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Everyone said it was easy, but I think I did so bad. Left the whole last question blank as I didnt understand what i had to do. already lost 23 marks. Wished I picekd another subject. Want to cry rn, but no tears are coming.
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Everything went well till the last question😥. I had no idea as I got a curve and they ask for gradient.
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I wouldn't worry too much Pabloscopy. Coming out of the exam it seems many were stumped with that final question, I'd say only the upper level students will get it.
I found most of the exam pretty typical for a physics exam. Maybe some of the multiple choice were slightly harder, and that final question with the graph probably threw people because there were two springs, rather than the typical one spring setup. A few very predictable linear acceleration/projectile questions. They used particle collisions to assess momentum conservation rather than vehicles, a nice question from an algebraic position (cancelling out the mass of the particles). No real tension questions. The loop vehicle question unexpectedly didn't assess minimum velocity but instead gave velocity at the top of the loop, making it an energy question. Relativity was pretty minor in this exam, I can only really recall a length dilation in multiple choice. Usual electricity transmission question, though perhaps a slight increase on comprehension and reasoning versus calculations.
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I wouldn't worry too much Pabloscopy. Coming out of the exam it seems many were stumped with that final question, I'd say only the upper level students will get it.
I found most of the exam pretty typical for a physics exam. Maybe some of the multiple choice were slightly harder, and that final question with the graph probably threw people because there were two springs, rather than the typical one spring setup. A few very predictable linear acceleration/projectile questions. They used particle collisions to assess momentum conservation rather than vehicles, a nice question from an algebraic position (cancelling out the mass of the particles). No real tension questions. The loop vehicle question unexpectedly didn't assess minimum velocity but instead gave velocity at the top of the loop, making it an energy question. Relativity was pretty minor in this exam, I can only really recall a length dilation in multiple choice. Usual electricity transmission question, though perhaps a slight increase on comprehension and reasoning versus calculations.
For the last multiple choice question, did you choose the option where it said initial momentum is absorbed
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I wouldn't worry too much Pabloscopy. Coming out of the exam it seems many were stumped with that final question, I'd say only the upper level students will get it.
I found most of the exam pretty typical for a physics exam. Maybe some of the multiple choice were slightly harder, and that final question with the graph probably threw people because there were two springs, rather than the typical one spring setup. A few very predictable linear acceleration/projectile questions. They used particle collisions to assess momentum conservation rather than vehicles, a nice question from an algebraic position (cancelling out the mass of the particles). No real tension questions. The loop vehicle question unexpectedly didn't assess minimum velocity but instead gave velocity at the top of the loop, making it an energy question. Relativity was pretty minor in this exam, I can only really recall a length dilation in multiple choice. Usual electricity transmission question, though perhaps a slight increase on comprehension and reasoning versus calculations.
Do you think I'm able to get 35< raw supposing im in the top 25% of my cohort? I guess I thought it was hard because it was quite different to previous years exams, including NHT imo, so I got caught by surprise. I have never seen a question like the last one, so got really panicked when I saw it and couldn't think clearly. This year's NHT exam was way easier imo, and got really high, so I guess I got caught offguard as I thought the exam would be of similar difficulty. Now I just can stop feeling disappointed of myself. Hard to forget about it
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Do you think I'm able to get 35< raw supposing im in the top 25% of my cohort? I guess I thought it was hard because it was quite different to previous years exams, including NHT imo, so I got caught by surprise. I have never seen a question like the last one, so got really panicked when I saw it and couldn't think clearly. This year's NHT exam was way easier imo, and got really high, so I guess I got caught offguard as I thought the exam would be of similar difficulty. Now I just can stop feeling disappointed of myself. Hard to forget about it
Yes you can. Please don’t stress
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For the last multiple choice question, did you choose the option where it said initial momentum is absorbed
I chose that one, so I hope it's right
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Yes you can. Please don’t stress
I wished I could, but I'm stressing even more now because I entered thinking I would do well, and be able to get really high based on my trial exam.
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For the last multiple choice question, did you choose the option where it said initial momentum is absorbed
I picked that kinetic energy is absorbed
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I picked that kinetic energy is absorbed
Yeah its KE - Anyone got solutions?
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What K values did people get? I got 150 for 'a' and 350 for 'b'. Everyone I've asked so far have had different answers. :-\
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What K values did people get? I got 150 for 'a' and 350 for 'b'. Everyone I've asked so far have had different answers. :-\
Thats what I got so hopefully others are wrong ;D
Edit: Actually I think I got 15, 35 oops
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I picked that kinetic energy is absorbed
How is it kinetic energy? During a crash, you try to decrease impact force which aims to increase time. Since this is related to impulse(which is change in momentum), I picked momentum should be absorbed.
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What K values did people get? I got 150 for 'a' and 350 for 'b'. Everyone I've asked so far have had different answers. :-\
Yeh I got those too!
Just using the gradients
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How is it kinetic energy? During a crash, you try to decrease impact force which aims to increase time. Since this is related to impulse(which is change in momentum), I picked momentum should be absorbed.
I would’ve thought that in order to crumble the material, work has to be done as the energy of the crash goes partly into the strain potential energy due to crumbling. Therefore energy was absorbed.
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I realised in the last minute left that I accidentally missed an entire page around 12 marks! Was hoping for 40+ raw so how many marks on the other questions do you think I could afford to lose to get a 40? :-\
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I would’ve thought that in order to crumble the material, work has to be done as the energy of the crash goes partly into the strain potential energy due to crumbling. Therefore energy was absorbed.
I chose momentum but after searching it up and your explanation I’d say it’s energy.
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I realised in the last minute left that I accidentally missed an entire page around 12 marks! Was hoping for 40+ raw so how many marks on the other questions do you think I could afford to lose to get a 40? :-\
Not many I think. Maybe 3-8 depending on the exam difficulty. Sorry to hear that :(
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How did you guys do the electron/x-ray diffraction question? I was absolutely clueless on that!
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How did you guys do the electron/x-ray diffraction question? I was absolutely clueless on that!
Was a kinda hard question, you just had to use the energy of the electrons to find the momentum, then find the wavelength from that, then change to frequency? Did it ask for frequency or wavelength i don't remember.
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How did you guys do the electron/x-ray diffraction question? I was absolutely clueless on that!
The first part was mainly explaining that electrons can have wavelike properties and can have a dobrolie wavelength similar to the wavelength of X-rays. Since diffraction is proportional to wavelenth/slitsize, thus xrays and electons with similar wavelength can produce similar diffraction patterns
The second part was calculating the velocity of the electron from its kinetic energy, using that to calculate the momentum, which you can then use to calculate the wavelength of the electrons. Knowing the wavelength you can determine the frequency of the electrons which will be the same as the frequency of the Xrays.
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Was a kinda hard question, you just had to use the energy of the electrons to find the momentum, then find the wavelength from that, then change to frequency? Did it ask for frequency or wavelength i don't remember.
Is the energy of electrons the same thing as the kinetic energy of electrons?
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The first part was mainly explaining that electrons can have wavelike properties and can have a dobrolie wavelength similar to the wavelength of X-rays. Since diffraction is proportional to wavelenth/slitsize, thus xrays and electons with similar wavelength can produce similar diffraction patterns
The second part was calculating the velocity of the electron from its kinetic energy, using that to calculate the momentum, which you can then use to calculate the wavelength of the electrons. Knowing the wavelength you can determine the frequency of the electrons which will be the same as the frequency of the Xrays.
Did u end up with an answer something like 1.3x10*19 Hz
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How is it kinetic energy? During a crash, you try to decrease impact force which aims to increase time. Since this is related to impulse(which is change in momentum), I picked momentum should be absorbed.
I think it's kinetic energy. Like the change in velocity is constant for both you and the car since you both go from x-0, with the car obviously having more momentum cause of it's mass, it's just you experience it differently. Even if the car softened the blow with airbags, you're still going from x-0, so you're right to say the force will be less because of the impact time. But it doesn't exactly absorb the momentum, like it isn't taking some of your momentum and somehow putting it into the surrounding objects. You are still after the collision. Your kinetic energy however, instead of being used to deform your bones is deforming the car, which makes more sense.
Or I might just be wrong entirely ;D. Let's see what the solutions are.
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Did u end up with an answer something like 1.3x10*19 Hz
10^19 seems familiar so yeah probably
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Is the energy of electrons the same thing as the kinetic energy of electrons?
They gave you kinetic energy (I am pretty sure) and that is what is used to calculate velocity/momentum
Did u end up with an answer something like 1.3x10*19 Hz
I unfortunately am horrible at remembering numbers so I have little clue. It seems right - but I am just taking a stab haha
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How did you guys do the electron/x-ray diffraction question? I was absolutely clueless on that!
I first converted the ev to joules by multiplying it by 1.6*10^-19 (charge of electron), then used the kinetic energy formula to find the velocity. Then, I used that velocity to find the momentum by multiplying by the mass of an electron then used the debroglie wavelength formula to find the wavelength. Then, since x-rays travel at C, I used C=f*lambda to find the frequency. The only commonality the x-ray and electron have are their debroglie wavelengths I think.
Maybe there's a shorter way but I'm sure this would work.
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Did u end up with an answer something like 1.3x10*19 Hz
Yeah I got that answer too so hopefully we both get some marks!!
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This is how to do it: (Frequency of X-rays using the kinetic energy of photons)
1. Ek=0.5mv2
V=sqr root (2*3000*1.6*1019/9.11*10-31)
V-3.2*107
2. lamda=h/mv
lamda= 6.63*10-34/9.11*10-31*3.2*107
lamda (de borglie wavelength)=2.2*10-11
3. C=lamda*F
F=C/lamda
F=3.0*108/2.2*10-11
4. F= 1.3*1019
The steps provided also show the areas where marks will be accounted for
Also a question: If you wrote the correct answer but only showed the first step would you get full marks?
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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
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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
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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I did spring B-spring A for k of b
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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!
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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)
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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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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 .
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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.
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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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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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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.
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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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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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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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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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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.
probably be around 88% since it was "easy" according to many people. Tho I guess q19 threw many people off so it could be around 86%.
If it is this high, SAC results are going to play a more crucial role in determining your study score according to my teacher.
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I was wondering, for the last question, if you drew the graph wrong (i even put g=9.8 :'( ), but i do the rest of the working out correctly according to MY graph, will i get consequential marks for those parts?
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I was wondering, for the last question, if you drew the graph wrong (i even put g=9.8 :'( ), but i do the rest of the working out correctly according to MY graph, will i get consequential marks for those parts?
Ye u will cze the nxt qns ans depended on that graph so ye
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Thats what I got so hopefully others are wrong ;D
Edit: Actually I think I got 15, 35 oops
Q19. Most part of the question 19 is ok. (total 18 marks)
Plotting the graph is straight forward (just to time mass in kg x 10 to convert to force, The compression in mm in x axis.
When you plot the graph you'll get two linear sections. Gradient of first linear graph is the K of spring A (which is 150 N/m). The gradient of the second liner section is the sum of K values of A and B which was 500 N/m. Hence the K of the spring B = 350 N/m. The scales need to be selected to spread your data more than 50% of the area. Uncertainty in compression was +/- 2.0mm . If you have selected scales appropriately, the error bars are like half a box left and right.
(c) (i) Area under the graph of spring A = 1/2*150*0.08^2 = 0.48J
C(ii) Spring Pot energy stored in the spring when the system is compressed by 80 mm. Need to calculate the area under both straight lines= 1/2*0.06*9 + (1/2(9+19)*0.02 =0.55 J
(iii) Work done only by spring B to compress to 80 mm = 1/2 * 350*0.02^2 (because Spring B only compresses from 60 mm to 80 mm). = 0.07 J
(b) How these type spring system can be used in small bumps and severe bumps.
Low K spring for small bumps and combination of Both springs for sever bumps to absorbs the shock.
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Q19. Most part of the question 19 is ok. (total 18 marks)
Plotting the graph is straight forward (just to time mass in kg x 10 to convert to force, The compression in mm in x axis.
When you plot the graph you'll get two linear sections. Gradient of first linear graph is the K of spring A (which is 150 N/m). The gradient of the second liner section is the sum of K values of A and B which was 500 N/m. Hence the K of the spring B = 350 N/m. The scales need to be selected to spread your data more than 50% of the area. Uncertainty in compression was +/- 2.0mm . If you have selected scales appropriately, the error bars are like half a box left and right.
(c) (i) Area under the graph of spring A = 1/2*150*0.08^2 = 0.48J
C(ii) Spring Pot energy stored in the spring when the system is compressed by 80 mm. Need to calculate the area under both straight lines= 1/2*0.06*9 + (1/2(9+19)*0.02 =0.55 J
(iii) Work done only by spring B to compress to 80 mm = 1/2 * 350*0.02^2 (because Spring B only compresses from 60 mm to 80 mm). = 0.07 J
(b) How these type spring system can be used in small bumps and severe bumps.
Low K spring for small bumps and combination of Both springs for sever bumps to absorbs the shock.
I realised afterwards I converted mm to m wrong like a dumbass, hopefully I'll get consequential marks since I did everything else right. Also was it really 18 marks?
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It was 16 marks
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I have some answers here. Please let me know if you agree/disagree with anything.
https://filebin.net/8tdp5nycdwy46qol
For 6c, I thought the answer was to move the step down transformer to the end of the circuit, after running the extention cord for 200m at 240v without changing anything, cause the question specifically said 'with existing equipment'
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For 6c, I thought the answer was to move the step down transformer to the end of the circuit, after running the extention cord for 200m at 240v without changing anything, cause the question specifically said 'with existing equipment'
This is what i wrote as well
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Both of them would be linear
Do you know exactly why it should be linear and not curvy for the second spring?
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Do you know exactly why it should be linear and not curvy for the second spring?
When you plot out the values and draw error bars and what not, the data points seemed at first look very linear to me and you can definitely draw a straight line that passes within the error bars of each point. I can't remember what the points looked like specifically, so perhaps you could have drawn a curved line that suited the plotted points well, however the fact that they wanted you to calculate the K value for the line indicated that the line had a constant gradient and thus must be a linear line.
Hopefully this makes sense, but don't stress if you drew a curvy line - i don't think you'd lose many marks for it and there's nothing you can do about it now. Enjoy your break!
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Do you know exactly why it should be linear and not curvy for the second spring?
Think of it from a physics perspective. You know Hooke's law is linear since F=kx, and if you imagine the system initially, there is only one spring active. Then, once the second one comes it should still be linear as both springs are now applying a force that still scales linearly. It doesn't make sense for it to be curved since Hooke's law is only ever linear, no matter the number of springs or their arrangement.