HSC Physics Question Thread

[mary123987]

Would appreciate clarification with the following:
11. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22192872_1340184369440416_505939024_n.png?oh=951d4909cc99aaa481a3dab26b06e4c1&oe=59D3E216
Answer is B, but not sure how this makes sense. Since wavelength is inversely proportional to energy, don't all low wavelength photons have high energy?
12. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22207574_1340181356107384_294692406_n.png?oh=c96794f9307d05aa00a60c4426ac77dc&oe=59D4C8C4
Answer is B, but I got A.
My method was:
F=qvBsin(theta)
F=ma
a=(qvBsin(theta))/m
Because from proton to alpha particle mass (electron mass is negligible) is doubling and so is charge, there would be no effect on the acceleration of the alpha particle. Can someone clarify this and explain why the answer is B?
21. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22207213_1340181849440668_499896216_n.png?oh=b29ca888cccec63415bb04481f6a0762&oe=59D393E1
Solution: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22184809_1340181916107328_1024926263_n.png?oh=6b585ab564eb36b31536643abce2f1d0&oe=59D39E15
I don't remember learning the third point: Superconductors only transmit DC. Can someone please explain this?
27. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22156703_1340182206107299_1451835380_n.png?oh=5f8f623e6e65c0da3a2a5b214d3b13d1&oe=59D3AD90
Solution: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22192973_1340182382773948_1424324833_n.png?oh=fe5852336b67e0da077f9b2559c0efa5&oe=59D3DD8F
Is it okay if from 10-40 minutes my gradient was a little sloped down (due to orbital decay)?
Can someone also give me a mark indication for the following question:
Explain how the adoption of AC as the dominant electricity supply benefits society in terms of the advantages of AC over DC. (6)
In the 18th century, Westinghouse triumphed over Edison for electrical distribution in Washington D.C, campaigning for the benefits over AC electricity over DC electricity. This is indicative of AC electricity's dominance over DC, greatly benefiting society as we know it today.
- AC electricity allowed for long distance transmission, through the use of transformers. Due to the large power losses associated with DC, power stations were created closer to the city, resulting in urban clutter. Transformers operate via Faraday's Law, which states that a conductor experiencing a change in magnetic flux will have an induced EMF. This is only achievable through AC, where oscillating charges produce a continuous change in flux. Transformers allow for the stepping-up of voltage, in turn reducing current due to the Law of Conservation of Energy (V1I1=V2I2), thus minimising power loss (Ploss=I^2R) and benefiting society through more efficient energy. Furthermore, this allowed power stations to be built further apart from major cities, reducing urban clutter and exposure of society to pollution.
- The use of AC and transformers had allowed for the use of a wide array of electronic devices, whilst minimising excessive cabling. If DC was used, a different supply cable would be required for each output. However through AC, simple transformers and rectifiers can be used to allow access to electrical devices such as TVs, phone chargers - all of which increase quality of life and benefit society immensely.
Quanta to Quarks:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22156994_1340194479439405_1611052904_n.png?oh=a50fb621978bb8f8b9290520cbb64c8f&oe=59D3CDE7
Part (ii).
Thanks in advance!

Hey I am answering the first question so far :
Well to answer your question intensity determines the number of photons I.e high intensity corresponds with high levels of photons and vice versa)
the question tells you it has a lower intensity and hence a lower number of photons from looking at this it is b but to further justify this :
E=hf
and c=fλ
and f=c/λ
by equating these two formulas we get E =h(c/λ)
if it has a very short wavelength as menitioned in the question you must consider the equation E =h(c/λ)
so think about h is a constant and c is a constant so the only thing determining energy is λ  so the wavelength is small sub in a value say 10 you get :
1.9878 x 10^-26
ok now sub in a high value say 10000 you get :
1,9878 x 10^-29
when you compare these the relationship is evident E is highe as λ decreases thus the answer must be B hope this makes sense let me know if you are lost

[justwannawish]

Could I please have a quick introduction to what to expect for medical physics?

[winstondarmawan]

Hey I am answering the first question so far :
Well to answer your question intensity determines the number of photons I.e high intensity corresponds with high levels of photons and vice versa)
the question tells you it has a lower intensity and hence a lower number of photons from looking at this it is b but to further justify this :
E=hf
and c=fλ
and f=c/λ
by equating these two formulas we get E =h(c/λ)
if it has a very short wavelength as menitioned in the question you must consider the equation E =h(c/λ)
so think about h is a constant and c is a constant so the only thing determining energy is λ  so the wavelength is small sub in a value say 10 you get :
1.9878 x 10^-26
ok now sub in a high value say 10000 you get :
1,9878 x 10^-29
when you compare these the relationship is evident E is highe as λ decreases thus the answer must be B hope this makes sense let me know if you are lost

Ohhh okay thank you I get it. I misinterpreted the answer and thought it meant that photons can have different energies given the same wavelength. Thanks

[Shadowxo]

12. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22207574_1340181356107384_294692406_n.png?oh=c96794f9307d05aa00a60c4426ac77dc&oe=59D4C8C4
Answer is B, but I got A.
My method was:
F=qvBsin(theta)
F=ma
a=(qvBsin(theta))/m
Because from proton to alpha particle mass (electron mass is negligible) is doubling and so is charge, there would be no effect on the acceleration of the alpha particle. Can someone clarify this and explain why the answer is B?
27. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22156703_1340182206107299_1451835380_n.png?oh=5f8f623e6e65c0da3a2a5b214d3b13d1&oe=59D3AD90
Solution: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22192973_1340182382773948_1424324833_n.png?oh=fe5852336b67e0da077f9b2559c0efa5&oe=59D3DD8F
Is it okay if from 10-40 minutes my gradient was a little sloped down (due to orbital decay)?

I'll just answer a couple of your questions
12. (Q17) The thing you missed is that the alpha particle has 4x the mass of the proton (it has 2 protons and 2 neutrons of similar mass, but the proton just has 1 proton)
27. I believe orbital decay would be negligible - it's only in the air for 40m (orbiting for 30m) so your graph should be horizontal.

[jamonwindeyer]

my school will likely select the pendulum experiment as our practical exam. I was wondering if someone could elaborate on their practical exam and how they went about it. E.g. Was a method provided and which you carried out? Or did u have to design and carry out your own.
Thx

At my school we had equipment, some background info and an aim provided and had to devise the method ourselves

Could I please have a quick introduction to what to expect for medical physics?

You'll learn, for various medical imaging techniques (ultrasound, endoscopy, X-rays, MRI, etc):

- The principles behind its operation
- How these are applied to obtain an image of the body
- The strengths and limitations of these images
- What sorts of ailments each technique is best at detecting and for what reason

[jamonwindeyer]

21. https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22207213_1340181849440668_499896216_n.png?oh=b29ca888cccec63415bb04481f6a0762&oe=59D393E1
Solution: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22184809_1340181916107328_1024926263_n.png?oh=6b585ab564eb36b31536643abce2f1d0&oe=59D39E15
I don't remember learning the third point: Superconductors only transmit DC. Can someone please explain this?

Yep, so superconductors can't really transmit AC effectively (this is massively simplifying and is perhaps a tad incorrect, but it's what we say at the HSC level). It's because cooper pairs can only form and travel without resistance if they are moving in a single direction. It doesn't work if they are zipping back and forth

Can someone also give me a mark indication for the following question:

I don't think you are quite at full marks yet, you spend a lot of time explaining the principles behind transformers, which isn't necessary here. Use that time to cover other advantages of AC, extrapolate your ideas further (allowing long distance transmission reduces urban pollution, allowing a larger range of electronic devices to be used has allowed increased technological innovation, increased efficiency makes energy more accessible, etc)

[Sukakadonkadonk]

Hi, could someone help me with part a) ?

Why would the line of best fit look like that? ie, what are the steps in determining the line of best fit for a graph again? I'm always getting conflicting answers.

Thanks.

[justwannawish]

Hi, could someone help me with part a) ?

Why would the line of best fit look like that? ie, what are the steps in determining the line of best fit for a graph again? I'm always getting conflicting answers.

Thanks.

The tips are:
-The line doesn't have to go through any particular data points or the origin
- should have equal number of data points on both sides except for obvious outliers, which I think are meant to be removed


Not sure if there is anything else besides always draw with a ruler

[Aaron12038488]

so for the dotpoint hsc physics Space book
for 1.6.2 why is the answer a +4.61x10^9J  instead of a -4.61x10^9J.
Also for simple questions such as F=mg and W=mg do i always give a direction?

[winstondarmawan]

Quanta to Quarks:
Can someone please check my answer for this question and give a mark indication.
A photon is incident on a hydrogen atom in the ground state.
Explain, using de Broglie's hypotehsis, why the photon is not absorbed by the hydrogen atom.
Answer: de Broglie's hypothesis stated that anything with momentum had a wavelength, also known as the wave-particle duality (i.e. all waves exhibited particle behaviour and vice versa). Bohr's second postulate stated that electrons must absorb/emit a quantised amount of of energy when performing quantum jumps up/down respectively in electron shells. Considering this in light of the de Broglie hypothesis, it is evident that the photon of energy was not an integral multiple of the ground state electron wavelength, and as such, was not absorbed.
Thanks in advance!

[Sukakadonkadonk]

The tips are:
-The line doesn't have to go through any particular data points or the origin
- should have equal number of data points on both sides except for obvious outliers, which I think are meant to be removed


Not sure if there is anything else besides always draw with a ruler

Hey, thanks for the answer. But I saw this as the suggested solution. Would it be accurate? It doesn't really have equal points on both sides.

*Edit: Sorry, needs to be rotated right for image to be right side up.

[hinakamishiro]

Hey guys could I have some help with these questions? I just don't understand the mathematical working out. Thanks! 

[sidzeman]

Hey guys could I have some help with these questions? I just don't understand the mathematical working out. Thanks! 

Not 100% sure if correct but,
For the motorcycle question, there are 2 forces you need to consider. For moving in a circular path, the first force you should immediately think about is centripetal - mv^2 / r.     The other force is Weight (mg)
Now just equate the 2 forces (mv^2 / r = mg), and solve for v. You should get an answer that rounds to 5.9

2nd Q:
Normally when you see 2 orbits and orbital periods you think of equating keplars law - however that is only when they are equating the same object.
For this, we have to stick with the usual speed = distance/time.  Rearrange to get T = D/S, then sub in the relevant equations. Distance is the circumference of the planets which is given by 2piR. Velocity is simply the orbital velocity. Rearrange that into a nicer form, and that will give you the orbital period for the Earth orbiting object. Now just sub in whats different about the object orbiting Xerus - you'll see its only the mass, and you'll get your answer of half the period.

[blasonduo]

Hey guys could I have some help with these questions? I just don't understand the mathematical working out. Thanks! 

Hey!

For the first one, sidzeman is 100% correct (if you need any clarification, I'll be happy to help)


2) For this question, we know the radius is the same while the planets mass changes, and we need to know how fast it is going.

Orbital velocity is the formula which pops to mind here.

As we know Xerus is 4 times the weight of earth, we will assume earth's mass = 1, and Xerus's mass = 4

For Earth:


As the other terms are constant for both equations, we can just ignore them





For Xerus;






From this, we can tell that the satellite around Xerus is going TWICE as fast than the satellite around Earth.

and as we know, If something is going twice as fast, it will take HALF the time. (IE its period is half)

So from this, it is B


EDIT: when posting, I was not alerted of sidzeman's edit Well done to you sir!

[justwannawish]

At my school we had equipment, some background info and an aim provided and had to devise the method ourselves

You'll learn, for various medical imaging techniques (ultrasound, endoscopy, X-rays, MRI, etc):

- The principles behind its operation
- How these are applied to obtain an image of the body
- The strengths and limitations of these images
- What sorts of ailments each technique is best at detecting and for what reason

Ooh 😀 I've heard that it was an unpopular option and was wondering why but it sounds good to me

Also could someone explain how the slingshot effect works? How do the satellites/rockets use the planet to gain velocity? I don't get it!!