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Section I - Multiple Choice (2 Viewers)

oh well

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20.A
calculate energy of photon. using around 10*10^-6m as wavelength. convert to ev get around 0.1eV which isnt enough energy to overcome the band gaps except for A
I did A too, but people here say I AM WRONG ! :(
 

cartoonmaiz

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This is what I got
1. D
2. A
3. C
4. C
5. C
6. C
7. B
8. A
9. B
10. D
11. D- this one was hard
12. A
13. B
14. D
15. D- i think the wording was a little off
16. A
17. B
18. A
19. D
20. B
 

IceDingo

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May as well throw these out there:

1. D
2. A
3. C
4. B - But fuuuu I dunno so just did final - initial
5. C
6. C
7. B
8. A
9. B
10. D
11. C - This I'm quite sure on, A would mean no net force on the rocket, B lolwut, D a rocket still works in a vacuum, launch pad doesn't matter
12. A
13. B
14. D
15. D - I think it says some because not all of the magnet's field would even be in the superconductor in the first place.
16. C
17. B
18. C
19. D
20. A - Wakarimasen lol
 

zhou

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This is what I got
1. D
2. A
3. C
4. C
5. C
6. C
7. B
8. A
9. B
10. D
11. D- this one was hard
12. A
13. B
14. D
15. D- i think the wording was a little off
16. A
17. B
18. A
19. D
20. B
4. B: higher altitude to lower is a loss in gpe so negative? (not too sure)
11. C: law of conservation
14: B: initial change in flux is positive gradient hence induced emf is initially positive
16: C: ideal motor no energy lost hence back emf = supply for constant rate
18: C: force is given by GM/r^2 hence i got ((1/3)^2)/((1/4.5)^2) = 2.25 (not too sure though)
 

cartoonmaiz

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4. B: higher altitude to lower is a loss in gpe so negative? (not too sure)
11. C: law of conservation
14: B: initial change in flux is positive gradient hence induced emf is initially positive
16: C: ideal motor no energy lost hence back emf = supply for constant rate
18: C: force is given by GM/r^2 hence i got ((1/3)^2)/((1/4.5)^2) = 2.25 (not too sure though)
Change in gpe is always positive.
Induced emf is in opposite direction. Lenz's law.
I think the back emf has to be smaller because there needs to be a constant supply emf for the motor to turn in a constant rotation.
For gpe the radius is not squared...
 

Doomader

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So question 20

How did people do it?

I converted the eV for each of the options to E, then used E=hf, (f=c/wavelength) to find the wavelength. I got on the graph:
A - before the turning point
B- At the turning point
C and D - way past the turning point

So i just put B :/

Thats wrong isn't it ):
 

Demise

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I think I got like 13-14/20 for MC, improved by 6 from my trial.
 

Automatia

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So question 20

How did people do it?

I converted the eV for each of the options to E, then used E=hf, (f=c/wavelength) to find the wavelength. I got on the graph:
A - before the turning point
B- At the turning point
C and D - way past the turning point

So i just put B :/

Thats wrong isn't it ):
That sounds exactly like what you had to do. How do convert the eV to E?
 

zeebobDD

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for 20 do you calculate it off the peak of the curve the wavelength? if so i got B pretty sure is right
 

zhou

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Oh no they were talking about force... i thought it was gpe...
yes multiple choice has been tricky this year. i probably got 14 wrong then hahaha.
for 19 i got the same as you, i assumed that 'low' resistance would cause minimal heating haha.
 

EinstenICEBERG

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yes multiple choice has been tricky this year. i probably got 14 wrong then hahaha.
for 19 i got the same as you, i assumed that 'low' resistance would cause minimal heating haha.

GAWD.. dat force question and. the cooktop.. so sad.... OUR BRAIN BEEN MINDFKED
 

mickstarify

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20 is A, We want a semi-conductor which is capable of detecting the whole graph, as such we try HgCdTe - 0.03 Ev, and multiply it by 1.602*10^-19 to find it in joules, then use E=hf to find f
use f in c=f*wavelength to find wavelength which is 4x10^-5 m or 0.4 micro-metres as oposed to InSb which is around 7 micro-metres. So that means A covers more of the scope of the graph than B, so A is the better choice

btw C & D are ridiculous
 
Last edited:

Sindivyn

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20.A
calculate energy of photon. using around 10*10^-6m as wavelength. convert to ev get around 0.1eV which isnt enough energy to overcome the band gaps except for A
10 wasn't the peak, ~8 was. If you use 8*10^-6 you get B.
 

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