Physics - Got Some Questions (1 Viewer)

[idkkdi]

Q
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Need some reasoning. I don't see how the voltage should appear constant in X when the switch is turned off.

A.
For the duration you close the switch, the circuit is connected and hence a current flows through, thereby a constant voltage as well.

 

[Life'sHard]

A.
For the duration you close the switch, the circuit is connected and hence a current flows through, thereby a constant voltage as well.

Answer was C. So what you're saying is that even though the switch is turned off (no voltage running through), there is still a current running since the circuit is still connected? But isn't this a voltage time graph not a current time graph? or does it not matter since V=IR and they are directly proportional and interchangeable?

 

[idkkdi]

Answer was C. So what you're saying is that even though the switch is turned off (no voltage running through), there is still a current running since the circuit is still connected? But isn't this a voltage time graph not a current time graph? or does it not matter since V=IR and they are directly proportional and interchangeable?

my bad, im retarded. typo.
it's C.

 

[Life'sHard]

my bad, im retarded. typo.
it's C.

OH NVM lmao I'm actually retarded. When it says closes the switch I read it as they turned off the voltage nvm i got this question ffs lmao.

 

[notme123]

I hate how you had to assume it's DC cuz then it's not a transformer. Hopefully, the other graph is clear. It's just Lenz's Law

 

[idkkdi]

I hate how you had to assume it's DC cuz then it's not a transformer. Hopefully, the other graph is clear. It's just Lenz's Law

I don't understand the curves for when the switch opens and closed at all in the second graph at all.


The second graph isn't even the first differentiated.
I'm thinking if I had to draw in an exam, I would be whipping out straight lines.

 

[notme123]

I don't understand the curves for when the switch opens and closed at all in the second graph at all.


The second graph isn't even the first differentiated.
I'm thinking if I had to draw in an exam, I would be whipping out straight lines.

It's because the immediate change in flux causes a temporary voltage in the other coil in the other direction (because the change in flux is 0 after a few seconds.)

I chose it because its the only one that makes sense

 

[ExtremelyBoredUser]

What's been the hardest topic for you guys in Year 12 Physics? Out of curiosity.

 

[notme123]

What's been the hardest topic for you guys in Year 12 Physics? Out of curiosity.

When i first went through the syllabus probs special relativity but after going over it again it has to be stuff like induction maybe. Sure, regurgitating the same explanation of Lenz Law is not hard but there are some hard qs out there. The easiest is probs projectile motion or the rest of mod 7 except for sr.

Mod 8 is just one big bore. Not hard by any means just tedious asf and theyre SUPER vague in terms of depth about whats in the syllabus. I haven't seen a single concrete explanation for what the weak force is in any hsc resource except for 'it interacts with nucleons' and crap about Z bosons or whatever. Meanwhile, I could just google a perfectly good definition online, mediates how a quark in a nucleon swaps its flavour for another quark. It even showed a Feynman diagram of beta decay which is extremely easy to understand. Why is this not mentioned in the syllabus? Who knows?

Don't get me started on the life cycle of a star. Most of the online resources I believe go into too much depth and there's just too much memorizing that will probs go to waste because stars aren't asked that much in the hsc.

 

[ExtremelyBoredUser]

When i first went through the syllabus probs special relativity but after going over it again it has to be stuff like induction maybe. Sure, regurgitating the same explanation of Lenz Law is not hard but there are some hard qs out there. The easiest is probs projectile motion or the rest of mod 7 except for sr.

Mod 8 is just one big bore. Not hard by any means just tedious asf and theyre SUPER vague in terms of depth about whats in the syllabus. I haven't seen a single concrete explanation for what the weak force is in any resource except for 'it interacts with nucleons' and crap about Z bosons or whatever. Meanwhile, I could just google a perfectly good definition online, where a quark in a nucleon swaps its flavour for another quark. It even showed a Feynman diagram which is extremely easy to understand. Why is this not mentioned in the syllabus? Who knows?

I see, I just got pearson year 12 textbook so I was flipping through the content. Mod 5 doesn't seem hard from first glance, feels more like a combination of Mod 1,2 with more stuff such as torques or keplers laws. Mod 6 seems pretty cool, the motor effect got my attention. I feel like I'll enjoy Mod 7 the most though since it goes through the quantum model and SR which I wanted to do in Mod 3. Mod 8 seems pretty nice too but I feel like m a bit of a disadvantage to chem students since the topics mostly the structure of the atom and quantum mechanical nature but interesting nonetheless.

 

[notme123]

I see, I just got pearson year 12 textbook so I was flipping through the content. Mod 5 doesn't seem hard from first glance, feels more like a combination of Mod 1,2 with more stuff such as torques or keplers laws. Mod 6 seems pretty cool, the motor effect got my attention. I feel like I'll enjoy Mod 7 the most though since it goes through the quantum model and SR which I wanted to do in Mod 3. Mod 8 seems pretty nice too but I feel like m a bit of a disadvantage to chem students since the topics mostly the structure of the atom and quantum mechanical nature but interesting nonetheless.

Looking back at it mod 7 was probably the most fun to learn. Mod 5 is pretty simple if you get mod 1 and 2 as you said. Motor effect isnt that bad and learning about motors and generators were fun. Just when it came to mod 8 (maybe until midway) it felt a bit slow but maybe you'll like it everyone's different.

 

[idkkdi]

It's because the immediate change in flux causes a temporary voltage in the other coil in the other direction (because the change in flux is 0 after a few seconds.)

I chose it because its the only one that makes sense

I'm referring to the concavity of the curves, not the curves themselves. why not the above drawn?

 

[notme123]

View attachment 31994
I'm referring to the concavity of the curves, not the curves themselves. why not the above drawn?

could have to do with start and ends being non-linear. I think the main point they were assessing was the negative part of faradys law and the fact no change in flux causes no induced voltage.

 

[idkkdi]

could have to do with start and ends being non-linear. I think the main point they were assessing was the negative part of faradys law and the fact no change in flux causes no induced voltage.

but if we were told to draw the graphs,
not sure what is expected.

 

[notme123]

but if we were told to draw the graphs,
not sure what is expected.

true but i think they would give us a more clear voltage graph to start off with. It wouldn't look weird like that one.

 

[Life'sHard]

Can someone decipher what this means. I know that mass defect of the nucleus is equivalent to the binding energy but what does the rest of the sentence mean?

Edit: Is it saying that mass defect <=> Binding energy of products - binding energy of reactants?

 

[notme123]

View attachment 32003
Can someone decipher what this means. I know that mass defect of the nucleus is equivalent to the binding energy but what does the rest of the sentence mean?

Edit: Is it saying that mass defect <=> Binding energy of products - Binding energy of reactants?

You could say it like that but remember they're different units. It's a bit ambiguous imo because I feel like no one points this out but when you compare the weight of constituent nucleons (broken up into individual protons and neutrons) to the actual mass of the nucleus, the former outweighs the latter. We aren't given really a reason why this occurs (could have to do with gluons and strong force perhaps) but the consequence is that this mass defect i.e. the difference of mass is converted into the binding energy that holds nucleus together. Now in relation to the statement. You can easily swap the 'binding energy gain' and 'mass defect' i.e. how much mass was lost in the reaction and converted into energy according to E=mc^2.
Say for example reactant A has 1u and reactant B has 2u mass and they form product C with 1.5u. The mass defect here is (1+2) - 1.5 = 1.5u. Convert this into MeV by timsing by 931.5 and you get the 'binding energy gain' of the product.
Since the mass defect of the product (1.5u + previous mass defect) is greater than the mass defect of the reactants (not given but inferred), there is 'binding energy gain' in the daughter nucleus

TL;DR 'binding energy gain' is mass defect (mass difference) between product and reactants, while binding energy is mass defect (difference between constituent mass and actual mass) of individual nuclei.

 

[Life'sHard]

Q)


Can someone provide a sample answer, I don't know how to answer this for 4 marks. I can only get up to 3 at best.

 

[idkkdi]

Q)
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Can someone provide a sample answer, I don't know how to answer this for 4 marks. I can only get up to 3 at best.

talk about back emf. etc. e&m -> relate to faraday's/lenz's

A=A0cos(wt)
dA/dt = -wA0sin(wt), dBFlux/dt = -wBA0sin(wt)
-> this part probably not needed.

As w increases, back emf increases.
At some point when back emf = voltage, the motor has a net voltage of 0.
By F = ILB, velocity will no longer increase.

 

[idkkdi]

Q)
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View attachment 32029
Can someone provide a sample answer, I don't know how to answer this for 4 marks. I can only get up to 3 at best.

send marking criteria.
I'm pretty sure they wanted Vnet = 0 for prediction, but even without friction I don't think this is actually true lol.