Physics: Discussion, Questions & Potential Solutions

[ralguy666]

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.

[Richard Feynman 101]

can someone send multiple choice answers through pls its been a day

IKR

[MrTeacherMan]

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.

[MrTeacherMan]

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.

[wanigara]

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 ( 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

[wanigara]

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.

[wanigara]

(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

[wanigara]

(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

[wanigara]

(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.

[Alexmaths]

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.

[aussiboi]

I think A+ cut off will be 125/130 considering how EASY the exam was

[wanigara]

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

[wanigara]

(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.

[wanigara]

(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

[wanigara]

(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.