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March 29, 2024, 03:03:47 am

Author Topic: VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions  (Read 15500 times)

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criminals for diminals

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Re: VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions
« Reply #30 on: November 13, 2019, 07:17:32 pm »
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Does anyone know what the graphs were supposed to look like in terms of line of best fit? like were the lines bendy or curved? That question was wack  :o

Erutepa

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Re: VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions
« Reply #31 on: November 13, 2019, 07:28:40 pm »
+1
Does anyone know what the graphs were supposed to look like in terms of line of best fit? like were the lines bendy or curved? That question was wack  :o
I think they wanted you to draw 2 lines of best fit: one for 0-60mm which was for just spring A and another from 60-80mm which was for spring A+B
Both of them would be linear
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Ninjamagics

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Re: VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions
« Reply #32 on: November 13, 2019, 07:41:55 pm »
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I did spring B-spring A for k of b
« Last Edit: November 13, 2019, 08:09:20 pm by Ninjamagics »

criminals for diminals

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Re: VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions
« Reply #33 on: November 13, 2019, 08:04:33 pm »
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I think they wanted you to draw 2 lines of best fit: one for 0-60mm which was for just spring A and another from 60-80mm which was for spring A+B
Both of them would be linear
Thank you!   ;D
Some of my smart friends said the line was curvy, and that you had to do it like a further maths least squares regression line or something.
Anyway, thanks for your answer :) , its made me feel a little better about the exam!

JohnWright

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Re: VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions
« Reply #34 on: November 13, 2019, 11:24:58 pm »
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Hey how did you guys do the double slit question to find the frequency? (Can’t remember the question very well but I was a bit stumped and ended up doing something, don’t know if it was right)

Erutepa

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Re: VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions
« Reply #35 on: November 14, 2019, 07:45:12 am »
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Hey how did you guys do the double slit question to find the frequency? (Can’t remember the question very well but I was a bit stumped and ended up doing something, don’t know if it was right)
I can't quite remember the question either but what the question was assessing was the knowledge that at local maximums (so the points labeled on the graph) the path difference is a multiple of the wavelength. At point p3 (I think that's what it asked) the path difference must be 3 times the wavelength in order for constructive interference to occur and since they give you the path difference you can solve for the wavelength and then obtain the frequency.
To clarify:
The path difference is the difference in distance light from each slit has to travel to one point. At each of the local maximums, the path difference needs to be a factor of the lights wavelength so that the waves constructively interfere (crests align with crests, troughs with troughs). At p0, the path difference is 0 (or 0*wavelength) and p1 the path difference is 1*wavelength and do on for all the values of p.
It then follows that local minima occur between these points and have path differences of multiples of half the wavelength so that troughs align with crests and crests align with troughs to result in destructive interference.

Hopefully with clears things up, but feel free point out anything I didn't explain very well.
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3086

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Re: VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions
« Reply #36 on: November 14, 2019, 09:04:44 am »
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The question was that from one slit the distance to p3 is 80.6 cm and from the other slit was 76.1 cm and so find the frequency of the wave.

So what I did was I first converted them to m so 0.806 and 0.761. Then I did the following

1. 0.806-0.761= 0.045 m (this is the path difference imo)

2. So pd=n*lamda ( since it said p0, p1, p2, p3 are all maxima which insinuates they are all bright bands)
Therefore 0.045=3*lamda (it asked from p3)
0.015=lamda

3. C=F*lamda
F=C/lamda
F= 3.0*108/0.015

So F= 2.0*1010 Hz

The numbers may not be exact. I don't remember exactly which numbers (0.761 maybe 0.723) but the method is correct .

« Last Edit: November 14, 2019, 10:00:58 am by 3086 »
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Theodric_Ironfist

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Re: VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions
« Reply #37 on: November 14, 2019, 09:20:26 am »
+4
I have some answers here. Please let me know if you agree/disagree with anything.

https://filebin.net/8tdp5nycdwy46qol

Update 1: Question 6c - Place the transformer at the end of the 200 m lead.
« Last Edit: November 15, 2019, 07:56:10 am by Theodric_Ironfist »

3086

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Re: VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions
« Reply #38 on: November 14, 2019, 09:42:56 am »
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Question three multiple-choice is B (so the arrow is towards the right). This is because a negative charge has arrows going into it not out of it, as illustrated by your diagram so it will be right.
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3086

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Re: VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions
« Reply #39 on: November 14, 2019, 09:49:59 am »
+5
I think for question 15 in multiple-choice the answer should be A (so fringe spacing increases as velocity increases). This is because velocity is directly proportional to the wavelength (so if you had c=F*lamda and you rearranged it would be lamda = c/F which means velocity is directly proportional to wavelength) and then the fringe spacing formula has x=lamda*L/d so lamda is directly proportional to that, therefore as velocity increases so does the fringe spacing.
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Ennezedo

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Re: VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions
« Reply #40 on: November 14, 2019, 10:21:43 am »
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I have some answers here. Please let me know if you agree/disagree with anything.

https://filebin.net/8tdp5nycdwy46qol
I think all of your MC are right, I'm pretty sure for 6c in the short answer they will be looking for people to say that the transformer should be moved to be used after the extension lead.

Joyn02

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Re: VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions
« Reply #41 on: November 14, 2019, 10:26:06 am »
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How do you guys remember what you got for multiple choice??

Also that last question absolutely massacred me 😥
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MubMurshed

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Re: VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions
« Reply #42 on: November 14, 2019, 10:47:31 am »
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Question three multiple-choice is B (so the arrow is towards the right). This is because a negative charge has arrows going into it not out of it, as illustrated by your diagram so it will be right.

I’m pretty sure the question asked what would be the force in the negative particle and not the electric field, hence it would be to the left, just by looking at the forces on it due to the other particles based on their charge.

I think for question 15 in multiple-choice the answer should be A (so fringe spacing increases as velocity increases). This is because velocity is directly proportional to the wavelength (so if you had c=F*lamda and you rearranged it would be lamda = c/F which means velocity is directly proportional to wavelength) and then the fringe spacing formula has x=lamda*L/d so lamda is directly proportional to that, therefore as velocity increases so does the fringe spacing.

V is inversely proportional to wavelength based on de Broglie wavelength of electrons, h/mv. Using c/f is not the way since c is a constant value, and speed is not dependent on wavelength or frequency, only the medium.
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Joyn02

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Re: VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions
« Reply #43 on: November 14, 2019, 10:53:43 am »
+1
Anyone have an estimate at what the A+ and A cutoffs will be this year?

It felt harder than 2018 for me but idk if that’s just cuz I was under more pressure due to exam conditions.
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3086

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Re: VCE Physics Exam - 13/11/19 - Discussion/Questions/Solutions
« Reply #44 on: November 14, 2019, 10:57:00 am »
+4
I’m pretty sure the question asked what would be the force in the negative particle and not the electric field, hence it would be to the left, just by looking at the forces on it due to the other particles based on their charge.

V is inversely proportional to wavelength based on de Broglie wavelength of electrons, h/mv. Using c/f is not the way since c is a constant value, and speed is not dependent on wavelength or frequency, only the medium.

It asked for what is the net force I'm pretty sure and by looking at all other forces you have to perform vector addition.

I don't think it mentioned what specifically we had to cover (photons or electrons) so either way would be correct and because it was a mathematical question (ie proportionality is a mathematical concept) you can refer to wavelength and frequency being directly proportional to the velocity. Furthermore, there was a question in the 2019 NHT exam that examined the wave equation clearly indicating that frequency and wavelength had an effect on velocity when we speak of it in a mathematical sense, rather than physical properties andf characteristics.
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