HSC Physics Question Thread

[jamonwindeyer]

Using S=D/t does pop around in the HSC I think. It's pretty rare though because it targets high band 5 in my opinion - have to know when to apply what formula, and in what order to apply them.

Definitely agree with that for standard projectile questions, know of a question like this (using it with relativity) appearing in any Trial exams? It's definitely never popped up in the HSC 

[RuiAce]

Definitely agree with that for standard projectile questions, know of a question like this (using it with relativity) appearing in any Trial exams? It's definitely never popped up in the HSC 

Aha ok maybe it wasn't there in the long run (pun intended but for wrong subject but you should know what i mean lol!)

I do remember seeing them out of somewhere though. My physics teacher had to go over several of them. Can't remember which paper though.

[jamonwindeyer]

Aha ok maybe it wasn't there in the long run (pun intended but for wrong subject but you should know what i mean lol!)

I do remember seeing them out of somewhere though. My physics teacher had to go over several of them. Can't remember which paper though.

Be careful mate, you are close to the long run shut down condition 

Ahh okay fair enough!! There you go, great question Loki, definitely one worth understanding if you are aiming high! 

[Loki98]

Thx for the help. It all makes sense now

[brontem]

What is photocurrent and why does it depend on intensity?? Or is it another deliberately brushed over/unimportant thing??

[RuiAce]

What is photocurrent and why does it depend on intensity?? Or is it another deliberately brushed over/unimportant thing??

Recall from the definition of current that it is a measure of flow of electric charge. They are related by the formula q=It or equivalently I=q/t.

Clearly, when more electrons are present, the charge will be greater. If the amount of charge is greater, then the rate that charge flows will also increase.

But what is intensity? The intensity of the photocurrent IS by definition of electrons that flow out. Since a greater intensity is equivalent to saying greater charge, it also means a greater current.
(We just call it photocurrent because we are interested in the photoelectric effect. This means we are interested in photoelectrons - those that are emitted off the surface of the metal.)

[jamonwindeyer]

What is photocurrent and why does it depend on intensity?? Or is it another deliberately brushed over/unimportant thing??

I think I interpret your question differently than Rui, do you mean dependent on the intensity of incoming light? Saying "intensity of current" is a little bit incorrect, current doesn't have an intensity per say, but if you did use that term for current it would be in reference to the size of said current, measured in amperes 

As to why the size of a photocurrent is dependent on the intensity of light, we turn to the photon model of light. A greater intensity of light means that more photons are striking the surface of our metal! This results in a larger amount of photo-electrons emitted from the surface, as each photon (with a high enough energy) will release a single photoelectron. Let me know if you need this explained a little more! 

From here, Rui has you covered! More emitted electrons means a larger photocurrent, by definition 

[smiley2101]

Hi! If v=IR why does increasing voltage decrease power loss if power loss =I^2R and according to v=IR voltage is proportional to current? Thank you!

[RuiAce]

Hi! If v=IR why does increasing voltage decrease power loss if power loss =I^2R and according to v=IR voltage is proportional to current? Thank you!

This mistakenly assumes that resistance is constant. In fact, power is what is constant!

Because P=VI, we have V=P/I. So there is actually an inverse proportionality between voltage and current.

[jamonwindeyer]

This mistakenly assumes that resistance is constant. In fact, power is what is constant!

Because P=VI, we have V=P/I. So there is actually an inverse proportionality between voltage and current.

Exactly, when we talk about increasing the voltage to reduce power loss, we are assuming that the same amount of power is being transmitted.

As an example, let's consider 1 Megawatt of power transmitted through a line resistance of 1 kilo-ohm at 100V and 1 MV respectively, and look at the power loss:




We can see that for 100V (top line), transmission is impossible. We can't transfer that much power through that resistance at that voltage, because too much power is dissipated in the line. The second line is much nicer, only 1000 Watts of power loss (this means the system is 99.999% efficient). This sort of calculation epitomises why we use high voltages for transmission 

[jakesilove]

Fucking lol

[conic curve]

Fucking lol

how did he do that?

[jamonwindeyer]

how did he do that?

Do what? 

[RuiAce]

Fucking lol

Maybe we need a "rule" on scientific-notation

[jamonwindeyer]

Maybe we need a "rule" on scientific-notation

If I want to emphasise the size of a number, I don't think the index does it justice. Like, 100 Billion FREAKING WATTS MATE! I like the "oomph"