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(https://i.imgur.com/4S8LMtJ.png)
This thread is for all exam-related discussion. Was it easy? Was it hard? What did you get for each question? Feel free to post any and all of your thoughts below.
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Yeah, spring compression was prety hard, probably lost 2 marks. Also that power loss question in transformers question was pretty dodgy. Everything else was pretty good.
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Yeah totally stuffed up second and third part of spring question
Think answers should have been 0 m/s^2 and 0.05 metres
One of the tranmission questions were dodgey but I think the main thing you had to recognise is that the power was the same as it was in the first scenario
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Yeah the answer for third part was 0.05 m. So stupid mistake. You can use kx = mg.
If the power was the same it would have been pretty easy. But I would have thought the power dissipated would be different from the original scenario. What did you choose for the definition for uncertainty in the multiple choice? I said it was the doubt in the measurement (choice A)?
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Apparently since it says in the question that the globe still operates the same as it did before (lol something like that) power was the same.
Got the same answer for mcq 👍
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Yeah totally stuffed up second and third part of spring question
Think answers should have been 0 m/s^2 and 0.05 metres
I reckon the answer for part b is 9.8 m/s^2 as that's the maximum acceleration due to gravity.
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what did you get for the powerloss questions
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Same as yesterday.
We'll do our best to get the solutions out as quick as we can!
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Yeah that’s what I had but it’s wrong
The question asked what is the acceleration when the ball had its max velocity
Once the ball comes into contact with the spring it is still acccelerating but the value for its acceleration decreases. Once the value of acceleration reaches 0 m/s^2 (when mg=kx) the ball reaches its max velocity
I’m guessing you did the same as me and assumed that as soon as the ball came into contact with the spring it would start slowing down. Unfortunately that’s not true Ahahah
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I feel like, logically, it has to be 0 - if the acceleration is positive, then it hasn't hit its max velocity. If it's still accelerating, it's increasing velocity, and it can't be at maximum velocity if its velocity is still increasing.
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What about part a of the spring question? Says to show that k = 392. I used to mg=kx and got k=39.2. Possible typo?
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Nah had to use energy equations
Mgh (top) = mgh (bottom +1/2kx^2 and solve for k
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No vcaa are never wrong :)
mg∆h = 1/2kx^2
2 x 9.8 x 2.5 = 1/2 x k x 0.5^2
rearranging it to find k gives 392 xD
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No vcaa are never wrong :)
mg∆h = 1/2kx^2
2 x 9.8 x 2.5 = 1/2 x k x 0.5^2
rearranging it to find k gives 392 xD
But it says when they come to rest. Then the forces would be balanced and mg=kx. Unless they meant "momentarily" at rest, in which case they should have specified.
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How did we meant to draw the 3 mark graph of emf ??? idk i drew my like sin
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How did we meant to draw the 3 mark graph of emf ??? idk i drew my like sin
Square wave; zero for the loop outside and inside sections, positive for loop entering, negative for loop exiting.
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do you guys think it was harder or easier than last years? Potential A+ cut offs?
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That exam was a welcome surprise. Explain type questions were my weakness all year, yet there were almost none of them, and none over 3 marks.
How far did the student between two speakers walk to get to the 2nd quiet band? I got the PD of 1.5m, then halved it to get 0.75m walked.
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The exam was probably a bit simpler than last years. Cutoff for A+ might be higher.
I do think that question 5d (second experiment with transformer at 8:1) and the whole of question 6 were badly written though.
Multiple choice was extremely easy.
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That exam was a welcome surprise. Explain type questions were my weakness all year, yet there were almost none of them, and none over 3 marks.
How far did the student between two speakers walk to get to the 2nd quiet band? I got the PD of 1.5m, then halved it to get 0.75m walked.
the wavelength was 1m?
Therefore 10 wavelengths in 10m. at centre it is quiet. walk along 1 wavelength to get 1m?
I got 1 meter from centre
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the wavelength was 1m?
Therefore 10 wavelengths in 10m. at centre it is quiet. walk along 1 wavelength to get 1m?
I got 1 meter from centre
Wavelength is 1 m.
They are standing at 2nd node, so PD = (2-0.5) x 1 = 1.5 m
If they are standing at point P, then AP - BP = 1.5, and AP + BP = 10.
Thus AP = 5.75 m and distance from center is 0.75 m
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i thought its 4th node because they gave you the PD already??
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the wavelength was 1m?
Therefore 10 wavelengths in 10m. at centre it is quiet. walk along 1 wavelength to get 1m?
I got 1 meter from centre
Wavelength is 1m
PD is 0 in the middle
2nd quiet means PD is 1.5*wavelength = 1.5m
Walk 0.75m to get closer to one speaker, and further from the other, creating a PD of 1.5m
0.75m
Could be wrong
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Wavelength is 1m
PD is 0 in the middle
2nd quiet means PD is 1.5*wavelength = 1.5m
Walk 0.75m to get closer to one speaker, and further from the other, creating a PD of 1.5m
0.75m
Could be wrong
I drew 10 sine waves connecting speaker a and b., then reverse for standing waves.
here is what i did. a bit messy Lol
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So the stupid spring question....
It states "the ball and spring come to rest when they are at a distance of 0.50 m below the uncompressed position of the spring."
To me, that sounds as if the ball and spring have stopped oscillating. Thus the forces are balanced, and mg = kx, and thus k = 39.2 N/m.
If it means the lowest point that the ball reaches, then it should have specified MOMENTARILY at rest. In which case you use an energy approach and you end up with k = 392 N/m. But the lack of the word MOMENTARILY changes the question completely.
Not to mention the rest of the question. Are we to assume that now the ball has stuck to the spring and is undergoing an oscillating motion? And we are to analyse the motion of the ball after it has stuck, disregarding the first part of the ball's motion? Because then the question makes sense. BUT NO WHERE DOES IT SAY TO DO THAT!
Rant over, sorry guys.
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My MC Answers please compare and give me a grade! (I am a C+ student)
1.) B
B
A
D
C
C
A
C
B
A
11.) D
D
D
B
C
C
A
B
B
B
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I got nearly the same just 5 qs different
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MC Answers:
B
A
A
D
C
C
B
C
B
D
D
C
D
B
D
C
A
A
B
B
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What did you guys get for the very last question? (if you remember lol)
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What did you guys get for the very last question? (if you remember lol)
This q will vary between students due to gradient calculation.
i god 6.x10^26. dont remember the decimal
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This q will vary between students due to gradient calculation.
i god 6.x10^26. dont remember the decimal
Alright cool, got something similar but thought it was way off hahahah
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Alright cool, got something similar but thought it was way off hahahah
How did u do the speaker question
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My first 5 questions:
1a.) = 5x10^4 Vm^-1
b.) = 1.3x10^6 ms^-1
c.) = 0.57m
2a.) emf = 6.4x10^-4
b.) GRAPH = y position = 0 , (neg) , 0 , (pos) , 0
3a.) Clockwise
3b.) right hand slap rule
4a.) (bit confusing) = 2.8V (Vrms/root(2))??
4b.) GRAPH = (amp and frequency) x2
5a.) no idea... with this question but 72W
5b.) 1.0V
5c.) 72W
5d.) 2W
5e.) she's safer
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My first 5 questions:
1a.) = 5x10^4 Vm^-1
b.) = 1.3x10^6 ms^-1
c.) = 0.57m
2a.) emf = 6.4x10^-4
b.) GRAPH = y position = 0 , (neg) , 0 , (pos) , 0
3a.) Clockwise
3b.) right hand slap rule
4a.) (bit confusing) = 2.8V (Vrms/root(2))??
4b.) GRAPH = (amp and frequency) x2
5a.) no idea... with this question but 72W
5b.) 1.0V
5c.) 72W
5d.) 2W
5e.) she's safer
Looking good so far mate! :P
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How did u do the speaker question
I made a mistake without that question (gotta love exam conditions) but I just said that the second quiet region is the second node which on a standard wave form would be one wavelength away, forgetting that I should've used the pd formula (pd=(n-1/2)lamda) which would've been 1.5m
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I made a mistake without that question (gotta love exam conditions) but I just said that the second quiet region is the second node which on a standard wave form would be one wavelength away, forgetting that I should've used the pd formula (pd=(n-1/2)lamda) which would've been 1.5m
I did that too.
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So the stupid spring question....
It states "the ball and spring come to rest when they are at a distance of 0.50 m below the uncompressed position of the spring."
To me, that sounds as if the ball and spring have stopped oscillating. Thus the forces are balanced, and mg = kx, and thus k = 39.2 N/m.
If it means the lowest point that the ball reaches, then it should have specified MOMENTARILY at rest. In which case you use an energy approach and you end up with k = 392 N/m. But the lack of the word MOMENTARILY changes the question completely.
Not to mention the rest of the question. Are we to assume that now the ball has stuck to the spring and is undergoing an oscillating motion? And we are to analyse the motion of the ball after it has stuck, disregarding the first part of the ball's motion? Because then the question makes sense. BUT NO WHERE DOES IT SAY TO DO THAT!
Rant over, sorry guys.
Difficultiy, confusion or frustration should never be confused with stupidity.
Spring question
(a)
k = 2mg∆h/(∆x)^2
= 2 • 2 • 9.8 • 2.5/ (0.5)^2
= 98/ 0.25
k = 392 N m^-1 (Q.E.D).
(b)
Common answer (reasonable) was 9.8 m s^-2 this is wrong however, actual and final answer is 0 m s^-2 since acceleration-mass system when max speed is reach is 0.
(c)
∆x for max speed. Max speed occurs at midpoint of compression, therefore, ∆x/2 = 0.5/2 = 0.25 m
PD was given so was wavelength, solve for n (n=4.0082),
Speaker
Wavelength = 1m (v=fwavlength).
Transmission questions were misleading since current in lines in normally what you would calculate first and then progress to the power loss questions. Anyhow, this would be answered terrible, overall, harder than last years as expected (first year study design) is always easier, but nevertheless 2018 class is all in this together.
Love, the zero gravity questions :o
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I did that too.
It was only like 2 marks right lolll rip
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what did everyone do for the "Find the Kinetic energy of the electrons" part of question 18?
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I made a mistake without that question (gotta love exam conditions) but I just said that the second quiet region is the second node which on a standard wave form would be one wavelength away, forgetting that I should've used the pd formula (pd=(n-1/2)lamda) which would've been 1.5m
Yeah mate, its quiet because of destrcutive interference (crest meet trough, ampltiude cancel out), therefore, node.
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what did everyone do for the "Find the Kinetic energy of the electrons" part of question 18?
Ek = (lorentz - 1) rest mass • c^2
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Difficultiy, confusion or frustration should never be confused with stupidity.
Spring question
(a)
k = 2mg∆h/(∆x)^2
= 2 • 2 • 9.8 • 2.5/ (0.5)^2
= 98/ 0.25
k = 392 N m^-1 (Q.E.D).
(b)
Common answer (reasonable) was 9.8 m s^-2 this is wrong however, actual and final answer is 0 m s^-2 since acceleration-mass system when max speed is reach is 0.
(c)
∆x for max speed. Max speed occurs at midpoint of compression, therefore, ∆x/2 = 0.5/2 = 0.25 m
PD was given so was wavelength, solve for n (n=4.0082),
Speaker
Wavelength = 1m (v=fwavlength).
Transmission questions were misleading since current in lines in normally what you would calculate first and then progress to the power loss questions. Anyhow, this would be answered terrible, overall, harder than last years as expected (first year study design) is always easier, but nevertheless 2018 class is all in this together.
Love, the zero gravity questions :o
[/quoin the transmission question, was the current given for the lines or for the current the ligh uses after transformer
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Yeah mate, its quiet because of destrcutive interference (crest meet trough, ampltiude cancel out), therefore, node.
Yeah that wasn't the bit I mixed up lmao
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the first part of question 5 was confusing for sure, but i found it easy once I figured out the trick.
You just need to use
I(2)/I(1)=N1/N2
with I1=3, N1/N2=4
then you get I2 (current going to globe) = 3*4->12 A
then just use P=IV to get a power of 48 watts
A quick way to test your answer is to see if the power at the globe and your power loss over the lines add up to your total power
(in this case, 48+24=72)
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Ek = (lorentz - 1) rest mass • c^2
Wasn't that a completely different question?
I meant the one with the two diffraction patterns
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Wasn't that a completely different question?
I meant the one with the two diffraction patterns
Idk if this was the right question but I think I used Ek=p^2/2m, found p from the de Broglie wavelength (which was the same as he wavelength of the x ray)
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think a + cutoff will be lower
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Guys what was the answer to the uncertainity principle mc qs pls tell by whole 20/20 depends on it :c
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think a + cutoff will be lower
Defeniately. I think uncertainity was B (cannot really remember only did it a couple of hours ago).
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Yo why for the two speakers question r people using path difference. That’s what I initially thought but then I went with the fact that a standing wave would have formed and the quiet regions would be where nodes exist
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Yo why for the two speakers question r people using path difference. That’s what I initially thought but then I went with the fact that a standing wave would have formed and the quiet regions would be where nodes exist
Same Did you get 1 meter form centre?!
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Lol I said 2 because when it said quiet region I wasn’t sure whether you counted where they were initially standing as the first quiet region. I think the answer is one metre and not two (according to my teachers)
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Lol I said 2 because when it said quiet region I wasn’t sure whether you counted where they were initially standing as the first quiet region. I think the answer is one metre and not two (according to my teachers)
Oh shoot that's what I did (I got 1m) but I read this thread and thought I was wrong. Because it's not diffraction right??? I might be completely off though
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Felt really good to have done it, but I think I did some janky methods to get to things...
What did ppl get for the radius of that zero gravity question?
Oh shoot that's what I did (I got 1m) but I read this thread and thought I was wrong. Because it's not diffraction right??? I might be completely off though
I said 2 as well, but I think 1 is the correct answer...
How does diffraction work in with path difference?
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the 0 gravity was honestly a troll question.
I guessed that and just used v = root(rg) and solved for r using g = 9.8 idk why it says 0 gravity when g = 9.8 :D
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Lol I said 2 because when it said quiet region I wasn’t sure whether you counted where they were initially standing as the first quiet region. I think the answer is one metre and not two (according to my teachers)
I said 2 as well, but I think 1 is the correct answer...
0.75m is the correct answer.
https://www.vcaa.vic.edu.au/Documents/exams/physics/2016/2016physics-amd-w.pdf
Take a look at Q7 from Detailed Study 6 for an identical question :)
You can get find solution using either path difference or standing wave theory.
Using standing waves:
Note that adjacent nodes (or adjacent antinodes) are spaced λ/2 apart. Median point is a maximum, so go λ/4 to the right to reach first minimum.
Therefore, the second minimum is at λ/4 + λ/2 = 0.75m to the right.
Using path difference:
P.D. = 3λ/2, half of it is 0.75m (as you move one unit either side, you are changing P.D. by two units)
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FWIW standing waves are special cases of constructive/destructive interference patterns that you generally see during diffraction. Antinodes are where constructive interference occurs, just like the antinodal lines in diffraction interference patterns. Which is why you can get the same answer assuming a standing wave or by using the path difference equation.
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What did people say about the "must the spaceship be an inertial reference frame" question?
The question mentioned that the ship was travelling at a constant speed, not velocity, so the ship may be an inertial reference frame if travelling in a straight line, but it doesn't have to be in the case of travelling in a circular path, which still has constant speed.
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Yeah had the same as you
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What did people say about the "must the spaceship be an inertial reference frame" question?
The question mentioned that the ship was travelling at a constant speed, not velocity, so the ship may be an inertial reference frame if travelling in a straight line, but it doesn't have to be in the case of travelling in a circular path, which still has constant speed.
It was not accelerating 8)
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It was not accelerating 8)
It doesn't say that. You can have constant speed and still be accelerating, ie. uniform circular motion.
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Yeah lol pretty sure it didn’t say that it wasn’t accelerating
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It doesn't say that. You can have constant speed and still be accelerating, ie. uniform circular motion.
For simplicity VCAA doesn't deal with that (or else componets of tangents are involved) constant speed simply no change in velocity. You are spot on though. an object moving in a circle at a constant speed can be accelerating since the direction of velocity changes tangential (more Uni stuff that) but full marks to u theodore.
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Yo why in the wave question is the midpoint between B and A a point of constructive interference? I thought completel destructive interference would occur here
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what was the experimental error thing on mc and what was uncertainity answer
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For simplicity VCAA doesn't deal with that (or else componets of tangents are involved) constant speed simply no change in velocity.
Do you have a source for that? Big if true
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Yo why in the wave question is the midpoint between B and A a point of constructive interference? I thought completel destructive interference would occur here
Think about it in terms of path difference. The midpoint has a P.D. of 0, so therefore the two waves constructively interfere.
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what was the experimental error thing on mc and what was uncertainity answer
https://www.vcaa.vic.edu.au/Pages/vce/adviceforteachers/physics/mis_uncertainity_errors.aspx
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Think about it in terms of path difference. The midpoint has a P.D. of 0, so therefore the two waves constructively interfere.
Yeah but if u think of it as a standing wave being formed, the mid point would be a node right?
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Yeah but if u think of it as a standing wave being formed, the mid point would be a node right?
Forget standing waves, it's not really the situation here. It's just a path difference question, and in the middle the path difference is zero corresponding to an antinode.
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Why did you guys have the force on A by B (or whatever it was)? Was it 8 N
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Why did you guys have the force on A by B (or whatever it was)? Was it 8 N
Yup
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Why did you guys have the force on A by B (or whatever it was)? Was it 8 N
I had nooooo idea but I put 40N for the first question
Then because of "action-reaction" 40N and to the left
So I (hope) because I recognized newton's third law and just repeated the answer (even if it's wrong) and changed the direction I get 2/3??
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I had nooooo idea but I put 40N for the first question
Then because of "action-reaction" 40N and to the left
So I (hope) because I recognized newton's third law and just repeated the answer (even if it's wrong) and changed the direction I get 2/3??
Pushing force on the whole 5 kg system is 40 N, giving the system acceleration 40/5 = 8 m/s/s.
Then just looking at block B (1 kg), it must have a net force acting on it of F = ma = 1 x 8 = 8 N. This net force is completely provided by the contact of block A. So F(A on B) is 8 N, and by N3, F(B on A) is also 8 N but to the left. (IIRC)
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Why wasnt it 32 N? I thought Block A exerts 4 x 8 N ??
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any thoughts on a+ cutoff?
higher or lower than last year?
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Why wasnt it 32 N? I thought Block A exerts 4 x 8 N ??
If that were true, then block B would have acceleration 32 x 1 = 32 m/s/s which contradicts the system acceleration of 8 m/s/s/.
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Anyone else struggle with 5d)?
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any thoughts on a+ cutoff?
higher or lower than last year?
I think it would be the same or slightly lower. Overall the questions werent hard like last year. Although the harder questions here were much harder than last years questions. Hopefully a+ cutoff is low 80s but with Physics thats unlikely :(
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Anyone else struggle with 5d)?
not sure if its correct this is what i did:
found voltage on transmission side and use V= IR to find current. Thus P = I^2 R not sure if that was correct
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Anyone else struggle with 5d)?
That was a malformed question. The reader had to assume things such as:
- is the power dissipated in the bulb the same?
- is the power generated by the variable voltage source the same?
The question didn’t have enough information to find an answer without one of these assumptions.
Personally, I think “operate correctly” means that the bulb has a voltage drop of 4 V while also having a current of 12 A as per part a.
Under that assumption, the current in the lines would have been 1.5 A and so power loss would have been 18 W.
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Same Did you get 1 meter form centre?!
i got 1 metre. i also thought this was a standing wave question as there was do diffraction/slits present in the diagram/question
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i got 1 metre. i also thought this was a standing wave question as there was do diffraction/slits present in the diagram/question
Heres what i did (in image)
hope i will get at least 1 mark?
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i got 1 metre. i also thought this was a standing wave question as there was do diffraction/slits present in the diagram/question
As was said before by someone, this is a special case of wave interference (of which standing waves are made from).
If the sound sources are coherent (which we can assume they are), then the peaks from A will reach the centre at the same as the peaks from B, similarly for troughs. Thus at the centre, the two waves should constructively interfere, forming an antinode.
Extend this idea along the line connecting the sources, and you should see that it really is just a path difference situation. Antinodes will occur every 0.5 m from the centre, with nodes in between.
Thus the second node will occur 0.5 + 0.25 = 0.75 m from the centre, in either direction.
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loool vcaa 'tricked' a fair few people with a interference question with sound last year and I reckon they have pulled it off again this year :P
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what did you guys get for the 20 hour quazer question?
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what did you guys get for the 20 hour quazer question?
I think I got like 14.2 h or something, have a feeling I'm off but
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I think I got like 14.2 h or something, have a feeling I'm off but
You’d be right. The 20 h period of the quasar is dilated time. If the Lorentz factor was 1.41, then proper time (period according to the quasar) is 20/1.41 = 14.2 h.
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I think I got like 14.2 h or something, have a feeling I'm off but
I believe ur right.
As from the quazers reference, the time for its emission is shorter (time/gamma) gamma was 1.4 or something, which works out to be around 14.8 :)
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You’d be right. The 20 h period of the quasar is dilated time. If the Lorentz factor was 1.41, then proper time (period according to the quasar) is 20/1.41 = 14.2 h.
Damn, thanks man :) was so sure I messed up hahaha
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Hey lads, ehat did you have as the undertainty for the voltmeter in the mcq?
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Hey lads, ehat did you have as the undertainty for the voltmeter in the mcq?
I had 0.5 A because the 'fineness' of the scale was 1A so I think the uncertainty would be +/- 0.5A
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What did people say about the "must the spaceship be an inertial reference frame" question?
The question mentioned that the ship was travelling at a constant speed, not velocity, so the ship may be an inertial reference frame if travelling in a straight line, but it doesn't have to be in the case of travelling in a circular path, which still has constant speed.
1.) The ship was not accelerating, so that rules out centripetal activity...
2.) The ship itself is not in an inertial frame of reference, but the people inside the ship were...
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1.) The ship was not accelerating, so that rules out centripetal activity...
2.) The ship itself is not in an inertial frame of reference, but the people inside the ship were...
The question does not state that the ship is not accelerating. It says that the person is stationary and that the ship is moving at a constant speed
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1.) The ship was not accelerating, so that rules out centripetal activity...
2.) The ship itself is not in an inertial frame of reference, but the people inside the ship were...
I disagree with the first point. The question didn’t specify that there was no acceleration, only that it was travelling at constant speed. Constant speed is still possible under uniform circular motion. Constant speed does not imply constant velocity.
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Multi Choice Answers:
(1) B (2) A (3) A (4) D (5) C (6) C (7) A (8) C (9) B (10) D (11) D (12) C (13) D (14) B (15)D (16) C (17) A (18) A (19) B (20) B
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Multi Choice Answers:
(1) B (2) A (3) A (4) D (5) C (6) C (7) A (8) C (9) B (10) D (11) D (12) C (13) D (14) B (15)D (16) C (17) A (18) A (19) B (20) B
i think 7 is b? it was just divided by 4. idk im doing this by memory
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i think 7 is b? it was just divided by 4. idk im doing this by memory
The question asked for the gravitational field strength when distance is increased BY 2R, not to 2R.
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what was the momentum mcq answer??
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The question asked for the gravitational field strength when distance is increased BY 2R, not to 2R.
yeah so divided by 4? idk what 2R, not to 2R. means
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The question asked for the gravitational field strength when distance is increased BY 2R, not to 2R.
It asked for the field strength at a distance 2R above Earth's surface
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Short Answers
Q1 (a) E = V/d = 5.0x10^4 Vm-1 (b) W= qV = 1/2mV2 = 1.4x10^6 ms-1 (c) r=mv/qB = 0.53 m
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It asked for the field strength at a distance 2R above Earth's surface
my teacher also divided by 4. so idk, but can you guys explain (ive done a trial mcq and the same concept was tested, i got it wrong so you guys are more likely correct)
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i think 7 is b? it was just divided by 4. idk im doing this by memory
Distance R (from the centre) is on the surface of the earth g= 9.76 N/kg. Then question asked the g value for a point 2R from the surface which is 3R from the centre. So new g is 1/9 * (9.76) = 1.08 N/kg .. using Inverse Square Law.
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Distance R (from the centre) is on the surface of the earth g= 9.76 N/kg. Then question asked the g value for a point 2R from the surface which is 3R from the centre. So new g is 1/9 * (9.76) = 1.08 N/kg .. using Inverse Square Law.
damn thought i was gonna get 50 :(
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what was the momentum mcq answer??
(Q8) C which is 4.0 ms-1 and (Q9) B Kinetic energy is not conserved but momentum is conserved.
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Spot on with Q9 (this has been tested before and done so badddd)
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(Q8) C which is 4.0 ms-1 and (Q9) B Kinetic energy is not conserved but momentum is conserved.
na the hisenbergs one
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When solutions coming :( we finish in 2.5 hours and they take whole day DONT TAKE THIS SRSLY
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Short Ans (Q2)
(a) emf = 6.4x10^-4 V
(b) Emf - time graph: Loop Outside Field = 0 V Loop Entering Field= -V, Loop Inside Field = 0V Loop Leaving Field =+ V Loop Outside field = 0 V.
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Short Ans (Q2)
(a) emf = 6.4x10^-5 V
(b) Emf - time graph: Loop Outside Field = 0 V Loop Entering Field= -V, Loop Inside Field = 0V Loop Leaving Field =+ V Loop Outside field = 0 V.
It was 3 marks so how are marks gonna be distrubuted
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na the hisenbergs one
(Q15 D. Uncertainty in their y- position affects the uncertainty in their y - momentum
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When solutions coming :( we finish in 2.5 hours and they take whole day DONT TAKE THIS SRSLY
I get that VCE is stressful and you want reassurance but the solutions are put together for free? It's a nice thing that people do to help out; not a service which you're entitled to. The people who do this also have their own responsibiliies
I'm pretty sure people have already posted MC answers, but if answers aren't up by tomorrow I'll write them :)
Edit: sorry, didn't see your "don't take this seriously" (despite the all-caps) but yeah, will do the solutions tomorrow if not sorted before then
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It was 3 marks so how are marks gonna be distrubuted
Most likely for calculating change in flux - 1 mark
Subbing no emf = - n*Change in flux / change in time - 1 marks
Final answer - 1 mark
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Most likely for calculating change in flux - 1 mark
Subbing no emf = - n*Change in flux / change in time - 1 marks
Final answer - 1 mark
Sorry i meant for that emf graph we had to draw cos i got it all wrong i drew a sin graph rippp
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I get that VCE is stressful and you want reassurance but the solutions are put together for free? It's a nice thing that people do to help out; not a service which you're entitled to. The people who do this also have their own responsibiliies
I'm pretty sure people have already posted MC answers, but if answers aren't up by tomorrow I'll write them :)
Chill bra. "Don't take is srsly". FFS.
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Q3 (a) Clockwise. Current flow in the coil H to G, magnetic field is to the right (N to S), using RH Force rule force on side HG Down and Force on Side EF is Up. Hence the rotational force (or torque) is clockwise, thus coil would rotate clockwise.
(b) Motor will not spin continuously in one direction with slip-rings. Because it dose not reverse the direction of the current in every half rotation as it does with spilt-ring commutator.
With slip rings coil will begin to spin. However after its passes the vertical orientation, it will spin in the anti-clock direction. After a few oscillations, coil will come to rest when plan of the coil is vertical.
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Q3 (a) Clockwise. Current flow in the coil H to G, magnetic field is to the right (N to S), using RH Force rule force on side HG Down and Force on Side EF is Up. Hence the rotational force (or torque) is clockwise, thus coil would rotate clockwise.
(b) Motor will not spin continuously in one direction with slip-rings. Because it dose not reverse the direction of the current in every half rotation as it does with spilt-ring commutator.
With slip rings coil will begin to spin. However after its passes the vertical orientation, it will spin in the anti-clock direction. After a few oscillations, coil will come to rest when plan of the coil is vertical.
keep em comin
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Here are the answers (q) 4 , 5 & 6
4(a) Vrms is the DC equivalent voltage = 4/ SQRT(2) = 2.83 V
4 (b) Doubling in frequency will double the emf and half the period. You should draw double amplitude (+/- 8 V).
(5) (a) Current in the Globe 4 x 3 = 12A, Voltage = 4.0 V, So Power dissipated in the globe = 12x4 = 48W
(b) V output = Vdrop in the line + V Primary coil = 3x8 + 16 = 40 V
(c) P loss = I^2 R = 72 W
(d) Assume Globe still operate at 48 W and the V = 4.0 V.
Current in the Globe = 12A. Therefore the current in transmission line = 12 / 8 = 1.5 A, Ploss = I^2 R = 18 W
(e) - For constant power increasing voltage would reduce current in the transmission line (P = VI). High current contributes to increasing Fire Risk, thus low current reduces fire risk.
- Also High voltage in the line or lower current in the line contributes to lower lower loss as Ploss = I^2R. So more efficient power transmission can be achieved by High voltage electric power transmission.
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Sorry i meant for that emf graph we had to draw cos i got it all wrong i drew a sin graph rippp
Most probably: 1 - Mark for overall shape of the graph, 1- Mark correct magnitude 1- Mark one constant V is negative and the other V is positive.
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For the record, VCAA exams are copyrighted. Do not share them here.
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For 5b i got same answer but I'm not sure if method is alright (cos I don't know anyone else who did it this way and part or working out I repeated in the following question)
Add power loss in transmission lines to power dissipated by globe to get total power (72+48=120). Then divide by 3A to get Voltage output of power supply. (120/3=40V)
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For 5b i got same answer but I'm not sure if method is alright (cos I don't know anyone else who did it this way and part or working out I repeated in the following question)
Add power loss in transmission lines to power dissipated by globe to get total power (72+48=120). Then divide by 3A to get Voltage output of power supply. (120/3=4A)
you mean 40 volts? i did the exact same thing :) so hopefully its right
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lol yeah thats what i meant :P
fixed it now
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Did anybody else find that power transmission question the hardest part of the exam? it was the only question I was very lost.
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Did anybody else find that power transmission question the hardest part of the exam? it was the only question I was very lost.
im with you
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Here are the answers (q) 4 , 5 & 6
4(a) Vrms is the DC equivalent voltage = 4/ SQRT(2) = 2.83 V
4 (b) Doubling in frequency will double the emf and half the period. You should draw double amplitude (+/- 8 V).
(5) (a) Current in the Globe 4 x 3 = 12A, Voltage = 4.0 V, So Power dissipated in the globe = 12x4 = 48W
(b) V output = Vdrop in the line + V Primary coil = 3x8 + 16 = 40 V
(c) P loss = I^2 R = 72 W
(d) Assume Globe still operate at 48 W and the V = 4.0 V.
Current in the Globe = 12A. Therefore the current in transmission line = 12 / 8 = 1.5 A, Ploss = I^2 R = 18 W
(e) - For constant power increasing voltage would reduce current in the transmission line (P = VI). High current contributes to increasing Fire Risk, thus low current reduces fire risk.
- Also High voltage in the line or lower current in the line contributes to lower lower loss as Ploss = I^2R. So more efficient power transmission can be achieved by High voltage electric power transmission.
for e ive seen so many different answers! i did the ploss one and that an alternative way of minimizing power loss is to use wires with less resistance such as gold, this impractical and expensive increasing voltage is the most efficient way to transmit power
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I put power loss and more efficient would that give me 2 marks???
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for e ive seen so many different answers! i did the ploss one and that an alternative way of minimizing power loss is to use wires with less resistance such as gold, this impractical and expensive increasing voltage is the most efficient way to transmit power
No moving parts and less capital cost.
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No moving parts and less capital cost.
is that what you said? i like it!
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(Q6) Spring
(a) Conservation of Energy mg (2.5) = 1/kx0.5^2, K = 392 N/m
(b) Initially object accelerate at 9.8 ms-2, and speed was increasing. When the ball hit the spring its acceleration begin to decrease as Net force = mg - kx.,, however still speeding up. Once mg = kx, it reaches the maximum speed and acceleration is zero as mg = F (spring force).
SO answer is 0 ms-2
(c) at Maximum speed F= -mg andf F= - kx
So x = mg/ K = 0.05 m
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Please let me know if you need answers to any particular question (s) first.
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Please let me know if you need answers to any particular question (s) first.
if you put 9.8 ms instead of 9 how many marks do you lose? im guessing both 2 marks
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if you put 9.8 ms instead of 9 how many marks do you lose? im guessing both 2 marks
which question ?
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which question ?
6 b
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6 b
I think Freddie means 9.8 instead of 0. Due to the difficulty of such a question (2 marks, 1 mark correct response (0) and exp) and maybe 2 marks (0 marks, incorrect answer but I would give 1 mark if your reasoning is correct but on your answer).
Also, not sure if 6c is correct, did it ask for the ∆x for max speed (if so, then ∆x = 0.25 m).
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Let's review the spring question (Q6).
The question states that "[t]he ball and the spring come to rest when they are at a distance of 0.50 m below the uncompressed position of the spring."
The wording of "come to rest" is slightly ambiguous. It may mean they completely come to rest, or momentarily come to rest.
However using the most obvious definition of "at rest", we can assume this means when velocity is equal to zero, which would then mean when the ball and spring are first momentarily at rest (at the lowest point of spring compression).
Using this definition, then part a. can be done using a conservation of energy approach and the correct answer can be obtained (in fact the question can only be answered if you assume this definition).
Part b. asks for you to find the acceleration at maximum speed. Many people assumed that the maximum speed occurs just before the ball hits the spring. This is not true. It is true that the maximum acceleration occurs just before the ball hits the spring, however as the ball begins to compress the spring, the acceleration will decrease, during which time the velocity is still increasing, just at a slower rate.
Thus the maximum speed will then occur as acceleration decreases to zero.
For part c., you are asked to find the compression at maximum speed. From b., the acceleration at this point will be zero, and hence net force will be zero. Thus mg = kx, and solving for x (using k = 392 N/m given in part a.), x = 0.05 m.
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Square wave; zero for the loop outside and inside sections, positive for loop entering, negative for loop exiting.
It asked for induced current, therefore zero for outside, negative gradient for travelling inside and positive gradient for travelling outside.
I think both would be accepted
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Have the solutions been released yet?
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Solutions Q7 - Projectiles
(a) d=v*t = 1.2 m
(b) S= ut+ 1/2at^2 = 0- 5 x0.4^2= -0.8 m or 0.8m
(c) There are two main methods (i) Conservation of energy (ii) find vertical velocity just before hit the ground and find the resultant speed (Pythag) Ans: 5ms^-1
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Q8 Motion
(a) Acceleration of the system 8.0 ms-2 , Using Fnet = ma where Fnet = 40 N and m= (4+1) kg
Then isolate B. Only horizontal foce on B is F on B by A = 1.0 * 8 = 8.0 N
(b) The magnitude of Fon A by B is same as Fon B by A (Newtons 3rd law). But in opposite direction. Answer: 8.0 N to the Left
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Q9. Gravity
(a) Gravitational Force = mg, where g = 3.0 Nkg-1 (from graph) so Fg = 4.50x10^3 N
(b) Magnitude of change in Gravitational potantial energy = area under the graph from 2.0x10^8 to 1.0x10^8 x mass of Juno
Count the square = 14
Value of one square = 0.5x10^8 x1 = 5.0x10^7 J/kg
Change in GPE = 1500x14x5x10^7 = 1.05x10^12 J or 1.1x10^12 J
(c) Use Kepler's equation and re-arrange it
T =sqrt( 4[PI]^2 R^3/GM) = 3.06x10^5 sec or 3.1x10^5 sec.
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Q9. Gravity
(a) Gravitational Force = mg, where g = 3.0 Nkg-1 (from graph) so Fg = 4.50x10^3 N
(b) Magnitude of change in Gravitational potantial energy = area under the graph from 2.0x10^8 to 1.0x10^8 x mass of Juno
Count the square = 14
Value of one square = 0.5x10^8 x1 = 5.0x10^7 J/kg
Change in GPE = 1500x14x5x10^7 = 1.05x10^12 J or 1.1x10^12 J
(c) Use Kepler's equation and re-arrange it
T =sqrt( 4[PI]^2 R^3/GM) = 3.06x10^5 sec or 3.1x10^5 sec.
1. I used GMm/r^2, the question did not explictly state to use the curve and got 4.75*10^3 N
2. You didn't need to calculate area under curve
You can use Change in mgh, however keep in mind, the value of g will change with height.
-> Change in mgh= Eg(f)-Eg(initial)
= m((13*1*10^8)-(3*2*10^8))
=1.05*10^12
It lead me to 1.1*10^12
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TWM send solutions its been past your organised time
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1. I used GMm/r^2, the question did not explictly state to use the curve and got 4.75*10^3 N
Yeah I ultimately did the same as you but I was unsure whether we were meant to use the graph cos you get two slightly answers depending on the method you use ???
They will accept both methods right? since the question didn't specifically ask us to use the graph??
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Q9. Gravity
(a) Gravitational Force = mg, where g = 3.0 Nkg-1 (from graph) so Fg = 4.50x10^3 N
(b) Magnitude of change in Gravitational potantial energy = area under the graph from 2.0x10^8 to 1.0x10^8 x mass of Juno
Count the square = 14
Value of one square = 0.5x10^8 x1 = 5.0x10^7 J/kg
Change in GPE = 1500x14x5x10^7 = 1.05x10^12 J or 1.1x10^12 J
(c) Use Kepler's equation and re-arrange it
T =sqrt( 4[PI]^2 R^3/GM) = 3.06x10^5 sec or 3.1x10^5 sec.
Just with the counting of the squares VCAA will have to accept a range for instance, I got 14 sqaures, however, 13 sqaures is perfectly reasonsable. Still don't know why integration is out of the scope but oh well.
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Yeah I counted 13 too (its closer to 13 if you do integration right?)
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Yeah I counted 13 too (its closer to 13 if you do integration right?)
Yes. Perfectly fine.
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can someone send multiple choice answers through pls its been a day
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can someone send multiple choice answers through pls its been a day
Why don't you put up your answers?
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Just wondering about the gravity question where you had to substitute 2R (radius) for R. There has been some talk at school that it said 2R from the surface meaning you had to sub in 3R. Was wondering if the exam said that.
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can someone send multiple choice answers through pls its been a day
IKR
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Just with the counting of the squares VCAA will have to accept a range for instance, I got 14 sqaures, however, 13 sqaures is perfectly reasonsable. Still don't know why integration is out of the scope but oh well.
While integration is technically correct, we can’t assume that students have undertaken the relevant subjects to be able to use it.
Counting the squares is the preferred method for Units 3 & 4.
However if you undertake physics in university, calculus plays a major role, so it’s a good idea to familiarise yourself with it.
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Just wondering about the gravity question where you had to substitute 2R (radius) for R. There has been some talk at school that it said 2R from the surface meaning you had to sub in 3R. Was wondering if the exam said that.
The question specified 2R above the surface of the planet. This means that you are 3R from the centre of the planet.
Substituting 3R then gives you an answer 1/9th of the original field strength.
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can someone send multiple choice answers through pls its been a day
Multi Choice Answers:
(1) B (2) A (3) A (4) D (5) C (6) C (7) A (8) C (9) B (10) D (11) D (12) C (13) D (14) B (15)D (16) C (17) A (18) A (19) B (20) B
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Solution Q(10)
(a) Passengers zero gravity implies to apparent weightlessness, where N= 0, hence Velocity at the top of the flight V= Root(rg)
r = 3.31x10^3 m
(b) No, Passenger is still under earths gravity at an altitude of 8000m. Zero gravity experiences means that normal reaction force acting on the passenger become 0 (N= 0) so he/she will experience the apparent weightlessness.
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(Q11)
(a) Lambda = 1.0 m v=f*lambda
(b) Path difference (PD) to second nodal position = 1.5 Lambda
PD = AX- BX = (AX - (10-AX)) = 1.5 m
AX = 5.75 m. So distance from centre to 2nd nodal point = 0.75 m
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(Q12) (a) C= f Lambda, f = 5.31x10^14 Hz
(b) n1 Sin Thetac = n2, Hence Critical Angel = 60.3 degrees
(c0 n = c/v, v = 3*10^8/ 1.67 = 1.80x10^8 ms^-1
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(Q13)(a) Energy of a single photon = E = hc/lambda = 3.26*10^-19 J
Hence number of photons leaving the laser in each second = 5.03x10^-3/ 3.26x10^-19 = 1.54* 10^16 Photons
(b) Point C at the centre, hence path difference is Zero. When path difference is zero it produces a constructive interference at the centre which gives a bright band.
(c) Path difference = 2.14x10^-6, Lambda = 610 nm
The order of the band at x = PD/ Lambda = 3.5 (4th Dark band) X is the 4th dark band to the right from the centre.
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What do you guys think A+ cutoff will be on this exam? Overall I felt it was close to last years but some questions definitely threw me off.
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I think A+ cut off will be 125/130 considering how EASY the exam was
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I think A+ cut off will be 125/130 considering how EASY the exam was
I would say around 87 - 88% which is around 115/130
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(Q14)
No. Its is uncertain whether its velocity is constant though the speed is constant. Spaceship could be in a circular orbit then will be accelerating radially or its direction could be changing. For ship to be an inertial frame of reference it should be non-accelerating.
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(Q15) Spaceship measures proper time (to) while the stationary scientist measures dilated time (t). So Gamma = 8.
Ek = moc^2(gamma-1) = 10000*(3x10^8)^2 (8-1) = 6.3x10^21 J
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(Q16) Scientist measures dilated time (t). Time measured in quasar's frame is the proper time (to).
to = t/gamma = 20/1.41 = 14.184 hrs = 14.2 hrs.
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Still don't know why integration is out of the scope but oh well.
If they did that, they'd probably have to make Methods 1/2 a required subject and they likely don't want to do that. It would also limit the number of students willing to take the subject if they make integration a requirement.
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If they did that, they'd probably have to make Methods 1/2 a required subject and they likely don't want to do that. It would also limit the number of students willing to take the subject if they make integration a requirement.
Have fun with the vomit of maths.
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(Q16) Scientist measures dilated time (t). Time measured in quasar's frame is the proper time (to).
to = t/gamma = 20/1.41 = 14.184 hrs = 14.2 hrs.
Great stuff mate. Good consensus on answers. :)
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wanigara you the real mvp
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Hey all, I'm sorry for this, but we can't put out the physics exam solutions until Saturday evening.
I realise most of you were waiting patiently for us to release them and we're so sorry to disappoint you.
We at TWM have had to deal with a few things and we're still getting accustomed to having new team members who have just joined us.
Again, we're sorry. I know just how nerve racking discussing exams and receiving results can be. Been there and done that, but we hope you can understand. To even be able to get them out, we have to deal with our own lives first :)
Special thanks to wanigara for sharing his answers. You're a legend.
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For the Photoelectric Effect Question, did you have to state Sai believes light is a wave and give the incorrect wave model answer?
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For the Photoelectric Effect Question, did you have to state Sai believes light is a wave and give the incorrect wave model answer?
That is what I did. She could only justify her position by using the wave model of light.. I think I ended my comments by confirming that this was the wrong line of thought but it is what the wave model would predict.
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I got 26.3m/s
Sorry mate got 5 as well.
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(Q17)
(ai) Kym
(aii) - Sai may assumed that energy delivered by the light will be accumulated over the time, thus total delivered energy will be increased.
- Using a Brighter light means more energy per unit time (greater power) can be delivered according to Sai
(b) h= Gradient = (5.6+/-0.3)x10^-15 eVs
(c) Work function = Y- intercept or X- intercept *h
W= 3.5 +/- 0.3 eV
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(Q18)(a) For the diffraction patterns produced by X-rays and electrons to be same they must have same wavelength and momentum.
Hence de Brogle Wavelength of electron can be found using Lambda = hc/E = 4.14x10^15*3*10^8/8000 = 1.55x10^-10m = 0.155 nm
(b) Momentum of Electrons p = h/Lambda = 6.63x10^-34/1.55x10^-10 = 4.277x10^-24 kgm-1
Hence Energy of electron E = p^2/2m = (4.277x10^-24)^2/(2*9.1*10^-31) = 1.0X10^-17 J
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(Q19) (a) RED
(b) Spectral lines corresponding to electron's transition from a higher to lower energy level. Its clear from the the emission spectra that energy levels are discrete.
- Electrons are only allowed to occupy in certain energy levels because for electrons to occupy a certain energy level they should be able to form standing waves around the circumference .
-This occurs when 2*pi*r= n*Lambda where n is a whole number (Principle quantum numbers).
- Hence electrons only orbit in discrete energy levels, therefore discrete spectral lines corresponding to electrons transition between these discrete energy levels. Otherwise emission spectrum should be continuous.
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(20a) (5 Marks total)
All the points correctly plotted - probably 2 marks
Selecting suitable x & Y scales to make use of the >50 % space - probably 1 mark
Fill the brackets with correct magnitudes 1010 S2 for T2 scale and 1024 m3 in R3 scale. - 1 mark
-Line of best fit - 1 mark (if you use the free hand to plot LOBF probably not going to get any marks)
(20b) Gradient = Rise / Run 1(15.2-0.66)*10^10/(146-04)*10^24 = 1.04*10^-15 S^]2 m^-3 +/- 3%
(c) Using Kepler's equation GM/4PI^2= R^3/T^2
Equation can be re-arranged in the form of y = mx where y = T^2 and x = R^3. Gradient of the graph = 4*PI^2/ GM
So M = 4*PI^2/(G*Gradient)
= 5.69x10^26 kg
I'm done. Thank you guys for the comments and feedback.
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Sorry mate got 5 as well.
I just realised I had a brain fart during the exam and I substituted wrong numbers.
Vx=20
Calculated Vy by using u^2+2as
Then used Pythagoras to solve for resultant velocity.
Would I still get 2 marks???
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I just realised I had a brain fart during the exam and I substituted wrong numbers.
Vx=20
Calculated Vy by using u^2+2as
Then used Pythagoras to solve for resultant velocity.
Would I still get 2 marks???
1 mark. Wouldn't stress over it. Enjoy your holidays.
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Hey guys, I changed my scale to x10^25 on the x axis and divided all the numbers accordingly so the numbers were closer to each other and the scaling was even, Will I lose a mark for doing that? I still got the gradient as 1.05 at the end of the day. Thanks
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Hey guys, I changed my scale to x10^25 on the x axis and divided all the numbers accordingly so the numbers were closer to each other and the scaling was even, Will I lose a mark for doing that? I still got the gradient as 1.05 at the end of the day. Thanks
No, that should not impact your marks. It's a logical change to the graph.
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= 5.69x10^26 kg
I'm done. Thank you guys for the comments and feedback.
I got 6.2x10^26, would I still gain 3 marks
- I used the gradient and substituted it into Keplers Law
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I got 6.2x10^26, would I still gain 3 marks
- I used the gradient and substituted it into Keplers Law
With such large values and the range since it is not exact values VCAA would look at your method if you use the correct method than your answer is CORRECT from the data.
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Multi Choice Answers:
(1) B (2) A (3) A (4) D (5) C (6) C (7) A (8) C (9) B (10) D (11) D (12) C (13) D (14) B (15)D (16) C (17) A (18) A (19) B (20) B
Looks like your question 8 accidentally got replaced with an emoji!