Physics Predictions/Thoughts (1 Viewer)

carrotsss

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yeah that's fair - so similar to chemistry this year? or chem last year?

anyway to get better at calculations?
maybe similar to chem last year? it’s nearly impossible to predict

for getting better at calculations the main this is just practicing and knowing all of the fundamental concepts, there are only really so many types of questions they can ask so if every time you can’t do one you revise it you’ll be confident in all of them in no time
 
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Do you think they could ask us to design an experiment like that weird 2023 independent question on a circular motion experiment
 

SB257426

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in the same vid science ready posted. he says it after saying the old one. the new is something like: a current of one ampere is one coloumb of charge passing through a given point in time
ah right.


the definition he gave is just another definition of the current but is not the SI definition of the ampere. Its the SI definition of the ampere that can be compared to newtons third law.
 

SB257426

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is it likely to be hard coz of calculations or coz hard theory?

What are some examples of "hard theory" q's?
calculations and theory both will be harder. Last years paper didn't delve into the concepts as deeply and calculations were too basic. NESA def gonna put some harder stuff for sure.

If they dont..... thats good
 

Trippzy

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do u reckon there will be an even split between calls and theory? or will 1 dominate
ah right.


the definition he gave is just another definition of the current but is not the SI definition of the ampere. Its the SI definition of the ampere that can be compared to newtons third law.
regardless I don't think they gonna ask such a niche part of the syllabus lol
 

Yueyue

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can someone please tell me what the mod 7’s definition of one second is I’m having trouble finding the proper wording for it
 

carrotsss

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can someone please tell me what the mod 7’s definition of one second is I’m having trouble finding the proper wording for it
I don’t think it’s really covered in the HSC syllabu but I think the current definition is based on the half life of caesium, and previous ones were based on the rotation of earth
 

WickedFawn

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How does Planck's theory solve the UV catastrophe. Like I get how the intensity of the light at set wavelengths is dependent on the number of quanta which have that energy but why is it that there's such a low amount of quanta/photons which have a super short wavelength thereby reducing the amount of radiation emitted by a BB at short wavelengths.
 

Rattlehead15

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How does Planck's theory solve the UV catastrophe. Like I get how the intensity of the light at set wavelengths is dependent on the number of quanta which have that energy but why is it that there's such a low amount of quanta/photons which have a super short wavelength thereby reducing the amount of radiation emitted by a BB at short wavelengths.
Ive wondered that too it was in a past hsc paper but the sample answer just said since waves are continuous energy transfer and quanta are discrete it solves the UV catastrophe. So idk if you have to go any deeper than that but yea Im also wondering what that actually means.
 

ExtremelyBoredUser

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How does Planck's theory solve the UV catastrophe. Like I get how the intensity of the light at set wavelengths is dependent on the number of quanta which have that energy but why is it that there's such a low amount of quanta/photons which have a super short wavelength thereby reducing the amount of radiation emitted by a BB at short wavelengths.
Planck's quantization of energy resolved the ultraviolet catastrophe problem because it prevented the infinite increase in energy at short wavelengths. When you quantize energy, basically it means that radiation cannot have just any energy level; it must have energies that are multiples of a fundamental quantum, which is determined by Planck's constant. As a result, at short wavelengths, the number of available energy levels (and, by extension, the number of photons with that energy) is limited. This leads to a decrease in the number of photons with very high energies at short wavelengths.

Think of stuff now like statistical mechanics, Q-physics and so on when you write about this, you can think of the transition from classical to quantum as from continuous to discrete. Think about the flaw of Rayleigh's law, the wave model of light etc. it will make much more sense how classical mechanics became unreliable for explaining blackbody emission and so on. Prior to this, the classical models were thought to be fool-proof, the whole "we almost know everything we need to know" but this kind of was the stepping stone for development of Q-physics.

This is from what i recall btw lol
 
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ExtremelyBoredUser

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Ive wondered that too it was in a past hsc paper but the sample answer just said since waves are continuous energy transfer and quanta are discrete it solves the UV catastrophe. So idk if you have to go any deeper than that but yea Im also wondering what that actually means.
Its crucial you have an in-depth knowledge of something as important as the quantisation of energy ESP. for discussion question, its literally a wild card in the sense you can weave it into so much Mod 7 questions while maintaining relevance/brevity due to how fundamental it is.

HSC answers aren't there to give you the "best answer" but moreso a vague/general descriptor of what consists of a good answer. You should make sure you're responses meet the following points:

- Concise (Don't waffle, this isnt Common Mod)
- Cites mathematical laws
- Utilises scientific jargon
- Clear (having clarity is just as important to being concise, be specific)
 

WickedFawn

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As a result, at short wavelengths, the number of available energy levels (and, by extension, the number of photons with that energy) is limited. This leads to a decrease in the number of photons with very high energies at short wavelengths.
Sorry if you actually explained this already but I feel like I'm still missing the point here. I understand that the number of available energy levels has to be limited because quantum theory means that there has to be an integer amount of photons with that energy, but why does it have to be low? Or is the fact that it's low more to do with the intensity of radiation and peak wavelength of radiation is dependent on the temperature of the radiator and bc it releases a continuous spectrum of light across all wavelengths the intensity has to shoot down dramatically from it's peak to zero?
 
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Planck solved it via a "mathematical trick" rather than energy being quantised as this would have been a deparature from classical phys (but he didn't know this was actually true). As to answer ur question, pretty much the probability of a quantum of energy being emitted would also approach zero as wavelength approaches zero. I dont think you need to go any further than that.
 

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