HSC Physics Question Thread

[jakesilove]

Thanks, Q8 makes sense. But for Q9 wouldn't there be an EMF induced due to the changing magnetic flux on the wire to vectorially add magnetic field lines? So how come it's detecting a decrease?

I don't think the question is asking about the CURRENT going through the wire, just the field WITHIN the wire. Note that they haven't added an Ammeter, but some sort of 'field detector'. Since none of the answers have to do with current, we know that it's not really a Lenz's law question

[FallonXay]

I don't think the question is asking about the CURRENT going through the wire, just the field WITHIN the wire. Note that they haven't added an Ammeter, but some sort of 'field detector'. Since none of the answers have to do with current, we know that it's not really a Lenz's law question

Yes, but wouldn't there be an induced current which creates extra magnetic field lines around/ within the wire?

[jakesilove]

Yes, but wouldn't there be an induced current which creates extra magnetic field lines around/ within the wire?

Ah okay, I see your point. There was a decrease (purely from the Magnetic field within the wire changing), however Lenz's law should result in an increase, to counteract the change!

The thing to know is that the field induced by Lenz's law will not make up for the original change. I'm not sure if you do Chemistry, but it's sort of like Le Chatelier's principle if you do. If you change the flux through a loop, the current will try to reverse that change, but not so strongly as to actually reverse it. Sorry, I understand your point now, and I hope you understand my explanation (well, it's less of an explanation, more of a 'this is just how it is')!

Jake

[FallonXay]

Ah okay, I see your point. There was a decrease (purely from the Magnetic field within the wire changing), however Lenz's law should result in an increase, to counteract the change!

The thing to know is that the field induced by Lenz's law will not make up for the original change. I'm not sure if you do Chemistry, but it's sort of like Le Chatelier's principle if you do. If you change the flux through a loop, the current will try to reverse that change, but not so strongly as to actually reverse it. Sorry, I understand your point now, and I hope you understand my explanation (well, it's less of an explanation, more of a 'this is just how it is')!

Jake

ahh ok, makes sense - thanks!

[jakesilove]

Hi!
Could someone please explain how to answer these questions?

(The Answer is A for Q9; and B for Q8; and C for Q14)

Thanks.

For your final question, I'm getting the answer of C when I divide the energy of the photoelectron by the distance between the plates. This would almost suggest that



Where E is the energy of the field required to exactly oppose the energy of the photo electron. This doesn't make any sense though, so hopefully someone else can step in and help us out! Either the multiple choice is wrong, or I'm not thinking of it correctly.

[jakesilove]

ahh ok, makes sense - thanks!

No worries! Sorry that I can't take a proper crack at that last one!

[jamonwindeyer]

Hi!
Could someone please explain how to answer these questions?

(The Answer is A for Q9; and B for Q8; and C for Q14)

Thanks.

Hey! Your last question is deceptive, you'd think it wanted Photoelectric Effect stuff, but do you remember this little formula from Year 11?



That's what we need, work is equivalent to the force applied over distance. The amount of work the voltage does will need to be equal to the kinetic energy of the photoelectrons (given in the last column. We know the distance over which it is done, 5mm. So:



Nasty question! Goes to show, it is important to know your Prelim results (this formula is on your reference sheet as well) 

Edit: So Jake's calculation was actually inadvertently correct

[FallonXay]

Hey! Your last question is deceptive, you'd think it wanted Photoelectric Effect stuff, but do you remember this little formula from Year 11?



That's what we need, work is equivalent to the force applied over distance. The amount of work the voltage does will need to be equal to the kinetic energy of the photoelectrons (given in the last column. We know the distance over which it is done, 5mm. So:



Nasty question! Goes to show, it is important to know your Prelim results (this formula is on your reference sheet as well) 

Edit: So Jake's calculation was actually inadvertently correct

No worries! Sorry that I can't take a proper crack at that last one!

mm, I see. Thanks a ton, your answers are very appreciated! 

[FallonXay]

Hello, again!

I've got a few more questions:

With Q9, I understand that the answer should be either A or B due to the nature of alternating current, however, how can you tell whether the current is negative or positive? (i.e why isn't that answer A; Answers say it is B)

and I don't understand how to derive the answer for Q3.

Thanks once again.

[Alexander23]

Anyone know why some lines in the hydrogen spectrum brighter than others? And why couldn't Bohr explain this?

Thanks!

[RuiAce]

Hello, again!

I've got a few more questions:

With Q9, I understand that the answer should be either A or B due to the nature of alternating current, however, how can you tell whether the current is negative or positive? (i.e why isn't that answer A; Answers say it is B)

and I don't understand how to derive the answer for Q3.

Thanks once again.

That generator question is very dodgy. My guess would be that if we incorporate Lenz's Law in, since the induced EMF always seeks to oppose the cause of induction using -sin(x) is more appropriate to +sin(x). It does say 'most accurately' represented.

[jamonwindeyer]

Hello, again!

I've got a few more questions:

With Q9, I understand that the answer should be either A or B due to the nature of alternating current, however, how can you tell whether the current is negative or positive? (i.e why isn't that answer A; Answers say it is B)

and I don't understand how to derive the answer for Q3.

Thanks once again.

Hey! Normally, we'd use the right hand rule to determine the direction of induced current in the wire, and that would tell us whether it was 'positive' or 'negative.'

However, that relies on the question defining what positive and negative actually is. This question doesn't do that! To me, either A or B could be correct and I don't see a way of distinguishing, you don't have enough information, you need to be told which way they define as positive current flow

[Spencerr]

Anyone know why some lines in the hydrogen spectrum brighter than others? And why couldn't Bohr explain this?

Thanks!

Bohr's atomic model consisted of electrons orbiting around the nucleus of an atom is discrete energy levels or shells that were a specific distance away from the nucleous. (this is his first postulate that electrons orbited in "stationary states"). His second postulate was that if an electron moved between discrete energy levels, (either moving to a higher one or to a lower one), an emission or absorption of a photon (E=hf) that has an equivalent energy to the energy difference between the two levels, is required. An emission of a photon of emr occurs when  electrons fall from a higher energy level to a lower energy level. Since E=hf, we can determine the frequency of the photons. The hydrogen spectra line series shows exactly this, the falling of electrons from different higher energy levels to the second lowest energy level. Their relative intensities suggest that some electron falls or electron jumps are more favoured than other electron jumps. (as more electron jumps = more identical photons=brighter/sharper lines on the hydrogen spectrum). Now this was a probabilistic nature describing the behaviour of electrons and Bohr's model did not factor this in. Hence he could not explain the relative intensities

[FallonXay]

Hiya, why is the answer for this question D?

[Spencerr]

Hey there,  since I'm online I'll give you a quick explanation. 

Weight is defined as the force experienced by an object when placed inside a gravitational field. As you know large celestial bodies like the moon and the earth have their own gravitational field that is directed towards the centre of the body. Midway between the earth and the moon, an object will experience both a force of attraction towards the centre of earth from earth's gravity and another force of attraction towards the centre of the moon from the moons gravity however these two forces oppose each other, acting in separate directions.  Hence the gravitational fields cancel out leaving the object near weightlessness (due to the absence of a NET FORCE). The other answer that might seem right is if the object is in orbit but remeber that at that point the object is still being subjected to a weight force or attractive force which forms the centripetal force keeping it in orbit. Feel free to ask any questions if you need me to clear things up

Ps is this a hsc paper? Or a schools paper. I've done this question before..