HSC Physics Question Thread

[winstondarmawan]

Hello again! Sorry for having so many questions, this paper has no solutions and just a vague marking criteria.
1. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19883468_1265097333615787_2044210444_n.png?oh=bcc97a8f4bd080378231c4a41352ea1d&oe=596225E2
Not sure how the answer can be A when the voltage needs to be stepped up... or does the orientation of the transformer not matter in this case?
2. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19904508_1265097566949097_1191709984_n.png?oh=6dbde37a3f515bad436a2a68ea64e3ce&oe=59632643
3. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19883825_1265098016949052_2064944303_n.png?oh=524fa20cdadf94347bdae0d48b2a68f9&oe=59624F62
Thought the answer was C but apparently it is B. Even after that, why is the lightbulb off?
4. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19894213_1265098310282356_1398760865_n.png?oh=e8c42473f96f19d88a151d6af7a12903&oe=59632482
5. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19866262_1265098473615673_1864449019_n.png?oh=23fcb08250bd8a080bd1b9228ee750a4&oe=59623D24

Again, sorry for the amount of questions! I really struggled with this paper and the solutions are not adequate. TIA

[JeffChiang]

Hi,
This question is question 15 from the 2013 HSC.
The answer is A and I really don't understand how.

[limtou]

Hi,
This question is question 15 from the 2013 HSC.
The answer is A and I really don't understand how.

When calculating torque, it is the angle between the magnetic field and the plane WXYZ, in this case the labelled 30. For the force on WX, it the angle between the direction of the wire and the magnetic field; in this case the wire WX is vertical while the magnetic field/flux line is horizontal, hence they are perpendicular (i.e. 90 degree).
Hope this helps

[kiwiberry]

4. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19894213_1265098310282356_1398760865_n.png?oh=e8c42473f96f19d88a151d6af7a12903&oe=59632482
5. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19866262_1265098473615673_1864449019_n.png?oh=23fcb08250bd8a080bd1b9228ee750a4&oe=59623D24

For 4), the semiconductor sensor needs to be able to detect radiation with wavelength of about 15 \(\mu m\) since the majority of radiation from Klingon is of this wavelength (from the graph). The semiconductor's band gap needs to be equal to the energy of a photon of Klingon radiation so that when the photon strikes the sensor, it gives an electron enough energy to promote it to the conduction band and make the sensor work (like a solar cell). So, calculating the energy of a photon:
Converting to eV,
So GaNj is the most appropriate!

For 5), \(u_x = V\cos 60 = \frac{V}{2}\) and \(u_y = V\sin 60 = \frac{\sqrt{3}V}{2}\). The time is takes for the projectile to reach \(\Delta x = 55\) will equal the time it takes for it to reach \(\Delta y = -34 \)
Not sure about the first 3, but hopefully someone else will step in! Hope this helps

[winstondarmawan]

For 5), \(u_x = V\cos 60 = \frac{V}{2}\) and \(u_y = V\sin 60 = \frac{\sqrt{3}V}{2}\). The time is takes for the projectile to reach \(\Delta x = 55\) will equal the time it takes for it to reach \(\Delta y = -34 \)
Not sure about the first 3, but hopefully someone else will step in! Hope this helps

Thank you! For this question, would it be appropriate to use maths methods?

[jamonwindeyer]

Thank you! For this question, would it be appropriate to use maths methods?

As in, from Extension 1 Math? nope, you'll want to stick with methods taught in the course so that you can get marks for working if you make any small errors

[jamonwindeyer]

Hello again! Sorry for having so many questions, this paper has no solutions and just a vague marking criteria.

Again, sorry for the amount of questions! I really struggled with this paper and the solutions are not adequate. TIA

No need to be sorry! Let me give you a rundown:

1) Orientation doesn't matter, you can plug the input into the right and get the output from the left, no issue there. So the voltage jump from input to output would be 2.5x, so 10V becomes 25V. Then at 90% efficient, that is 22.5V - That is the best option to make sure you get your 22V without going ridiculously far over

2) Recall that induced emf is proportional to the rate of change of flux. What this means, since you do 2U and will understand the terminology, is that the induced emf is proportional to the derivative of flux. Flux is a sine curve, so we'd expect the induced emf to be a cosine curve. The answer is (B). We could do this without the knowledge of derivatives (namely that \(\frac{d}{dx}\sin{x}=\cos{x}\)), but this is way easier, since we have it in our toolbelt

3) Brutal question this one, the issue is that the voltmeter is in series, where it is supposed to be connected in parallel. Voltmeters have a very large internal resistance (essentially an open circuit) to prevent impacting the circuit when connected in parallel. In series, pfft, you done f-ed up. The huge resistance stops any current from flowing, so while you'll register the 12V in your circuit, the circuit stops working - No current (0A on the ammeter) and no light 

Hope this helps

[winstondarmawan]

No need to be sorry! Let me give you a rundown:

1) Orientation doesn't matter, you can plug the input into the right and get the output from the left, no issue there. So the voltage jump from input to output would be 2.5x, so 10V becomes 25V. Then at 90% efficient, that is 22.5V - That is the best option to make sure you get your 22V without going ridiculously far over

2) Recall that induced emf is proportional to the rate of change of flux. What this means, since you do 2U and will understand the terminology, is that the induced emf is proportional to the derivative of flux. Flux is a sine curve, so we'd expect the induced emf to be a cosine curve. The answer is (B). We could do this without the knowledge of derivatives (namely that \(\frac{d}{dx}\sin{x}=\cos{x}\)), but this is way easier, since we have it in our toolbelt

3) Brutal question this one, the issue is that the voltmeter is in series, where it is supposed to be connected in parallel. Voltmeters have a very large internal resistance (essentially an open circuit) to prevent impacting the circuit when connected in parallel. In series, pfft, you done f-ed up. The huge resistance stops any current from flowing, so while you'll register the 12V in your circuit, the circuit stops working - No current (0A on the ammeter) and no light 

Hope this helps

Thank you so much! For the 3rd question, are we required by syllabus to know that? Because I never remember learning that in my life.

[jamonwindeyer]

Thank you so much! For the 3rd question, are we required by syllabus to know that? Because I never remember learning that in my life.

Yeah that's a bit of a stretch imo - You learn it a bit in Year 11, but I doubt the HSC would assess you on it

[JeffChiang]

When calculating torque, it is the angle between the magnetic field and the plane WXYZ, in this case the labelled 30. For the force on WX, it the angle between the direction of the wire and the magnetic field; in this case the wire WX is vertical while the magnetic field/flux line is horizontal, hence they are perpendicular (i.e. 90 degree).
Hope this helps

Yes. Thanks limtou. But does this mean that for F=BILsin(w), w=90 regardless of the wire's rotated position? so the force would be constant. I never really understood this part of the syllabus.

[jakesilove]

Yes. Thanks limtou. But does this mean that for F=BILsin(w), w=90 regardless of the wire's rotated position? so the force would be constant. I never really understood this part of the syllabus.

Yep, that's totally right! Imagine the direction of the arrows indicating the magnetic field, and the direction of the current in the wire. As the wire moves, it stays perpendicular to the arrows of the field lines. So, the angle between the field lines and the current doesn't change, and the force stays the same! Does that sort of make sense?

[JeffChiang]

Yep, that's totally right! Imagine the direction of the arrows indicating the magnetic field, and the direction of the current in the wire. As the wire moves, it stays perpendicular to the arrows of the field lines. So, the angle between the field lines and the current doesn't change, and the force stays the same! Does that sort of make sense?

Yes, it makes sense now. Thanks Jake!

[justwannawish]

Hey,
This is more of a general question. Hope that's okay. I think our school is a bit behind and we've just started the third module for prelim physics and our teacher has estimated that we would have to gloss over Cosmic Engine at this rate. How relevant are the concepts taught in the module for HSC physics (is space an extension of cosmic engine)?

[jakesilove]

Hey,
This is more of a general question. Hope that's okay. I think our school is a bit behind and we've just started the third module for prelim physics and our teacher has estimated that we would have to gloss over Cosmic Engine at this rate. How relevant are the concepts taught in the module for HSC physics (is space an extension of cosmic engine)?

Hey!

In reality, year 12 content doesn't really build on year 11 knowledge. So, when you reach year 12, you could effectively forget the content you learned in Year 11, and still be absolutely ok.

The more important thing that you should get out of year 11 is the skills, and your study techniques. Practicals, writing the answers to extended response questions, using physics terminology; those are the sorts of skills that are invaluable when you get to Year 12. Similarly, if you get into good study habits now, and learn the ways in which you learn the best, you'll get a massive jump on Year 12.

However, overall, I wouldn't worry too much if you speed through some Year 11 content. You won't at all be at a disadvantage

[jennifer.le11]

Hi!

I'm having trouble pinpointing exactly why the answer to this is Q, is it because of the right hand grip rule? Could you explain in detail how the force acts towards Q?

Thank you very much!