating until I find one that matches). This usually works on the simpler questions where you don't need to find out much.
So, elimination:
It cannot be 10.2 eV (E), so it cannot be (F). A few quick calculations show that it can't be 1.4eV (B) - (the smallest I found was 1.6). This also allows us to eliminate (A).
We are left with (C) and (D) to check.
My method seems a bit inefficient, I might be missing a few numbers.
A better way might be to systematically punch in numbers (e.g. 10.4 - 8.8, then 10.4 - 6.7 etc. and when you do all the 10.4 calculations, drop down to 8.8 - 6.7 (you won't need to do 8.8 - 10.4, since you've already done it at the start - only difference is the answer will be negative).
I don't know why I wasn't doing that before. I went through all the possible discrete energies in about 1 minute. Which raises a problem, none of the energies matched the energies in the list. I'm confused, is there anything I'm forgetting?
Oh of course. Emerging electrons. The energies listed are the energy after the electron has been emitted. The electrons are left with a bit of extra kinetic energy. I should have realised this earlier, it would have made calculations quicker.
To quickly show you all the possible calculations, I just used excel:
(http://i.imgur.com/wvQmv.png)
So we can get B - 1.4, C - 3.5, D - 5.3
Sorry for the almost stream of consciousness style, I can't be bothered editing the post to just focus on the correct working out.
Post by: HarveyD on October 29, 2011, 10:54:34 am
lol
but its TSFX, so I'm inclined to believe you over them
thanks
Post by: Lasercookie on October 29, 2011, 11:05:58 am
answer is F :/I'm not sure how 'E - 10.2eV' could possibly be the excess kinetic energy given to an electron. 10.2 eV is the energy being bombarded, an electron cannot emerge without at least some of that energy being used up.
lol
but its TSFX, so I'm inclined to believe you over them
thanks
I'm inclined to believe myself over TSFX as well :P
Post by: cranberry on October 29, 2011, 11:25:00 am
Post by: xZero on October 29, 2011, 11:53:41 am
Don't expect this type of question in a vcaa exam, way too tricky.
Post by: Lasercookie on October 29, 2011, 12:47:44 pm
Post by: Bozo on October 29, 2011, 02:03:25 pm
Post by: Lasercookie on October 29, 2011, 02:27:32 pm
Wait so just to clarify, it is ONLY WITH PHOTONS, that for energy to be omitted the energy of the photon must match the discrete energy level exactly - otherwise nothing will happen. (disregard ionization)Yeah, photons can only exist as discrete amounts of energy. For the electron to move to a higher/lower state, the energy of the photon must match the difference between the energy levels exactly.
Post by: HarveyD on October 29, 2011, 02:31:34 pm
i.e. cant go from n=2 to n = 3?
Post by: Lasercookie on October 29, 2011, 02:38:52 pm
and it can only go from ground state up yeah?edit: removed incorrect stuff, refer to xZero's post
i.e. cant go from n=2 to n = 3?
Post by: xZero on October 29, 2011, 03:08:35 pm
Post by: Lasercookie on October 29, 2011, 03:15:35 pm
umm i dont think it can jump from n=2 to n=3, it must go up from ground state but when it drops down it can go from n=3 to n=2 etc.:|
I better get back to studying in that case.
Sorry for telling you the wrong thing Harvey.
edit: fixed annoying typo
Post by: Bozo on October 29, 2011, 07:57:52 pm
Post by: xZero on October 29, 2011, 08:04:33 pm
Post by: Lasercookie on October 29, 2011, 09:26:17 pm
It can jump from n=2 to n=3, but in VCE physics, we assume it does not purely because its a 1 in 1000 chance. That's what the chief assessor at TSFX physics said.
according to my lecturer, it cant because there's no atom in the n=2 state, or simply atoms in n=2 state transitions back to n=1 too fast to absorb enough energy to jump up a state :S im all confused
I don't think those two statements contradict too much. If it transitions too fast to absorb enough energy to jump up the state, there is still a small tiny chance that it might actually absorb enough energy? I'm not sure if 1 in 1000 is the actual probability or not.
Also, I didn't realise that Bruce Walsh does stuff for TSFX :| Or did they get a new chief assessor this year?
Post by: Bozo on October 29, 2011, 09:37:35 pm
Post by: Bozo on October 29, 2011, 11:15:07 pm
Post by: paulsterio on October 29, 2011, 11:19:18 pm
Also, I didn't realise that Bruce Walsh does stuff for TSFX :| Or did they get a new chief assessor this year?
Yeah, I thought Bruce was the Chief Assessor as well? :S
Post by: cranberry on October 30, 2011, 07:37:18 pm
Who here has done Lisachem 2011. I don't get sound q2 and q3. Can anyone clarify?
attach it up and ill be glad to have a ping :)....im redoing phys exams cause i got no new ones
Post by: HarveyD on October 31, 2011, 12:34:43 am
for this question, in the table, we would give the peak voltage yeah?
Post by: Lasercookie on October 31, 2011, 07:24:00 am
just wondering:Yeah, X and Y are located at the locations of the peak voltages.
for this question, in the table, we would give the peak voltage yeah?
Post by: HarveyD on November 01, 2011, 12:47:10 pm
with a question like this
where it asks for the force direction acting on side X-Y
I know its clear that its into the page when you use the right rule etc
but if you said the force was acting "down", would that be wrong?
because thats how it works in real life? since they tell you its a motor....
Post by: iroflmfao on November 01, 2011, 12:55:24 pm
Printing at school = free.
Half pages = bonuses prints.
Post by: Lasercookie on November 01, 2011, 05:51:01 pm
another quick questionI think down might refer to when - how do I explain this? uhh, so fingers pointing into the page, thumb pointing left and your palm faces 'down' (in other words, along X to Y). I think that direction would be considered down in this case.
with a question like this
where it asks for the force direction acting on side X-Y
I know its clear that its into the page when you use the right rule etc
but if you said the force was acting "down", would that be wrong?
because thats how it works in real life? since they tell you its a motor....
If it was one of those drawings where it's not flat, but it's like a perspective drawing (I hope you know which ones I'm talking about, e.g. like the usual VCAA ones) - I think then it's alright to use the word 'downwards'. To be safe, I use 'vertically downwards' - I saw it used somewhere (assessor reports possibly, don't quote me on that).
edit: Don't forgot about the direction key/little compass thing if they give you one. Always go by the wording given, if provided.
Post by: HarveyD on November 01, 2011, 05:52:36 pm
just have to judge from the perspective they give i guess
Post by: HarveyD on November 04, 2011, 02:55:57 pm
Post by: Lasercookie on November 04, 2011, 04:57:01 pm
Any ideas on how to do these?For the first one, it's simply right hand grip rule. I'm guessing that the answers will be either out of the page, or into the page (judging from the fact we are asked about inside and outside the loop). Take your hand (with your fingers in the correct position, thumb starts pointing right and your fingers are curled as if it's gripping the wire), and follow the current direction.
A - Out of the page
B - Out of the page
C - Into the page
D - Into the page
For the second one, you have a wave moving into an area of normal air pressure.
Straight off, remember that sound is a longitudinal wave. It can only be left or right.
The first part of the wave to reach the particle is a rarefaction.
A compression pulls particles, and a rarefaction pushes particles.
So if the wave is moving to the right, a compression would pull the particle with it - the particle would move to the right.
However, a rarefaction is the opposite, so it'd push the particle left.
A - Left.
That was an interesting sound question.
Post by: HarveyD on November 04, 2011, 05:02:07 pm
since the current is going in the same direction?
Edit: All your answers are correct btw, thanks!
Post by: Lasercookie on November 04, 2011, 05:23:33 pm
with 2, just was confused as to why A and D are differentSo the field starts out of the page. It gets to the loop. Once you're out of the loop,
since the current is going in the same direction?
Edit: All your answers are correct btw, thanks!
Like I said, try to follow the wire with your hand. It's kind of like a rollercoaster going round a loop.
Maybe some badly drawn scribble over the diagram will help:
(http://i.imgur.com/Judh0.gif)
Post by: HarveyD on November 04, 2011, 05:26:14 pm
never knew that :S
thanks ^^
Post by: Lasercookie on November 04, 2011, 05:29:37 pm
The reason why A and D are different: The field curls around the wire.
A is above the wire. The field is poking out of the page.
D is below the wire. The field at this point is into the page.
Where the 'I' is, the field would be out of the page - the same as A. If you had a point below the wire (directly below A), it would be into the page.
Perhaps thinking about it in three-dimensions will help. Picture an actual wire with a field around it.
Post by: cranberry on November 04, 2011, 06:16:40 pm
Post by: HarveyD on November 04, 2011, 06:18:01 pm
Post by: Lasercookie on November 04, 2011, 06:23:17 pm
weird questions, must be 2011 trials ay ?Looks like an Insight exam to me.
I didn't think the field question was too odd, just right hand grip rule. My explanations aren't that good, it's much easier to explain right hand grip rule stuff in real life lol.
For the sound stuff:
http://www.physicsclassroom.com/class/sound/
Lesson 1 explains the rarefaction/compression stuff a bit.
I just googled that website, but it looks like it pretty much covers the most of the VCE Physics sound stuff (minus the stuff about microphones and speakers). I might read through all that actually.
Post by: iroflmfao on November 04, 2011, 07:38:49 pm
Post by: Lasercookie on November 04, 2011, 07:48:47 pm
Probs a stupid question how you know its a rarefactionWell here's the picture of the wave (with my annotations):
(http://i.imgur.com/EC2cK.png)
Here it's not represented with lines, it's represented with dots.
I assumed the evenly spaced out dots were representing normal air pressure. I think this is reasonable, it does say that Particle X is originally at normal air pressure.
A compression is a point where there's a lot of pressure (compressed pressure if you will?), right? So when the dots are close together, that point is a compression.
When they're spaced out a bit, that's a rarefaction
I liked this diagram:
(http://www.physicsclassroom.com/class/sound/u11l1c2.gif)
You might be used to those graphs that look a bit like a sine wave. Those a graphs of change in air pressure vs time. The line at 0 represents normal air pressure usually. As you would know, a compression is a crest (point of higher air pressure), a trough is a rarefaction (point of less air pressure).
If the wave was moving to the right, then the rarefaction dots would be the dots to reach the particle first right?
Post by: iroflmfao on November 04, 2011, 08:01:34 pm
what happens then ?
Nothing ?
Post by: Lasercookie on November 04, 2011, 08:14:29 pm
What if x was on a compression pointYou mean like this: (sorry about the rough editing :P)
what happens then ?
Nothing ?
(http://i.imgur.com/XCvkb.png)
What's actually happening is interference (since you have waves interacting). If a trough and crest meet, then they destructively interfere and nothing would happen.
What we were getting before was basically "nothing" and a trough - leaving just a trough (point of low pressure) - would you call that interference at all? (if anything, would it be considered constructive interference?).
If you meant that if X was originally at normal pressure and a compression came along, then it should move to the right initially?
Of course, if you had a rarefaction and rarefaction meet up, you'd get constructive interference and up with a point that has an even lower pressure than before.
Post by: iroflmfao on November 04, 2011, 08:16:19 pm
Thanks man your so helpful
Post by: Lasercookie on November 04, 2011, 08:25:48 pm
Thanks man your so helpfulNo problem dude, I'm glad to try and help.
Post by: iroflmfao on November 05, 2011, 11:58:21 am
2/3 < 1
So i wrote little or no diffraction
Is that the same as > There would be a diffraction pattern because the wavelength is of the same order of magnitude as the interatomic
spacing.
Post by: Lasercookie on November 05, 2011, 12:54:10 pm
Q 5 of 2007 VCAA of light and matterNo it's not the same. You wrote 'no diffraction', while that states there is diffraction.
2/3 < 1
So i wrote little or no diffraction
Is that the same as > There would be a diffraction pattern because the wavelength is of the same order of magnitude as the interatomic
spacing.
There is always diffraction (even if it is below or above 1), what varies is how significant/noticeable it is.
I would most definitely avoid the use of the term 'no diffraction will occur'.
That ratio wavelength/slit width is for the 'extent of diffraction'. If wavelength is larger than the slit width then you'd get a lot of diffraction. If the wavelength is smaller than the slit width and you get a low level of diffraction. But there is still diffraction.
I'd tend to write something like that diffraction would occur because of the order of magnitude stuff, but due the wavelength being smaller than the slit width a weak pattern may be observed.
The Australian Institute of Physics gave two possible answers in their solutions:
Quote
Yes (1), the wavelength is comparable to the interatomic spacing (1).
The answer ‘No’ (1) could be argued for by saying the wavelength is smaller than the spacing and little diffraction will be noticed (1).
So you are correct going by the AIP. If the VCAA would accept the answer, I don't know.
edit: fixed wording
Post by: HarveyD on November 05, 2011, 12:59:34 pm
Post by: Lasercookie on November 05, 2011, 01:14:04 pm
is there any difference between sound and light regarding that^I saw that on the report, I am not sure what they were talking about though.
The high and low frequency stuff for sound is just derived from that same ratio:
high frequency = smaller wavelength = less diffraction
low frequency = larger wavelength = more diffraction
Was there anything else particular about sound diffraction?
Post by: HarveyD on November 05, 2011, 01:23:51 pm
i normally assume that when its above one theres more diffraction though
but if you were given two options
1) Almost 1
2) Greater than 1
With both sound and light, would you choose 2 for more significant diffraction?
Post by: Lasercookie on November 05, 2011, 01:50:25 pm
hmm not sureYeah I assume that as well, I would choose 2. Larger than 1 means that extent of diffraction is greater blah blah blah
i normally assume that when its above one theres more diffraction though
but if you were given two options
1) Almost 1
2) Greater than 1
With both sound and light, would you choose 2 for more significant diffraction?
Below 1 and it's less significant. But it's still there.
Reading what VCAA wrote again:
Quote
Many students confused the diffraction requirements from the Detailed study ‘Sound’ with what is necessary to form a useful diffraction pattern in this topic.What are the definitions for useful diffraction pattern? Are they saying that the other AIP option is incorrect?
0.66 is still some level of diffraction I guess. Like obviously it wouldn't be a lot of diffraction, but you'd still notice it. Like in water waves, if the wavelength is smaller, you can still sort of see the circular bits at the edge. With light, you'd probably notice a blurred edge as well.
(http://esfscience.files.wordpress.com/2009/03/no-diffraction-water-waves1.gif?w=432&h=218)
I don't believe that diffraction is vastly different for sound. I guess in sound we really only deal with where is the sound going and where will it be heard (e.g. you have to be in front of the source to hear high frequency sounds the best). Or it's role in loudspeakers etc. We don't consider the fact that it is still being slightly diffracted - really only given the two extremes. Perhaps this is what VCAA was referring to.
Post by: iroflmfao on November 05, 2011, 02:09:00 pm
Post by: Plan-B on November 05, 2011, 06:17:04 pm
is there any difference between sound and light regarding that^
I believe that a lambda/slit width ratio of around 0.5 or greater is already significant for sound. Where as for light, it is only significant once it approaches or is greater than one. Don't take my exact word for it though, but I remember something like that, unless i'm wrong. :s
Another question:
http://images.tutorvista.com/content/feed/tvcs/series20parallel20resistor20circuits.JPG
Just found a simple parallel circuit. Would you say that current through R3, is from B to C, or C to B? Many of these practice exams have many inconsistencies, so I was wondering how you guys do it.
Although it's not so bad since VCAA makes interpretation explicit when they ask for direction.
Just wanted to clear it up just in case.
Post by: Lasercookie on November 05, 2011, 06:45:38 pm
Another question:We use conventional current, so it's the flow of positive charge. (so starts from positive end of the battery). The positive end of the battery is the long end.
http://images.tutorvista.com/content/feed/tvcs/series20parallel20resistor20circuits.JPG
Just found a simple parallel circuit. Would you say that current through R3, is from B to C, or C to B? Many of these practice exams have many inconsistencies, so I was wondering how you guys do it.
Although it's not so bad since VCAA makes interpretation explicit when they ask for direction.
Just wanted to clear it up just in case.
In that picture, just use the arrows on the wire? So it'd be B to C. You can't really tell since the symbol of the battery isn't given.
http://images.tutorvista.com/content/electricity/parallel-resistors-energised-battery.jpeg
So here, current is A to B. (again, the arrows are there, but they give the battery symbol this time)
If it's in a generator/alternator, then current direction will be given by the right hand rules. With AC, the current will reverse according to the frequency e.g. 50Hz AC will change 50 times a second.
If you want more diagrams, I googled and found this website:
http://www.allaboutcircuits.com/vol_2/chpt_1/1.html
It seems to cover most of the circuit stuff we do in Unit 4.
Post by: Plan-B on November 05, 2011, 09:50:00 pm
Post by: HarveyD on November 06, 2011, 01:14:56 pm
heres a similar question I found similar to the magnetic field question
try your method and check if it works here
Post by: Lasercookie on November 06, 2011, 02:41:00 pm
hey laserddWas this about this thread: TSFX '08 ?
heres a similar question I found similar to the magnetic field question
try your method and check if it works here
That question is a slightly different situation actually. You have a current carrying wire generating the magnetic field. The field generated by this current will circle around it - I don't think the field in the TSFX 2008 question would curl around like that, as that was an external field.
Anyway, I would just use the regular right hand grip rule (fingers curled and all). The current is facing right, so your thumb points out right. Your fingers grip the wire.
Above the wire your fingers are pointing out of the page (so the little dots). If you extend that in your head, it would be point into the page (the crosses) below the wire.
This means it can be either A or D. The magnetic field in D is uniform. Since the field is due to the current-carrying wire, it'd get weaker as you move out from it.
So I'd say the answer is A.
Is that correct?
Post by: HarveyD on November 06, 2011, 02:45:11 pm
but yeah that's correct
so above the wire, the direction will be the same as what's indicated by the right-hand grip rule, while below it, it will be the opposite?
Post by: Lasercookie on November 06, 2011, 02:56:36 pm
nah the loop one i posted beforeYeah, that was the case with this one. I'm not going to say that it'll always be opposite, since a situation might came up where it is isn't. It's easier for me to just think of it in terms of the RH Grip rule.
but yeah that's correct
so above the wire, the direction will be the same as what's indicated by the right-hand grip rule, while below it, it will be the opposite?
I tend to think of it as just imagining your fingers are longer. If your fingers are curled, and if they were longer they'd end up forming a full circle around the wire, right?
Here's another diagram (you can probably tell I like diagrams):
(http://upload.wikimedia.org/wikipedia/commons/thumb/3/3e/Manoderecha.svg/220px-Manoderecha.svg.png)
That other "method" (this post: Re: Bozo's queries) was just me screwing up my explanation, I wasn't thinking straight. My other two posts about that question is pretty much the same thing I did with this question.
Post by: HarveyD on November 07, 2011, 02:00:12 am
which option would you select for this:
Post by: Lasercookie on November 07, 2011, 07:17:50 am
ah yeah, i kinda see nowMy initial assumption is blue - but I'm not too sure.
which option would you select for this:
Quote
For a given metal and frequency of incident radiation, the rate at which photoelectrons are ejected is directly proportional to the intensity of the incident light.
The frequencies of the two light are different. I think this would mean that you can't really directly compare the two intensities. I don't know of any other way of figuring out the intensity.
Since there's only two options, I'd pick blue - I think that's the "most correct" option.
edit: didn't notice unable to determine.
Post by: HarveyD on November 07, 2011, 12:13:20 pm
cause they had unable to determine lol
Post by: Lasercookie on November 07, 2011, 04:31:16 pm
yeah think tsfx may be wrong?"Unable to determine" is correct as far as I know, as intensity is proportional to photocurrent "for a given metal and incident frequency".
cause they had unable to determine lol
I didn't realise that "unable to determine" was an option :/
Hmm... these TSFX questions seem interesting - I might have a go at the exams later.
Post by: iroflmfao on November 07, 2011, 04:49:31 pm
They kinda teach you everything at their masterclasses. Even if its not on the study design.
Post by: Lasercookie on November 07, 2011, 05:16:07 pm
I've been intending to ask about this question for a while now, I kept forgetting to:
(http://i.imgur.com/UaDxj.png)
It's from STAV 2011, Q11 from the Light and Matter section.
I sort of understand the reasoning in the solutions, but the explanation feels kind of dense to me.
Post by: xZero on November 07, 2011, 05:22:36 pm
Post by: Lasercookie on November 07, 2011, 06:35:38 pm
current = amount of charge flowing per second right? find the charge then divide that by the electrical charge of an electron to get the number of electrons/second and that should be the minimum number of photons per secondYep, that gives the correct answer. Thanks xZero :)
Post by: iroflmfao on November 11, 2011, 06:56:30 pm
Post by: Lasercookie on November 11, 2011, 07:16:39 pm
Is AC voltage RMS ?(http://www.kpsec.freeuk.com/images/wave.gif[img]You can be given AC voltage inPeak (value of the amplitude), peak to peak or RMS.[img]http://www.kpsec.freeuk.com/images/rms.gif)
RMS voltage is the "effective voltage" - this also happens to be the amount that the voltage provides if it were equivalent to smooth D.C.
If the type of voltage isn't stated (it will be most of the time), then you assume it is RMS. This is what occurs in every day life as well. It'll say that the voltage in our houses is 240V AC. This is an RMS value.
Post by: Lasercookie on November 11, 2011, 07:18:01 pm
Here they are (sorry for double posting)
(http://www.kpsec.freeuk.com/images/rms.gif)
(http://www.kpsec.freeuk.com/images/wave.gif)
Post by: Aurelian on November 11, 2011, 07:27:40 pm
Seeing that we've turned "Bozo's queries" into "Everyone's queries", might as well continue posting here :P (I hope you don't mind Bozo).
I've been intending to ask about this question for a while now, I kept forgetting to:
(http://i.imgur.com/UaDxj.png)
It's from STAV 2011, Q11 from the Light and Matter section.
I sort of understand the reasoning in the solutions, but the explanation feels kind of dense to me.
Basically all you need to realise is that there is a one-to-one interaction between the incident photons and the ejected photoelectrons. As a result of this, the number of photons contained in the incident light beam per second is *the same* as the number of ejected photoelectrons per second, which you can figure out from the current and the charge on one electron =)
Post by: iroflmfao on November 11, 2011, 07:43:38 pm
Whats the unit of Work function LOL
Post by: Lasercookie on November 11, 2011, 07:53:21 pm
Thanks LASERWork = Energy, so we'd be given it in either Joules or eV.
Whats the unit of Work function LOL
Most of the time we are given it in eV - makes things simpler.
The SI unit for work is Joules (derived from Newton metres e.g. W=Fx).
Post by: HarveyD on November 12, 2011, 01:00:48 pm
Post by: Lasercookie on November 12, 2011, 01:52:16 pm
Any ideas on how to do this :/Field lines go from north to south. So the middle bit is north and the outer rim is south.
Post by: iroflmfao on November 12, 2011, 04:41:30 pm
Any ideas on how to do this :/
LOL i just did this exam before.
Lines go from North to South.
Post by: Bozo on November 12, 2011, 08:34:23 pm
Post by: HarveyD on November 13, 2011, 01:29:44 am
SHould be extrapolate it down to the work function?
Or just leave it at the threshold frequency
Post by: dc302 on November 13, 2011, 03:27:39 am
Post by: Bozo on January 04, 2012, 01:33:00 pm
Post by: Lasercookie on January 04, 2012, 01:39:28 pm
This thread was legendary, laserred you're a mad dog.Haha thanks, Bozo :D Don't discredit all the other people that answered questions as well.
But yeah, you're right, this thread was legendary, I learnt a lot from it.