Bazza's 3/4 Physics question thread

[pi]

They advised pen (as the exam is scanned), but as I didn't have much confidence in physics, I used a 2B pencil for both (or for at least one of them) and still got a SS. But I'd advise pen to be on the safe side

[Phy124]

4) is yeild point dependent on cross sec area ? (is it related directly to stress or strain, or both?)
6) is strain energy toughness x new volume ( after extension  ) or old volume?

4) Didn't do Structures

6) Didn't do Structures

4) The yield point of a material is the maximum amount of stress it can withstand before it ceases acting elastically i.e. it enters plastic deformation and can't return to its original form.

As for it's relationship to cross sectional area and strain...

The yield point or yield stress of a material is not reliant on the cross sectional area of its application, but  rather whether the yield stress is met.

As the smaller the cross sectional area, the larger the stress per given force. Therefore it is more likely that the yield stress of a material will be met for a lower cross sectional area. e.g. if you were slowly increasing the amount of force on the object, it would plastically deform earlier with a lower cross sectional area.

Similarly for strain, as , when the strain increases, so does the stress on the material. This means the larger the strain the more likely it is that the yield stress will be met and the material will start performing plastically.

6) I'll get back to you on this because I don't want to tell you the wrong thing (although, IIRC, it was final)

[paulsterio]

thanks for the help

(what is the diff bewtween func earth and protective earth?) and how can we tell if current is flowing through it? I thought that since V=IR if R -> oo then I approaches zero?

with 5) i know energy is positive, but the CHANGE in energy can be negative (can it :S?). The negative there is to indicate when one increases the other decreases
with 7) i thought the current is the other way around ? http://mbed.org/media/uploads/harrisjunaid/photo_diode_amp_circuit.png
like the current is reverse as opposed to the diode, but is thus normal compared to the battery?

and pen or pencil in the exam? (is pencil allowed?)

5) You can have a decrease in energy, so yes, you can have a negative CHANGE in energy but I would avoid going near this, it's too controversial and to be honest, you might as well say that "Kinetic Energy decreased by 100J" rather than saying "deltaEk = -100J". It's just so much more safer - considering VCE

7) I think you're a bit vague, what do you mean by reverse current? Current that is in the opposite direction to the normal current, or current that is flowing against the diode?

[WhoTookMyUsername]

Ah dw i think ive got it thanks paul


...
Are photo transistors on study design? Ive heard some conflicting information so im not sure ( along with other transistors, capacitors etc.)


And is it true if you get the right numerical answer for any question you automatically get full marks on the exam?

[Lasercookie]

Are photo transistors on study design? Ive heard some conflicting information so im not sure ( along with other transistors, capacitors etc.)

Yes, phototransistors (usually called photodiodes) are on the course. However, transistors and capacitors aren't apart of AOS2 (they were on the old study design: 2008 and earlier). You don't need detailed understanding of how diodes, photodiodes actually work (no need to worry about p-n junctions and all that). You'll need to understand what a photodiode does, it's basic functions, how to interpret the graphs etc.

apply the concepts of current, resistance, potential difference (voltage drop) and power to the operation of electronic circuits comprising diodes, resistors, thermistors and photonic transducers including light dependent resistors (LDR), photodiodes and light emitting diodes (LED), (V=IR, P=VI);

And is it true if you get the right numerical answer for any question you automatically get full marks on the exam?

Huh? Are you asking if you get one question right, then you win the entire exam?

Or are you asking if you get the right answer on a question, despite flawed working out, do you get full marks? I'm assuming you meant this, because it sounds a bit more sensible. In short, no.

For questions worth 2 marks or greater, you'll be given marks for working. There's also consequential marking for multi-part questions.

[WhoTookMyUsername]

Oh wow... i didn't realise that photodiodes and phototransistors are the same thing until now :\ (but now that i think about it it's so obvious they are :X)

yeah i meant if you put the write answer you get full marks for the q (although it would be nice to get full marks on exam)

thanks laseredd

[pi]

And is it true if you get the right numerical answer for any question you automatically get full marks on the exam?

Not always, depends on the examiner. An examiner (like Mr K from MHS) will scrutinise every letter in your working whilst others will just give marks for the answer in the box.

In general, do full working and don't make up bs working, especially if you think that you may make a mistake in the exam, then your working will be vital.

[WhoTookMyUsername]

yeah seems like the safer way to go :X

http://www.vcaa.vic.edu.au/vce/studies/physics/pastexams/2007/2007physics1.pdf

pg 28

how do i do the last Materials and structures question here?

thanks

and also generally how many marks can you lose per exam to still get a 50 ? (assuming good sacs)

[pi]

Can't answer that M&S question (I doubt I even did that paper tbh) but you don't want want to lose more than probably 4 marks over both exams assuming exams like last year.

[paulsterio]

Photodiodes and phototransistors are not the same component!

Photodiodes are like diodes, except they allow current to travel in the reverse direction (called photocurrent) when there is light shone upon it - so it is installed in reverse bias in order to get this effect.

Phototransistors are like transistors - so I know you guys haven't done transistors, but think of them as amplifiers, because a change in voltage across two of its three terminals will cause a change in voltage across the other pair of terminals. A phototransistor gets rid of one of the pins and uses the light shone onto it as the changing input voltage.

However, phototransistors are no longer on the course. The only two receiving devices still on the course are LDRs and Photodiodes.

The issue with working is always complex - but VCAA do state that the answer in the box is enough - but the truth is, why wouldn't you show working, so just do it anyways.

pi, I think probably less than 4 marks per exam. I lost 11 marks in total, 10.5 on Exam 1 and 0.5 on Exam 2 - with 100 for SACs and I ended up getting 46. But granted, I lost a lot of marks on one exam rather than spread evenly, but it's hard to say

Moderator action: removed real name, sorry for the inconvenience

[WhoTookMyUsername]

D: lol (so we don't need to know about phototrans :X)

http://www.vcaa.vic.edu.au/vce/studies/physics/pastexams/2007/2007physics1.pdf

pg 28

how do i do the last Materials and structures question here?

thanks

do you know how to do this question?

[Phy124]

D: lol (so we don't need to know about phototrans :X)

http://www.vcaa.vic.edu.au/vce/studies/physics/pastexams/2007/2007physics1.pdf

pg 28

how do i do the last Materials and structures question here?

thanks

do you know how to do this question?

Sum forces in x direction (Tension in AB & horizontal component of compression in BC) and y direction (weight force of lamp & vertical component of compression in BC) or alternatively you could take moments (torques) about point C (remember to take perpendicular distances). Either should work, I believe 

edit: left out a word

edit2: after thinking about, the latter option is definitely easier

[pi]

Paul, I meant 4 over BOTH exams

[paulsterio]

Paul, I meant 4 over BOTH exams

Oh right - I misread, but that sounds about right - although it's probably a bit little - I think Thushan lost more and got a 50 :p

[Phy124]

but there's no distances D:

You have two angles, so you know the ratios of the distances, that's all you need