Physics: Discussion, Questions & Potential Solutions

[number6603]

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.

[Richard Feynman 101]

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.

[Richard Feynman 101]

(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. 

[KiNSKi01]

wanigara you the real mvp

[AlphaZero]

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.

[G-Fr3sh]

For the Photoelectric Effect Question, did you have to state Sai believes light is a wave and give the incorrect wave model answer?

[mybigtoehurts]

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.

[Richard Feynman 101]

I got 26.3m/s

Sorry mate got 5 as well.

[wanigara]

(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

[wanigara]

(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

[wanigara]

(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.

[wanigara]

(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.

[G-Fr3sh]

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???

[Richard Feynman 101]

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.

[Alexmaths]

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