HSC Physics Question Thread

[winstondarmawan]

From the 2016 HSC:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/23113597_1363981670394019_1360459847_n.png?oh=c3f94f6c6d29cbbf6ba0dc38a8752be5&oe=59F726B1
A: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22894985_1363980263727493_1683153853_n.png?oh=820e57fbde8c1b68d7cea1ed59384be5&oe=59F82BC1
How many plots should their be until it is considered a valid relationship?

[CyberScopes]

From the 2016 HSC:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/23113597_1363981670394019_1360459847_n.png?oh=c3f94f6c6d29cbbf6ba0dc38a8752be5&oe=59F726B1
A: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22894985_1363980263727493_1683153853_n.png?oh=820e57fbde8c1b68d7cea1ed59384be5&oe=59F82BC1
How many plots should their be until it is considered a valid relationship?

I dont think theres a certain amount, but I believe the question wanted to find a flaw/benefit of each data set, and the flaw of A would be that it didnt have as much data as that of B.

[sidzeman]

Ok so we estimate the wavelength is roughly 8um. Thats 8 x 10^-6m, we sub that into E = hf = hc/wavelength and that gives us our energy in joules. Now eV is in joules so we simply divide by the value given on the formula sheet - 1.602x10^-19. And that gives us our value of eV.

Note: the electron needs more eV than what is required by the band gap. So it has to be less than the 0.155eV we get from our working. So it has to be A... ok I'm confused. Where's this from?

The 2012 HSC - but you're correct I'm pretty sure the answers are wrong, especially since the Excel 1 Books have A written down as the correct answer as well

[Checkmate123]

Can someone pls help with Q2Q reactions:

Since the products have higher mass, does this come from the fact that the products have less binding energy so the energy is converted to mass? Also how is energy released in nuclear reactions?

Thanks

Not quite sure about the reactions you are talking about. But, I can  answer on why energy is released in nuclear reactions.
If you look at the mass number vs average binding energy graph, you can see that average binding energy increases up until around Iron, then decreases.
In fission, you are splitting a nucleus into smaller nuclei, such as breaking down uranium into barium and krypton. From the graph, baryon and krypton have higher binding energy than uranium. Since binding energy is related to mass defect, the products of this reaction have higher mass defect than uranium. So, if you were to actually measure the masses of baryon, krypton and uranium, you would see that there appears to be a loss of mass. This 'missing mass' is released in the form of energy by energy-mass equivalence. Similarly, fusion of smaller nuclei to form a large nuclei can also release energy, provided the average binding energy of product is higher than reactants.

Note: Nuclear reactions always release energy provided the binding energy of products is greater than binding energy of reactants.

[Mymy409]

Can someone please help me with Q12 from the 2014 HSC? I don't quite understand what's going on.
https://www.boardofstudies.nsw.edu.au/hsc_exams/2014/pdf_doc/2014-hsc-physics.pdf

[blasonduo]

The 2012 HSC - but you're correct I'm pretty sure the answers are wrong, especially since the Excel 1 Books have A written down as the correct answer as well

Hey, are you able to show me what the excel book said? I believe the answer is B, but I'd like to see why Excel said A

Can someone please help me with Q12 from the 2014 HSC? I don't quite understand what's going on.
https://www.boardofstudies.nsw.edu.au/hsc_exams/2014/pdf_doc/2014-hsc-physics.pdf

Hey! This was answered a few pages back!

https://atarnotes.com/forum/index.php?topic=164552.msg996472#msg996472

Here it is!

[sidzeman]

Hey, are you able to show me what the excel book said? I believe the answer is B, but I'd like to see why Excel said A

 

Hey! This was answered a few pages back!

https://atarnotes.com/forum/index.php?topic=164552.msg996472#msg996472

Here it is!

Sorry I don't have the book on me anymore, so I can't show you (

[pikachu975]

The 2012 HSC - but you're correct I'm pretty sure the answers are wrong, especially since the Excel 1 Books have A written down as the correct answer as well

Should be B! 0.155 eV doesn't mean the semiconductor wont work, as moving slightly away from the peak would give 0.155 eV but still have a pretty high intensity on the curve. Using the semiconductor from A, with like 0.03 eV, there are barely any photons with that energy hence it wouldn't be suitable.

[sidzeman]

Can anyone help me with 19 - should the electrical resistance be high for rapid heating, as high resistance = greater heat by resistive heating? Therefore answer should be D, but is in fact C

[blasonduo]

Sorry I don't have the book on me anymore, so I can't show you (

Just as pikachu said, but i'll use calculations
I say this because of this;

[winstondarmawan]

Would appreciate help with this question (2015 HSC):
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22894854_1364012660390920_933578363_n.png?oh=b87e2f32349fb86ce6953e8bc9eaba08&oe=59F85C98
TIA!

[Zainbow]

Can anyone help me with 19 - should the electrical resistance be high for rapid heating, as high resistance = greater heat by resistive heating? Therefore answer should be D, but is in fact C

Hey! The pot gets heated up when eddy currents are formed in the base. The larger the eddy currents, the more resistance THEY will form and hence the more heat produced. The pot base itself doesn't require high resistivity - in fact, the more electrical resistance it has, the less likely eddy currents will form, knowing that the more electrically conductive the material is, the larger these eddy currents will be.

[blasonduo]

Can anyone help me with 19 - should the electrical resistance be high for rapid heating, as high resistance = greater heat by resistive heating? Therefore answer should be D, but is in fact C

For this we must take in account power loss, P = I^2 x R

Yes, resistance is there, but the current is a much larger factor than resistance for power loss, but how do we increase current?

V = IR

I = V/R

For current it be high, we need Voltage to be HIGH and resistance to be LOW

And that's where we find our answer! to optimise our heat loss, we need current to be at its highest, and it is at its highest when RESISTANCE is low

EDIT

Would appreciate help with this question (2015 HSC):
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22894854_1364012660390920_933578363_n.png?oh=b87e2f32349fb86ce6953e8bc9eaba08&oe=59F85C98
TIA!

19) This is MUCH easier if we rule out the incorrect ones, For A, the force of gravity is NOT negligible, as that's what keeps objects in orbit! so it is wrong. While B is true, this is not answering the question, it is just a statement.

We now have it to C or D, for C, IF the forces were to be the same, (ie F = ma) since, the question claims F is the same, while mass is different, their accelerations must be different, and well, if their accelerations are  different, how on earth can they travel at the same speed? So C is incorrect, leaving only D as the answer (if you would like me to explain why i'll be happy to!)

[khadeeja_]

Hey, are you able to show me what the excel book said? I believe the answer is B, but I'd like to see why Excel said A

 

Hey! This was answered a few pages back!

https://atarnotes.com/forum/index.php?topic=164552.msg996472#msg996472

Here it is!

I checked the excel book to help you guys out..
so it says
A - Max. energy output from human = E=hf
=hc / wavelength
= (6.626x10^-34) x( 3x10^8) / (9x10^8)
= 0.138 eV
Therefore the band gap should be less than 0.138 eV. This makes HgCdTe the best choice.

[Dante1091]

What stuff about Westinghouse and Edison do we have to know? Since I think I know too much irrelevant information about this dotpoint