HSC Physics Question Thread

[jakesilove]

Hi, could someone explain the differences between the four options for the HSC 2010 Q20 Multiple choice? I can understand how to identify which way the wire would turn given both the direction of the current and the direction of the magnetic field, but I cannot differentiate at this point how the diagrams differ.  THanks

https://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2010exams/pdf_doc/2010-hsc-exam-physics.pdf

Yeah, that was a fucking terrible diagram hey! Basically, you need to use the right hand rule to figure out the direction of the electromagnets (follow the current around; will there be a north pole on the right, or the left?). Then, establish the direction of the current flow (again, just follow the positive to the negative, as that will be the direction of current). Then, compare these two things (ie. draw where N and S are, and the direction of current flow), and figure which one will turn as desired! Does that make sense?

[jamonwindeyer]

Hi, could someone explain the differences between the four options for the HSC 2010 Q20 Multiple choice? I can understand how to identify which way the wire would turn given both the direction of the current and the direction of the magnetic field, but I cannot differentiate at this point how the diagrams differ.  THanks

https://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2010exams/pdf_doc/2010-hsc-exam-physics.pdf

Hey! It's a tough one to spot hey. What differs is:

a) Which side of the coil that each terminal is connected to. You can see that in half of them, the '+' is connected to the left, and the other half, it is connected to the right. Follow the wire, you'll spot it!
b) The direction of rotation around the pieces of metal. Look carefully, some wrap OVER and so cause what you could call a clockwise rotation, others wrap UNDER and cause a rotation in the opposite direction

[katnisschung]

the surfing answers may be wrong or i am yet again 
11) A car, travelling 30m/s drives over the edge of a cliff into the water
58m below.
t=3.46s
a) calculate the time it takes the car to hit the water

12) A group of lemmings run over the edge of a 200m cliff at 0.6 m/s
t=4.52s

[jakesilove]

the surfing answers may be wrong or i am yet again 
11) A car, travelling 30m/s drives over the edge of a cliff into the water
58m below.
t=3.46s
a) calculate the time it takes the car to hit the water

12) A group of lemmings run over the edge of a 200m cliff at 0.6 m/s
t=4.52s

Hey! For all questions where there is only horizontal velocity, we can completely ignore pretty much all the information except for height, and work it out in the y direction. So, using







Pretty close to the given answer, which means our values are just a little different somewhere! Maybe they used a=10. Can you get the second one using this working?

[katnisschung]

for 11 i got 3.46 thought i got it wrong becos i wrote √12 
but for 12 i got 6.4 secs

[MarkThor]

With my logic I got C, but the answer is A. Does this have something to do with the ring being split, or am I just stuffing up induction?

[proficles]

Hi,  why is the answer C?

[jakesilove]

for 11 i got 3.46 thought i got it wrong becos i wrote √12 
but for 12 i got 6.4 secs

Yep, I definitely agree that 12 is 6.4s! Great job, looks like the book is just wrong

[jakesilove]

With my logic I got C, but the answer is A. Does this have something to do with the ring being split, or am I just stuffing up induction?

Definitely a seriously tricky question! What you have to remember is that a current will only be induced where there is a change in flux, or basically a 'change in the shit going on between magnetic fields'. In this case, where does the actual CHANGE occur? Only when entering, and leaving, the field! When the magnet swings entirely into the field, nothing is in essence 'changing'. Change only occurs when the number of field lines going through the magnet changes. Thus, there should be two peaks (one for the magnet swinging in, one for the magnet swinging out). Naturally, they will have different signs.

[jakesilove]

Hi,  why is the answer C?

In this question, the only current is an induced current. Stupid question, but there isn't a supplied current; you're causing the wheel to turn through pedaling, not by supplying current! So, the 'induced' current is the current created by the electromagnets. When that current is a maximum, you've essentially go the maximum back-emf possible, thus reducing the movement of the wheel. Very stupid question, which required you to think about what was actually going on

[MarkThor]

Ahh cool thanks Jake, I had thought that even though the magnetic field was constant it was still cutting field lines and thus was moving relative to the magnetic field 

[Neutron]

Hey Neutron!

Synchronous motor isnt in the HSC syllabus so you dont really need to know about it. Clearly by the way Im speaking here I dont really know what it is (hehe soz soz ). Jake and Jamon are a bit busy right now so they can come help later.

In terms of that question on whether or not AC slip ring motors exist, I can tell you that yes it does but not in that form. AC slip ring motors can only work with induction motors (something I learnt from engineering studies, not in the syllabus). And if in the exam they want you to provide a motor that works on AC, just talk about universal or induction motor.

I'm not too capable here, I will inform Jake/Jamon about your question.

Best Regards
Happy Physics Land

Yeah but the thing is they wanted us to talk about how the MOTOR EFFECT (sorry for capitalising, I can't bold on mobile rip) is used in the AC motor haha and if the motor I posted a picture of isnt viable, then what is :O

[Goodwil]

Hey! Check out my answer here

Thanks, makes sense now

[imtrying]

Hey could I get a hand with question 9 from the 2015 paper? Sorry can't upload the photo but here's the link https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-physics.pdf

[Goodwil]

Hi, could someone explain (c) for this question. Specifically, I don't understand why they're dividing by d=30cm (30 x 10^-2)