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Physics Exam Thoughts (1 Viewer)

Drdusk

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Would you not just derive orbital velocity from centripetal force = gravitational force? Then get 1480ms^-1?

See overall, I thought the paper was conceptual and actually difficult, but if you are a "good student" you wouldn't have gotten fucked over like 2017 last q would've fucked most students regardless of your ability, study preparation etc. This paper really favoured students who studied, as you'd struggle otherwise ~ Gravitional vs electric definition, photocell and solar cell, electricity in the home and the projectile error and velocity questions were all quite conceptual, and I'd argue they are difficult. Although I think I got them all quite well, except the electricity one xP
It said the projectile is projected upward, nothing about it orbiting. Remember throwing a projectile doesn't always mean it will achieve orbit.
 

782341

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Would you not just derive orbital velocity from centripetal force = gravitational force? Then get 1480ms^-1?

See overall, I thought the paper was conceptual and actually difficult, but if you are a "good student" you wouldn't have gotten fucked over like 2017 last q would've fucked most students regardless of your ability, study preparation etc. This paper really favoured students who studied, as you'd struggle otherwise ~ Gravitional vs electric definition, photocell and solar cell, electricity in the home and the projectile error and velocity questions were all quite conceptual, and I'd argue they are difficult. Although I think I got them all quite well, except the electricity one xP
I dont think it mentioned it was orbiting
 

Stev3n

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What final velocity did everyone get for the 20kg mass at 500km question?
 

782341

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4EEF4C81-EAB1-42BA-AC36-BB8A05A9E1D2.jpegIm seeing a common answer 426ms too this isnt my working out but saw it somewhere else
 

maxcf212

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It relies on the particle ionizing the water vapor around it. Alpha particles are really highly ionizing hence they don't make it much past where they're generated(hence their path is short).

Beta particles are much less ionizing and hence similarly their paths are long
My understanding was that alpha was thicker and shorter (as you said) but beta was zig-zaggy/bendy. I put gamma as source Y since it was straight lines and also faint.
 

Drdusk

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My understanding was that alpha was thicker and shorter (as you said) but beta was zig-zaggy/bendy. I put gamma as source Y since it was straight lines and also faint.
Yeah beta is technically zig-zaggy, but it complied more with beta than Gamma. Gamma doesn't produce continuous tracks. Rather they are very short dots which are at a distance away from the source.
 

supR

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Yeah I've realised my mistake now guys about orbital velocity grrrr ~ Hopefully I get a mark for it ://
 

Shadowfaxx

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The question said to ignore orbital velocity and rotational of the moon. And over the 500km altitude the acceleration change would be to large to be a projecitle motion question. So the kinetic energy which was equal to the work done could be derived from part i). 1.26x10^7= 1/2mv^2
hence v=1122m/s thoughts?
 

supR

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Yeah beta is technically zig-zaggy, but it complied more with beta than Gamma. Gamma doesn't produce continuous tracks. Rather they are very short dots which are at a distance away from the source.
Also gamma has barely any impact on the nuclei, and surely the 4 marks would have wanted some info on alpha affecting nuclei and beta affecting nuclei
 

maxcf212

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Also gamma has barely any impact on the nuclei, and surely the 4 marks would have wanted some info on alpha affecting nuclei and beta affecting nuclei
Yea probably. I just put that energy is released as gamma and the nuclei isn't affected.
 

asiansensation46

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The question said to ignore orbital velocity and rotational of the moon. And over the 500km altitude the acceleration change would be to large to be a projecitle motion question. So the kinetic energy which was equal to the work done could be derived from part i). 1.26x10^7= 1/2mv^2
hence v=1122m/s thoughts?
You have to consider the original velocity as well.
So you work out initial GPE and initial KE, giving you total energy of the system. (1)
Then you work out final GPE at 500km altitude. (2)
Conservation of energy means that (2)+final KE is equal to (1)
Therefore you get value of final KE (at 500km altitude)
Equate to 1/2mv^2 and solve for v :cool2:
You end up with 426.83 ish for v
 

moosa7021

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The ruler wasn't held against the wall so the camera would have picked up errors in reading.
 

Drdusk

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Haha holy crap I just realized I wrote Quanta to quarks on the extra writing booklet and not Physics.
smh
 

Joshfotho

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I said for the projectile that the camera wasn't centred and would therefore distort the image. Reckon that is a valid response?
 
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impulsebro89

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I said for the projectile that the camera wasn't centred and would therefore, distort the image. Reckon that is a valid response?
I was about to write down something about the camera but then I read the question closely and it said the camera captured the whole trajectory, so I scrapped that. So i'm not sure
 

123qqxxzz

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nah u have to do change in KE, so change in GpE + change in KE = 0, to get about 400 i believe
 

asiansensation46

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For induction cooktops in multiple choice. Was the frequency 50kHz or 50Hz because isn't power supply for AC limited to 50Hz?
 

FieryFX

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I was about to write down something about the camera but then I read the question closely and it said the camera captured the whole trajectory, so I scrapped that. So i'm not sure
it captures the whole trajectory but parallax error takes into affect because it has an angled view so its a valid response
(checked with my physics teacher)
 

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