HSC Physics Question Thread

[Fergus6748]

The first step is to figure out the direction of the magnetic field (either left to right of right to left)

From the positive terminal you can see that the magnets are going to be electromagnets, you can see that it wraps around the left block in clockwise way, You know it has a SOUTH facing pole (I use the term Nanti-Socks to remember if clockwise is South or North )

To double check, we'll do the other side, from the negative terminal, this time, it is travelling in an ANTI-clockwise way (remember to look at it standing in the middle) This means that the RIGHT block is the NORTH pole. So the magnetic field lines follow from right to left.

The next part is the simple right hand palm rule, as you know the + and - terminals, follow the current lines until the right option works. In this case, B is the only one that rotates clockwise. (Left side coil current goes up (so thumb up!) and magnetic field right to left (so fingers to right) shows the palm upwards and thus it rotating clockwise.

Basically, this question is asking about your ability to figure out electromagnets, As B and D are almost the same, where D's right block is coiled the wrong way!

Hope this helps!

Nice Mnemonic!!

[Blissisignorance]

Heya, so basically the loudspeaker is being used in reverse, as the cone of the speaker is being pushed and that is moving the coil. So therefore it is acting as a generator, because it is transfering mechanical energy to electrical energy. The movement of the coil is creating current in the circuit, as according to Lenz's Law. Then you use right hand grip rule to find that current travels from X to Y. Hope this helps!!

But since the current is induced to oppose motion of speaker, shouldn't it be from Y to X then?

[Ralopsi]

What is the necessary explanation for why doping changes the electrical properties of semiconductors? Do i need to talk about donor/acceptor levels or is that unecessary?
Thanks for your help

[Fergus6748]

What is the necessary explanation for why doping changes the electrical properties of semiconductors? Do i need to talk about donor/acceptor levels or is that unecessary?
Thanks for your help

Heya, so it would depend on the question. If it's a 4-6 marker, you would need to describe what doping is and what p-type and n-type semiconductors are and how that it affects the electrical properties of semiconductors. You should also talk a bit about Undoped semiconductors as a comparison. You will need to talk about valence electrons and how that creates 'holes' or adds electrons which boosts the movement of electrical charge. Hope this helps!!

[fkkiwi]

Is there a good explanation for this? The HSC solutions don't really help

[Divayth Fyr]

Is there a good explanation for this? The HSC solutions don't really help

We have to look at the effects of the two fields on the proton, and then look at how they will combine.

Electric Field
Remember that the electric field lines show the direction that a positive charge will move in them. As the particle is a proton, it will move in the direction of the field (to the left).

Magnetic Field
This part is a little bit more difficult. We need to use the right hand palm rule to see how it will interact with the field (right because it is positive). Using this, we can see that it will move out of the page, meaning that it is accelerating out of the page. This means that it will have a velocity out of the page. BUT, it already has a velocity going up the page. This means that the total velocity will be a combination of these two vectors, meaning it will be sort of diagonal. However, this is a charged particle moving in a magnetic field. So, this diagonal velocity will have a force applied on it causing it to gain an even greater diagonal velocity. No matter where it is, the force is pointed towards the center; it's a centripetal force. This means that the particle will undergo uniform circular motion (a fancy way of saying it will go in a circle).

It's kind of difficult to explain in a paragraph, so you should look at this picture: https://files.mtstatic.com/site_4539/8298/0?Expires=1540985526&Signature=Lwcy7Hau7aSkOnD1W1Pqon~3uyalcn4MNngMGnmRi~SNUc~HvHA41Z9mHW-JfC48UanNzky5D7rFMEHIkQ4pwxAANbACbIKCtMrHWivfUt8nXDH7s9-UlIYEgdfEkvlqk-rpzC78EUSF7qvrul-fTlrpmekHTEXnq22~pfHNopg_&Key-Pair-Id=APKAJ5Y6AV4GI7A555NA

Keep in mind that the picture shows an electron in the magnetic field, so you need to use the left hand palm rule in order to find the force on it.

Both Of Them Together
So, we have a particle that is moving to the left and going around in circles (starting with going out of the page). The velocity going to the left has no effect on the circular motion, as that is parallel to the magnetic field: This means that it will create a helix shape, as the circles are being made as it moves forward.

Once again, it is difficult to explain in a paragraph, so you should look at this animation of it: https://www.youtube.com/watch?v=a2_wUDBl-g8

I hope this helps.

[phunky]

haha thanks for the nansocks analogy!!
I have no idea how to do this question??

[Divayth Fyr]

haha thanks for the nansocks analogy!!
I have no idea how to do this question??

The solar cell uses the photoelectric effect, so you're looking to find the energy value for which electrons will be able to jump the gap and complete the circuit. This means you're using:



Remember that as intensity increases, more electrons will be able to bridge the gap. So, you want to pick an intensity around a wavelength of 10 micrometers. We need the frequency of that, so we use:



Put frequency into the equation, and then divide by \( 1.602\times10^{-19} \) and you end up with a value of \( 0.12 \) . However, we chose a value that was close to the maximum intensity. So this means that our answer is slightly off. Thus, we chose the closest one so the answer is B.

[not a mystery mark]

Hey, coming at you with a Projectile Motion question: Why is this happening?

"A projectile is fired from the top of a 120m high cliff at 25m/s. It lands on the ground 6.4s after firing."
a) find the Intial Vertical Velocity (u_y)

So my working is as follows:



...Which is not correct. Infact - that's the final vertical velocity... what the heck.
Even weirder. When I do this the answer is "correct"?!?!?


My mind is McBlown. It seems that 'v' and 'u' have switched? Pls explain what is up. Thank you, I love you.

[jamonwindeyer]

Hello my friend you are firing from a 120m high cliff, and landing on the ground. Your vertical displacement is downwards - It is negative. I think if you make \(s=-120\) in both those equations, you'll be MindUnBlown

(the reason this swap happens is interesting, and I can discuss it if you like!

[not a mystery mark]

Hello my friend you are firing from a 120m high cliff, and landing on the ground. Your vertical displacement is downwards - It is negative. I think if you make \(s=-120\) in both those equations, you'll be MindUnBlown

(the reason this swap happens is interesting, and I can discuss it if you like!

I hate that it was that simple ahahah thank you heaps. Also a discussion would be hella neato and interesting

[jamonwindeyer]

I hate that it was that simple ahahah thank you heaps. Also a discussion would be hella neato and interesting

Sweet! Okay, so it's pretty much about symmetry. Picture the path as defined in the question - It spends a certain amount of time (call it \(T_1\)) being accelerated downwards on its way up, which sets the velocity to 0 at the peak of motion. Then it spends another amount of time (call it \(T_2\)) being accelerated downwards to reach a final velocity.

In your mistake, you swapped the scenario - You started at the bottom, and you wanted to land on the cliff. To do that, you would be tracing your path in reverse. But to make sure you get to the same peak, you'd need to start with the vertical velocity you finished with in the above scenario, since you know that is how much will be taken away on the ascent. You spend \(T_2\) time getting accelerated the same as before, but now it acts against you - So to end up at \(v_y=0\), you need to start with that higher value of velocity. Then it is similar on the way down - Before, you were stopped after \(T_1\) seconds, so now doing it in reverse, you end up with the same velocity you started with in the first scenario.

Okay, that was actually tough to explain without a diagram - Maybe it made some sense? Maybe not... Haha if anyone wants to jump in and clarify feel free

[jamonwindeyer]

Hello everyone! I know there have been a couple of questions already, but just putting this here as a marker for the new syllabus which kicks in for 2019 exams and beyond. Everything before this is still useful - There is actually a lot of overlap with the old course (Physics lucked out and got the least changes in content out of the three sciences), but just keep in mind the course is a little different now!

Keen to see this thread be just as much of a collaborative resource for the new course as it was for the old

[david.wang28]

Hi guys,
I'm having trouble with a kinematics question which I have posted in the link. Can I get a detailed reply please? Thanks

[fun_jirachi]

5. a) Remembering that s=ut, and converting 75 km/h to SI units, you get that he travelled

b) Since we're given that the driver comes to a stop a few millimetres away from the hazard, the assumption is that this is negligible, and so we subtract our answer from part a) from 50 since that's the distance he spends braking after his reaction. So we get 43.5m
c)
Sub in your known values for s, v and u (which should be 43.5, 0 and 75/3.6 respectively) to get your t, which should be 4.2s.
d) You can use either s=ut+(at^2)/2 or v=u+at, since acceleration is constant, but use the second one, its so much easier. You should get -5.0m/s/s.

Hope this helps