HSC Physics Question Thread

[beau77bro]

Ahhh I see thank you very much!

Could someone also help me with part d of question 20 from the 2006 hsc?

i attempted it by multiplying the mass of the rod by 9.8 to give to force that should act on the balance. Then using a given current and the mass at that force i calculated the actual force that was acting on the balance (e.g. at I = 2.8, reading = 0.5485 - times that by 9.8 to give force on balance). Subtract the original weight from this = force due to parallel conductors = equate to lkII/d to find d - however this does not work

ok so the force is actually the difference between the weight and a measurement taken, so make sure to go weight - measurement then x by 9.8. wait u did that. ill give it a go

[beau77bro]

I think ur (or my graph) may be off because when I graphed it I got 3.2 for 0.5485 not 2.8 - is that the answer?

[mary123987]

Hey guys so was just watching the revision video for Motors and gen https://atarnotes.com/hsc-revision-physics/ just wondeing at 10:13 shouldnt the answer be 0.07071......... due to sin 45 ? thanks in advance

[banway]

What sort of bottom raw mark do you reckon you'll need for a band 5 or 6 in physics?

[beau77bro]

What sort of bottom raw mark do you reckon you'll need for a band 5 or 6 in physics?

depends on how hard/easy the paper is so who knows yet - this gives and indication tho http://rawmarks.info/wiki/Main_Page

[aryak]

volts are equal to eV since the units are 1:1 (im pretty sure - jamon can help here), so what you can do is create a linear y = mx + b, with y = eV, x = frequency, b = - 4.1 eV and M = the gradient from the line. then u can sub in the eV = 1.2 and rearrange to find the frequency, or just do it by graphing the new line by making eV = 0 and finding the threshold frequency whilst keeping the gradient the same as the first graph.

[/quote]

How did you find the value of b?

[sidzeman]

(Image removed from quote.)

I think ur (or my graph) may be off because when I graphed it I got 3.2 for 0.5485 not 2.8 - is that the answer?

Yup around there I'm pretty sure the solutions were just wrong hahaha thank you! Also (this is probably a common question but) what is the difference between relativity and special relativity?

[beau77bro]

Yup around there I'm pretty sure the solutions were just wrong hahaha thank you! Also (this is probably a common question but) what is the difference between relativity and special relativity?

well i think relativity can either just be in general physics where things are relative. or its the general theory of relativity which im pre sure includes non-inertial frames of reference (so like i think it relates alot to gravity like acceleration due to gravity - which apparently isn't real lmao), whilst special is just inertial.

How did you find the value of b?

oh because in the original graph there is an additional 4.1V helping them pass, or 4.1eV, so if u minus that it will give you the graph with a 0 voltages, or 0 eV, applied

Mod edit: Removed duplicate post. Do not double post unless necessary. Edit existing post instead

[beau77bro]

BUMP

no their answer is correct, because the side AB is always perpendicular. the other side may not be, but AB will always be.

[beau77bro]

hey so why do we use positron emitters rather than just gamma emitters?

BUMP

[sidzeman]

well i think relativity can either just be in general physics where things are relative. or its the general theory of relativity which im pre sure includes non-inertial frames of reference (so like i think it relates alot to gravity like acceleration due to gravity - which apparently isn't real lmao), whilst special is just inertial.

Hmmm I still dont fully understand sorry. Also, could someone explain why the answer to this is B and not A?

[beau77bro]

Ok so we estimate the wavelength is roughly 8um. Thats 8 x 10^-6m, we sub that into E = hf = hc/wavelength and that gives us our energy in joules. Now eV is in joules so we simply divide by the value given on the formula sheet - 1.602x10^-19. And that gives us our value of eV.

Note: the electron needs more eV than what is required by the band gap. So it has to be less than the 0.155eV we get from our working. So it has to be A... ok I'm confused. Where's this from?

[blasonduo]

BUMP

My understanding was that positron was just as effective as gamma emitters, as, in the end, gamma radiation is always produced. For example, Technetium 99m is a gamma emitter, and it is very useful in the medical world!

[pikachu975]

Can someone pls help with Q2Q reactions:

Since the products have higher mass, does this come from the fact that the products have less binding energy so the energy is converted to mass? Also how is energy released in nuclear reactions?

Thanks

[khadeeja_]

I always have difficulties in maximising my marks in multiple choice, is there any  tips that i could use?

Are there any good sources that summarise the impacts of medical physics on society, for x-rays, ct scans, ultrasound etc.. like in tables?

Mod edit: Removed duplicate posting. Please edit your previous post in future.