HSC Physics Question Thread

[bsdfjnlkasn]

Hello and I'm back 

In the Space topic, do we use the formula v= ωr often in practical applications? Is angular speed often tested?

2. Could a definition for orbital velocity be: the speed at which an object orbits another

3. Is an object's linear speed it's instantaneous speed at a particular point? As all objects describing a circle are actually travelling tangentially to the centre of the circle, can we take the proposed definition to be true? (Textbook just name drops 'linear speed' in the context of centripetal force and circular motion, which to me doesn't make much sense)

[jamonwindeyer]

Hello and I'm back 

In the Space topic, do we use the formula v= ωr often in practical applications? Is angular speed often tested?

2. Could a definition for orbital velocity be: the speed at which an object orbits another

3. Is an object's linear speed it's instantaneous speed at a particular point? As all objects describing a circle are actually travelling tangentially to the centre of the circle, can we take the proposed definition to be true? (Textbook just name drops 'linear speed' in the context of centripetal force and circular motion, which to me doesn't make much sense)

Welcome back!

Angular speed is never tested (I didn't know what \(\omega\) was until university), so no stresses there. That formula is never used.

I'd say that's a good definition! I normally prefer to define it mathematically, \(v=\frac{2\pi R}{T}\), but that is a solid worded definition.

Annnnnd yep, pretty much agree with you there. I never use the term 'linear speed' because it is confusing in that regard. In my opinion, just say instantaneous velocity, which in an orbit is always tangential to the orbit

[Aaron12038488]

I'm going to begin Yr 11, and my first test is a Prac Test. Is there any chance it would be the 'The World Communicates Topic'.

[bsdfjnlkasn]

I'm going to begin Yr 11, and my first test is a Prac Test. Is there any chance it would be the 'The World Communicates Topic'.

Hey Aaron,

The test will probably be on the topic you're studying in term 1. The only prac exam I had was in my prelim exams, the main focus being on the Electrical Energy unit.

[bsdfjnlkasn]

Welcome back!

Angular speed is never tested (I didn't know what \(\omega\) was until university), so no stresses there. That formula is never used.

I'd say that's a good definition! I normally prefer to define it mathematically, \(v=\frac{2\pi R}{T}\), but that is a solid worded definition.

Annnnnd yep, pretty much agree with you there. I never use the term 'linear speed' because it is confusing in that regard. In my opinion, just say instantaneous velocity, which in an orbit is always tangential to the orbit

Hey Jamon!

Great job on the maths lecture it was really well paced and thorough 

Would it be safe to say that we don't need to know the definition for angular velocity then? Because I still don't really have a strong grasp on what it is.
Or, does it come up in longer response questions?

What does the R in your formula for orbital velocity stand for?

Also, do we need to know the specifics (as in the method) of the Michelson-Morley experiment or just the result of it?

Thank you for the rest of the answers, though! Will add them to my notes now 

[bsdfjnlkasn]

I was wondering if someone could explain to me the last few lines on the following screenshot. I attached the entire explanation for context, but my confusion arises from the last bracketed statement "(since d is still equal to d')."

If the train is moving and as a result the observer is as well, how is the distance that the light has to travel to the observer still the same and only the time different? It's probably got something to do with the different frame of reference but I'm not really seeing how it connects to the outside world (where the fireworks are). But I do get why the time taken is longer for the tail of the train, I just don't understand why this doesn't apply to the distance as well.

Also in the equations relevant to this scenario, how does vt arise (in the numerator?)

 

[kiwiberry]

I was wondering if someone could explain to me the last few lines on the following screenshot. I attached the entire explanation for context, but my confusion arises from the last bracketed statement "(since d is still equal to d')."

If the train is moving and as a result the observer is as well, how is the distance that the light has to travel to the observer still the same and only the time different? It's probably got something to do with the different frame of reference but I'm not really seeing how it connects to the outside world (where the fireworks are). But I do get why the time taken is longer for the tail of the train, I just don't understand why this doesn't apply to the distance as well.

Also in the equations relevant to this scenario, how does vt arise (in the numerator?)

I think the time for the light from the rear to the observer is supposed to be \(t'=\frac{d'+vt}{c}\), not sure though

\(vt\) is the distance that the train moves in the time that it takes for the light to reach the observer (distance = speed x time, so distance=vt since the train is travelling at velocity v). So the light from the front will travel \(vt\) less than it would at rest, ie \(d-vt\)

[jabuibui]

Just wondering with the G-force equation (a + g/ g), do you change both the 'g' values according to the planet's gravitational acceleration?
So if they ask you to calculate the g-force the astronaut experiences during launch off on Mars ( g.a = 3.71 m/s^2), do you also change the bottom g from 9.8 to 3.71?

[jakesilove]

Just wondering with the G-force equation (a + g/ g), do you change both the 'g' values according to the planet's gravitational acceleration?
So if they ask you to calculate the g-force the astronaut experiences during launch off on Mars ( g.a = 3.71 m/s^2), do you also change the bottom g from 9.8 to 3.71?

Really great question! I got asked the same thing in my lecture the other day. I'm fairly sure that you do have to change one of the 'g's to the acceleration on that planet, but I think it's the g on top of the equation. ie.



Think about it this way. Say acceleration on a given planet was 1m/s instead of 9.8m/s. Presumably, you feel heaps 'lighter'. Thus, if you were travelling UPWARDS at 8.8m/s, you would definitely feel 'heavier' than normal, but you would probably only feel as heavy as you feel on earth. ie



However, this will never come up in the HSC!

[bsdfjnlkasn]

I think the time for the light from the rear to the observer is supposed to be \(t'=\frac{d'+vt}{c}\), not sure though

\(vt\) is the distance that the train moves in the time that it takes for the light to reach the observer (distance = speed x time, so distance=vt since the train is travelling at velocity v). So the light from the front will travel \(vt\) less than it would at rest, ie \(d-vt\)

Hey that makes much more sense! I was just a bit hesitant to question the textbook because obviously you wouldn't expect for things to be written incorrectly. I'll make sure to clarify this with my teacher when we cover it in class.

Thanks for your help 

[bsdfjnlkasn]

Hey there,

I'm just a bit confused by time dilation, specifically how it can be 'observed' on Earth like in the example of looking at a rocket travelling at c.
My textbook explanation is this "time dilation can be summarise as a 'moving clock appears to run slower'"

I find it a bit vague especially since all the previous explanations were about observing a light on a train by two people, one on the train and one outside on Earth. Also, just to quickly clarify with this example; time dilation is best observed by considering distances and how their apparent length varies with different frames of reference. Is it right to draw a connection between the two and hence say that if the distance appeared to increase then time dilation was observed (as t = d/c in both instances)?

Now back to the textbook definition, Is it just an accepted fact that when looking at something closely, time goes slower? I just don't understand how this definition helps with an understanding of time dilation (like at all  ). So when, I try and apply this definition (key word being try haha) to the twin paradox I don't understand how the Earth twin is observing time dilation by seeing the 'spacecraft ticking more slowly than time on Earth.' Does the rocket appear to be moving slower than it actually is? If that's the case then I really don't understand time dilation at all.

IDK I guess i'm just looking for a clear explanation of time dilation so that I can better 'imagine' it /understand it practically through thought experiments.

[Whitey]

Hello!

I have had a look for the answer to my question on the forum, but was unsuccessful. I do apologise if my question has already been answered and I have not been able to locate it.

I was looking through my space notes to see if they are up to date for the start of school when I noticed an inconsistency with my notes and the syllabus. I have written in my notes for the Law of Gravitation "F = G(mM/r^2)" when the syllabus says "F=G(mM/d^2)". I would just like to know why there is a differences between the too. I'm not sure if I am completely missing something.

I'd also like to say that looking at the HSC Physics slideshow from the Term 3 holidays is extremely helpful for a quick revision refresher!

[Iminschool]

Hello!

I have had a look for the answer to my question on the forum, but was unsuccessful. I do apologise if my question has already been answered and I have not been able to locate it.

I was looking through my space notes to see if they are up to date for the start of school when I noticed an inconsistency with my notes and the syllabus. I have written in my notes for the Law of Gravitation "F = G(mM/r^2)" when the syllabus says "F=G(mM/d^2)". I would just like to know why there is a differences between the too. I'm not sure if I am completely missing something.

I'd also like to say that looking at the HSC Physics slideshow from the Term 3 holidays is extremely helpful for a quick revision refresher!

r ----> Radius of planetary body
d ----> Distance from the centre of a planetary body

Therefore they're the same thing

[Iminschool]

r ----> Radius of planetary body
d ----> Distance from the centre of a planetary body

Therefore they're the same thing

P.S 'd' may also include altitude if your point of reference isn't on the surface

[jamonwindeyer]

Hello!

...

I'd also like to say that looking at the HSC Physics slideshow from the Term 3 holidays is extremely helpful for a quick revision refresher!

Welcome to the forums! Let us know if you need a hand finding stuff I'll let Jake know his slides are being put to good use!!