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January 21, 2020, 11:20:35 am

### AuthorTopic: VCE Physics Question Thread!  (Read 273646 times) Tweet Share

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#### lzxnl

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« Reply #2250 on: March 10, 2019, 08:00:29 pm »
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Hey goiz, I got a quick question,
Why is the electric field between two parallel charged plates uniform? I mean I get that their field lines are parallel, but why are they even parallel?
Also why does the field line curve outwards at the ends of the plates?

Hey there. It's actually not. The electric field between two INFINITE parallel plates (in both the x and y dirextions) is uniform. However, real plates have finite area, so you'll find the field isn't constant (near the middle is certainly different to being near the edge of the plates).

In the infinite case, there are a few reasons why the field is constant.
1. There is no sense of distance. Your plates are infinitely sized. Getting closer to a plate is like zooming in on the plate...which, being infinitely sized, looks the same (try zooming in on an infinitely sized square. It will look the same). Hence, the field strength can't change.

2. If the plates are infinite and oriented horizontally, then if you move across horizontally, you find that the plates look the same. This means the field lines have to look the same as you move between the plates.

3. If you rotate the infinite plates about an axis perpendicular to both plates, you find the plates still look the same. Thus, the shape of the field lines can't be affected by such a rotation, and they must ALL be perpendicular from one plate to another (think about it)

This is an introduction to how symmetry can greatly simplify a physics problem. Ask if you're still stuck! This explanation isn't needed for VCE, don't worry.

Note: none of the above holds for finite-sized parallel plates. Infinite plates are an approximation and a model only.
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#### juntyhee

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« Reply #2251 on: March 11, 2019, 08:40:04 pm »
0
How do you explain why the overall magnetic flux direction of a solenoid reverses when a falling magnet passes the midpoint of the coil?

2018 - Biology 

#### dream chaser

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« Reply #2252 on: March 18, 2019, 09:48:47 pm »
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Hi Guys,

What forces are acting on you when you are feeling zero gravity. Is it just the weight force meaning there is no reaction force. And is experiencing true weightlesness and zero gravity the same thing?

Thanks. All help will be much appreciated.
« Last Edit: March 18, 2019, 09:52:07 pm by dream chaser »

#### lzxnl

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« Reply #2253 on: March 19, 2019, 10:13:08 pm »
+1
Hi Guys,

What forces are acting on you when you are feeling zero gravity. Is it just the weight force meaning there is no reaction force. And is experiencing true weightlesness and zero gravity the same thing?

Thanks. All help will be much appreciated.

From a general relativistic perspective, gravity isn't a force, so when you're free falling, there are no forces acting on you, which makes sense because if you let go of something, that object moves with you.

From a classical/VCE perspective, when you're free-falling, the net force acting on you is mg, and that's it. Zero gravity is, ironically, when you move under the full force of gravity, with NO reaction force. The 'no reaction force' bit leads to apparent weightlessness.

To experience true weightlessness, you need to have zero net gravitational force on you. This is achieved by either moving really, really, really far away from any matter, or by being in a location between the Earth, Moon and the Sun such that the gravitational attractions from all three cancel (for instance).
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2013
English Language (50) Chemistry (50) Specialist Mathematics (49~54.9) Physics (49) UMEP Physics (96%) ATAR 99.95

2014-2016: University of Melbourne, Bachelor of Science, Diploma in Mathematical Sciences (Applied Maths)

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#### dream chaser

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« Reply #2254 on: March 20, 2019, 05:35:33 am »
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From a general relativistic perspective, gravity isn't a force, so when you're free falling, there are no forces acting on you, which makes sense because if you let go of something, that object moves with you.

From a classical/VCE perspective, when you're free-falling, the net force acting on you is mg, and that's it. Zero gravity is, ironically, when you move under the full force of gravity, with NO reaction force. The 'no reaction force' bit leads to apparent weightlessness.

To experience true weightlessness, you need to have zero net gravitational force on you. This is achieved by either moving really, really, really far away from any matter, or by being in a location between the Earth, Moon and the Sun such that the gravitational attractions from all three cancel (for instance).

Thanks lzxnl. Really appreciate it. One question. So like apparent weightlessness, does zero gravity mean when the only force acting on you is the weight force?

#### captainbobted

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« Reply #2255 on: March 25, 2019, 08:47:10 pm »
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Hello!

So for DC voltages, I understand that as you had more armatures the voltage smoothens out even further. Is this the reason why there are DC-emf graphs that have a 'consistent bumpy' emf AND also there are DC-emf graphs of a linear horizontal line (y=...say 8V), where the 'linear horizontal line' DC-emf graphs are DC generators with so many armatures that it basically smooths out perfectly?

Thank you   #### julia_atarnotes

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« Reply #2256 on: April 15, 2019, 04:47:02 pm »
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Hello!

So for DC voltages, I understand that as you had more armatures the voltage smoothens out even further. Is this the reason why there are DC-emf graphs that have a 'consistent bumpy' emf AND also there are DC-emf graphs of a linear horizontal line (y=...say 8V), where the 'linear horizontal line' DC-emf graphs are DC generators with so many armatures that it basically smooths out perfectly?

Thank you   Yes! That sounds about right! The emf graphs from DC generators are consistently bumpy as they are the gradient of the flux graph - as dictated by Faraday's Law. If there is a generator creating a straight line for the emf graph though, it would be because there are many armatures because all the voltages from each armature would add up so that there would be a straight line. However, this scenario is not generally encountered in VCE Physics so I would not worry about it too much!

#### tayzerface

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« Reply #2257 on: May 01, 2019, 10:24:12 am »
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Hey guys, first post on here.

Question from Unit 1 Physics, does anyone know find the effective resistance of this circuit?

TIA!

#### AlphaZero ##### Re: VCE Physics Question Thread!
« Reply #2258 on: May 01, 2019, 12:50:35 pm »
+2
Hey guys, first post on here.

Question from Unit 1 Physics, does anyone know find the effective resistance of this circuit?

TIA!

Hey there and welcome to the forums!

Just for clarity, I'm going to label the resistors from left to right as $R_1,\ R_2,\ R_3\$and $R_4$. The effective resistance of the circuit is given by $R_T=R_1+\left(\frac{1}{R_2}+\frac{1}{R_3+R_4}\right)^{-1}$ since $R_3\$and $R_4$ are wired in series, which is wired in parallel to $R_2$. The result can then be added to $R_1$. Plugging in values, we obtain $R_T=10+\left(\frac{1}{10}+\frac{1}{5+5}\right)^{-1}=15\ \Omega$
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#### tayzerface

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« Reply #2259 on: May 01, 2019, 01:41:29 pm »
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Thanks so much!

#### randomnobody69420

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« Reply #2260 on: June 01, 2019, 06:39:54 pm »
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any good physics EPI ideas? kinda desperate

#### Bri MT

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« Reply #2261 on: June 03, 2019, 11:51:28 am »
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any good physics EPI ideas? kinda desperate

There are some ideas from vicphysics

My main advice would be to go for something relatively simple and think about what topic you want to revisit or study in more depth 2018-2021: Science Advanced - Global Challenges (Honours) @ Monash

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Want QCE help? Leave a post here #### Jackson.Sprigg

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« Reply #2262 on: June 03, 2019, 05:20:49 pm »
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Hey guys for the first page question 5a. (attached) I'm not really understanding what voltage drop is exactly? Does it not say that there was a drop of 5.5V and thus that correlates to how much power lost? Why would it not just say that the voltage across the globe was 5.5V if that's what its going for?

And then for Q 8 why can it be said that Hubble's orbital speed is faster?? Wouldn't the one further out move faster? In the answers it says velocity is proportional to 1/sqrt(radius) but if I rearrange v^2/r = 4pi^2r/T^2 I get v is proportional to r and a different proportionality for a = v^2/r.

i just realised as writing this that I can rearrange Newton's law of universal gravitation to arrive at their answer by replacing F with mv^2/r but why do I have to do this? Is it because g isn't constant?

Thanks for any help it's greatly appreciated!

Sorry for all the questions but I'm quite confuddled and not sure if I'm even doing proportionalitys right.

#### S200

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« Reply #2263 on: June 03, 2019, 10:18:22 pm »
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And then for Q 8 why can it be said that Hubble's orbital speed is faster?? Wouldn't the one further out move faster? In the answers it says velocity is proportional to 1/sqrt(radius) but if I rearrange v^2/r = 4pi^2r/T^2 I get v is proportional to r and a different proportionality for a = v^2/r.

i just realised as writing this that I can rearrange Newton's law of universal gravitation to arrive at their answer by replacing F with mv^2/r but why do I have to do this? Is it because g isn't constant?

Thanks for any help it's greatly appreciated!

Sorry for all the questions but I'm quite confuddled and not sure if I'm even doing proportionalitys right.

I'm really terrible on the RMS side of things, but for Gravity you sorta have it.

Obviously the force of gravity is inversely proportional to radius, and therefore radius and velocity cannot be proportional, because as radius increases, the the gravitational force relative to the earth decreases. This means that the closer something is to a large body, the more gravitational force it is experiencing (relative to mass of course). So the closer object (the Hubble) must travel be traveling faster to stay in orbit (velocity being tangential), whereas the other satellite is experiencing less gravitational "pull" and hence if it was traveling as fast as (or faster than) the hubble telescope it would break orbit and hammer off into space.

I believe your first error was that you were focusing on acceleration, and as gravity is the Force created by the mass undergoing the centripetal acceleration (not the dictionary definition, I know... ), you needed to equate forces rather than accelerations. It sound silly, 'cause the masses are the same so only the accelerations change, but see where i'm coming from?
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5233718311 #### JeKnYan ##### Re: VCE Physics Question Thread!
« Reply #2264 on: June 10, 2019, 08:18:43 pm »
0
What's the recommendation on decimal places/sig figs in the exam? I've always been taught it's two, just asking for your two cents
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