HSC Physics Predictions / Thoughts (2 Viewers)

[Salad Man]

2020 HSC Physics
Multiple Choice Answers
1 - D, 2 - A or D both correct, 3 - C, 4 - B, 5 - B, 6 - A, 7 - D, 8 - C, 9 - A, 10 - B, 11 - C, 12 - D, 13 - A, 14 - C, 15 - D, 16 - A, 17 - D, 18 - B, 19 - A, 20 - A

It will be interesting to see what the head of marking decides to do about Q.2 They must give a mark to either A or D.

How is 19 (A)? EMF is induced at the ends of the points perpendicular to the axis of the rod and hence no emf between p and q.

 

[homeworkatedog]

If I have time I will post some more answers. Here is a sample ...
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Question 25. Describe the hydrogen atom in terms of the Standard Model of matter. (4 marks)
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The hydrogen atom consists of one proton ( a hadron composed of three quarks and carrying one unit of positive charge) and one electron (a lepton carrying one unit of negative charge). The proton is not fundamental, it has a quark composition of u,u,d . The electron is fundamental, and it has never been observed to decay into any smaller units. Both the electron and the proton have a net magnetic moment, hence they have spin. The spin of the proton and the electron are both h-bar/2 or spin-1/2, so they both obey the Pauli Exclusion Principle. (that should be enough to get you 4 marks). The electron and the proton in the hydrogen atom are bound by the electromagnetic force, and hence they exchange virtual photons, the photon being the force-carrying particle (boson) that mediates the electromagnetic force. The quarks in the proton are bound by the gluon field, and hence the u,u,d quarks are exchanging virtual gluons (bosons).

what about colours?

 

[wizzkids]

Question 32.
The mass will stop accelerating when the power output of the motor equals the rate of work being done to the mass.
Let the terminal velocity of the mass be "v". Let the tension force in the rope be "T".
(v) x (T) = power (watts)
The power output of the motor is the first limitation (obviously).
Now, the question does not state that friction should be ignored, so friction should be considered in the next step.
Friction = µmg
The tension force T in the rope consists of (inertial mass x acceleration) + (the normal force x coefficient of friction).
Hence, increase of mass will obviously decrease the acceleration (Newton's Second Law), plus increase the horizontal friction force.
Hence inertial mass will limit the acceleration, but greater inertial mass alone will not limit the terminal velocity. A friction-less mass could still be accelerated to high velocity; it just takes longer to reach that velocity. The friction force is the constant force, opposing motion.
A greater friction force will limit the acceleration AND the terminal velocity. Friction = µmg so any increase in either µ or m will limit the speed at which the mass can be pulled along the horizontal surface.

 

[wizzkids]

homeworkatedog wrote:
How is 19 (A)? EMF isn't induced at any positions of the B-field when rotated .
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The motional e.m.f. definitely will be produced when the conductor cuts lines of magnetic flux.
We don't know the area that the conductor presents to the flux, but let the thickness of the conductor be "y"
The flux passing through the conductor Phi = y L B cos(theta). Faradays Law of Induction says the induced e.m.f. e = -(d.Phi/d.theta)
d.Phi/d.theta = - y L B sin(theta) At theta = 0 the rate of change of flux is zero, but at theta = 90 degrees, the rate of change of flux is maximum.
Now at t=0 the flux is decreasing, so the e.m.f. is increasing. (A) is the correct answer.

 

[wizzkids]

homeworkatedog wrote:
what about colours? (in reference to Q.32)
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You could include quark chromodynamics, but that is probably a bit beyond what the syllabus requires. For example, none of the recognised textbooks for the new syllabus include quark chromodynamics. You could mention that the only stable triplets of quarks require one red, one blue and one green quark, or one anti-red, one anti-blue and one anti-green, but this is starting to get away from a discussion of the hydrogen atom, and may be considered irrelevant detail.

 

[Maurice69]

Ok here's my logic for 20 and why I think its D not A. When the ball reaches the top, its at its absolute slowest speed. This means that by Fc = mv^2/r, Fc is at a minimum (i.e. NOT CONSTANT)

 

[homeworkatedog]

Ok here's my logic for 20 and why I think its D not A. When the ball reaches the top, its at its absolute slowest speed. This means that by Fc = mv^2/r, Fc is at a minimum (i.e. NOT CONSTANT)

yes, but the velocity component tangential to the curve is still constant right

 

[Fabrizio]

Are the solutions out yet?

 

[Fabrizio]

Ok here's my logic for 20 and why I think its D not A. When the ball reaches the top, its at its absolute slowest speed. This means that by Fc = mv^2/r, Fc is at a minimum (i.e. NOT CONSTANT)

My teacher said D

 

[Speed’o’sound Sonic]

Are the solutions out yet?

Don’t think full solutions are out yet I’ve seen a few multis though

 

[tkempy02]

View attachment 29427
What we saying?

So Faradays law is:

And magnetic flux is



Therefore emf is proportional to cos(theta)
B

 

[Speed’o’sound Sonic]

So Faradays law is:
View attachment 29430
And magnetic flux is

View attachment 29431

Therefore emf is proportional to cos(theta)
B

Yeah but emf has the delta implying rate of change so differentiate to get sin

 

[idkkdi]

My teacher said D

yes, but the velocity component tangential to the curve is still constant right

Yeah but emf has the delta implying rate of change so differentiate to get sin

Ok here's my logic for 20 and why I think its D not A. When the ball reaches the top, its at its absolute slowest speed. This means that by Fc = mv^2/r, Fc is at a minimum (i.e. NOT CONSTANT)

can someone tell me what q20 is?

 

[Speed’o’sound Sonic]

can someone tell me what q20 is?

I said A

 

[idkkdi]

I said A

what the actual question was......

 

[NN03]

Yeah but emf has the delta implying rate of change so differentiate to get sin

For Q19, when the conductor is parallel to the magnetic field, could there be no force on the charges in the conductor, therefore there is no emf in the conductor?

 

[大きい男]

For Q19, when the conductor is parallel to the magnetic field, could there be no force on the charges in the conductor, therefore there is no emf in the conductor?

But when it is parallel to the B field, the change in flux is at a maximum, so emf is at a maximum

 

[Speed’o’sound Sonic]

what the actual question was......

Oh yeah it was the one with the curved wall and a ball being thrown up and across it. The full paper is a few pages back if u want it too.

 

[Time&moretime]

So are there any predictions of what the band 6 cut off will be? Will it be higher or lower than last years?

 

[Speed’o’sound Sonic]

So are there any predictions of what the band 6 cut off will be? Will it be higher or lower than last years?

Predictions seem to be around low 80’s MAYBE high 70’s at the lowest