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June 20, 2019, 03:21:36 pm

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#### blasonduo

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##### Re: HSC Physics Question Thread
« Reply #3585 on: November 30, 2018, 11:23:44 pm »
+6
Hi guys,
Sorry for being a pest, but I'm having trouble with the kinematics questions which I have posted in the link. Can I get a detailed reply please? Thanks

Hey! Sorry for the late reply!

5) I get the same displacement as you to 2 s.f, which is fantastic! However, our angle in which this displacement occurs from is different (Mine is around N0.4W). By trialling your angle, it seems to me that you claim that the boat has moved a net distance of 3.13km in the x-axis. Which from intuition, seems way to large (since the boat basically only went upwards as it travelled East then West.) Try this part again, and let me know if you have any trouble with it. (For all we know, it could've been a mistake on my end )

6) Follows the same method. It seems to me that your complicate your working more than I, which is fine. I tend to draw individual right-angled triangles for each "event". You get the same answer in displacement as I do for the x-direction. There is a little in the y-direction (0.0386km North) but since you round to 2 s.f, it makes you assume that it is directly west, when in fact I get N78W; which is pretty west, but does show the 12 degree error, which I'm not a fan of, but your answer is still good! c) looks great!

5(ball) seems correct, follows exactly what I would do, might've been an error I don't see, but it's all good!

Your relative velocity seems perfect! Great job!

For section 10, we need to point to completely negate the velocity of the stream and have an x-axis of exactly 10m/s (the distance of the river over time) use these velocities to find the net velocity and thus the angle in which it needs to travel at, make your own coordinate system.

For section 11, this is somewhat tricky!

a) m*g*sin(theta) = m*a

b) same as a) include the force of friction, use projectile motion formula.

c) m*g*sin(theta) needs to equal frictional force (ie net force is 0)

d) same as above

e/f) bit tricker as you need to find the net force of the two systems as a whole system. Remember that one of the forces will change with the angle while the other will not, once you find the net force, use methods used for Diagram A.

Come back with some working out with these, and I'll happily help out where needed! This should give you a slight boost to the questions

If you notice a problem with any of my comments, feel free to ask!
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#### david.wang28

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##### Re: HSC Physics Question Thread
« Reply #3586 on: December 02, 2018, 04:54:39 pm »
0
Hey! Sorry for the late reply!

5) I get the same displacement as you to 2 s.f, which is fantastic! However, our angle in which this displacement occurs from is different (Mine is around N0.4W). By trialling your angle, it seems to me that you claim that the boat has moved a net distance of 3.13km in the x-axis. Which from intuition, seems way to large (since the boat basically only went upwards as it travelled East then West.) Try this part again, and let me know if you have any trouble with it. (For all we know, it could've been a mistake on my end )

6) Follows the same method. It seems to me that your complicate your working more than I, which is fine. I tend to draw individual right-angled triangles for each "event". You get the same answer in displacement as I do for the x-direction. There is a little in the y-direction (0.0386km North) but since you round to 2 s.f, it makes you assume that it is directly west, when in fact I get N78W; which is pretty west, but does show the 12 degree error, which I'm not a fan of, but your answer is still good! c) looks great!

5(ball) seems correct, follows exactly what I would do, might've been an error I don't see, but it's all good!

Your relative velocity seems perfect! Great job!

For section 10, we need to point to completely negate the velocity of the stream and have an x-axis of exactly 10m/s (the distance of the river over time) use these velocities to find the net velocity and thus the angle in which it needs to travel at, make your own coordinate system.

For section 11, this is somewhat tricky!

a) m*g*sin(theta) = m*a

b) same as a) include the force of friction, use projectile motion formula.

c) m*g*sin(theta) needs to equal frictional force (ie net force is 0)

d) same as above

e/f) bit tricker as you need to find the net force of the two systems as a whole system. Remember that one of the forces will change with the angle while the other will not, once you find the net force, use methods used for Diagram A.

Come back with some working out with these, and I'll happily help out where needed! This should give you a slight boost to the questions

If you notice a problem with any of my comments, feel free to ask!
Here is my working out for section 10 and section 11, I have no idea what to do for Q5 and Q6 for section 11 (maybe it's because I haven't learnt the dynamics module yet). Can you at least check if my working out is correct? Thanks
« Last Edit: December 02, 2018, 08:55:13 pm by david.wang28 »
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#### worldno1

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##### Re: HSC Physics Question Thread
« Reply #3587 on: December 25, 2018, 04:36:54 pm »
0
hi guys!
can someone help me for this circular motion question? i'm quite lost:

1. The maximum sustained acceleration that humans are subjected to in a machine (radius of 8.84m) is 12.5g. What is the difference between the acceleration of his head and feet if the person inside is 2m tall? Assume his head is at r = 8.84m.

#### redpanda83

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##### Re: HSC Physics Question Thread
« Reply #3588 on: December 25, 2018, 06:18:17 pm »
+1
hi guys!
can someone help me for this circular motion question? i'm quite lost:

1. The maximum sustained acceleration that humans are subjected to in a machine (radius of 8.84m) is 12.5g. What is the difference between the acceleration of his head and feet if the person inside is 2m tall? Assume his head is at r = 8.84m.
first get the centripetal acceleration. 12.5 g = 122.5 m/s2. With a radius of 8.84m, Use a = v2 /r, to calculate the  32.91 m/s as the velocity.
Second step
use angular velocity (defined as rate of change of angle), which can be expressed as w = v / r
r1=8.84, r2 depends on if the person is inside or outside of the machine, so r2=6.84 if inside or 10.84 if outside
So v1 / r1 = v2 / r2
Should work
actually r2=6.84, as the max acceleration was 12.5g. So the person must be inside the machine obviosly. if you solve with r2=10.84, a2=15.32g, which is greater than 12.5g

hi guys!
can someone help me for this circular motion question? i'm quite lost:

1. The maximum sustained acceleration that humans are subjected to in a machine (radius of 8.84m) is 12.5g. What is the difference between the acceleration of his head and feet if the person inside is 2m tall? Assume his head is at r = 8.84m.

Derivation of a_1_/r_1_ = a_2_/r_2_
Since tangential velocity v is given by
v = ω r
we can show that the equation for acceleration you have already mentioned becomes

a = ω^2 r

solving for the value we know is constant, ω , we get

a/r = ω^2

As ω is constant, then  ω^2 is a constant as well.

Now that we have our equation solved for a constant, we can connect the two radii like so

a_1_/r_1_ = a_2_/r_2_

Mod Edit: Merged posts. Please refrain from double posting.
« Last Edit: December 25, 2018, 07:42:19 pm by beatroot »

#### 006896

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##### Re: HSC Physics Question Thread
« Reply #3589 on: January 05, 2019, 07:26:32 pm »
0
Hi,
I am wondering whether for HSC sciences, is the Prelim content able to be tested by NESA in the HSC exam? I know for Maths, apparently 30% of Prelim content will be tested in the HSC exam. Is this true for Physics (and Chemistry)? As in, should I be studying Prelim topics during these holidays?
Thanks

#### RuiAce

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##### Re: HSC Physics Question Thread
« Reply #3590 on: January 05, 2019, 08:08:48 pm »
+3
Hi,
I am wondering whether for HSC sciences, is the Prelim content able to be tested by NESA in the HSC exam? I know for Maths, apparently 30% of Prelim content will be tested in the HSC exam. Is this true for Physics (and Chemistry)? As in, should I be studying Prelim topics during these holidays?
Thanks
To my knowledge that is only for mathematics and not the sciences. It was how it was with the old science syllabuses but I don't see anything that hints at "prelim content is suddenly examinable for the new science syllabuses" either

#### r1ckworthy

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##### Re: HSC Physics Question Thread
« Reply #3591 on: January 09, 2019, 07:31:21 pm »
0
Hey guys. Hope y'all are having a wonderful holiday. Just a question: I recently bought the Dot Point workbook (where they have multiple questions under each dot point) for both physics and chemistry. Is it a good idea to completely have this as my study resource? ie. where the majority of my revision is spent doing these problems and questions.
Currently, I have been summarising the information under each dot point, but I'm not too sure whether it is as effective as answering questions. What do you guys think?

PS. the dot point workbook is by Brian Shadwick
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#### fun_jirachi

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##### Re: HSC Physics Question Thread
« Reply #3592 on: January 09, 2019, 09:06:41 pm »
+4
Hey there!

My school prescribes the Dot Point book, so I guess you could say I have experience using it. I think it definitely helps; you're not going to be rattling off your knowledge to the examiner for the majority of the exam. Even if the questions are redundant slightly in that they are pretty easy (in the fact that the use of concepts you learn is pretty straightforward, but dont really prep you for the much harder questions), it's still practice to make sure you learn everything and get question answering practice - definitely better than just summarisation. There a few stupid questions here and there though that my class tends to skip, but for the most part it thoroughly covers all the dot points (hence the name ). Also, be mindful of the answers, they're not always right! If you've thoroughly done a concept, and you do a question, and the answer is different, do it again, and if the answer is different still, you're probably right.

That's about Dot Point and how it could help, but in terms of having it 'completely as my study resource' is not a great idea; doing timed tests (resources may be few right now, but its worth doing whatever you can find), and of course you do need to rehash content every now and then. It's pretty good, but only when used in conjunction with a few other things.

Hope this helps
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#### louisaaa01

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##### Re: HSC Physics Question Thread
« Reply #3593 on: January 10, 2019, 04:39:20 pm »
0
In the HSC/Trials, if a question were to ask you to outline a first-hand investigation you conducted on a certain topic/area, what information would you need to include in your response in order to get full marks?

What parts of my practical investigations do I need to include in my study notes?

#### jamonwindeyer

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##### Re: HSC Physics Question Thread
« Reply #3594 on: January 10, 2019, 07:27:24 pm »
+1
In the HSC/Trials, if a question were to ask you to outline a first-hand investigation you conducted on a certain topic/area, what information would you need to include in your response in order to get full marks?

What parts of my practical investigations do I need to include in my study notes?

Hey louisaaa! To outline means to "Sketch in general terms; indicate the main features of", that's according to NESA. So I'd be probably including:

- The aim (what was being measured/investigated?)
- Rough description of method
- Key result/trend identified

Would obviously depend a bit on the context of the question - For my pracs, I would just write a super brief dot point experimental report for each thing I did! So that's aim, method, results, discussion in super brief dot points for each practical

#### david.wang28

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##### Re: HSC Physics Question Thread
« Reply #3595 on: January 11, 2019, 12:20:55 pm »
0
Hello,
I'm having trouble with question 2.2 and 2.3 that are in the attachments. Some I have done working out, some not. Can anyone help me with those questions please? Thanks
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#### jamonwindeyer

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##### Re: HSC Physics Question Thread
« Reply #3596 on: January 11, 2019, 07:37:17 pm »
+3
Hello,
I'm having trouble with question 2.2 and 2.3 that are in the attachments. Some I have done working out, some not. Can anyone help me with those questions please? Thanks

Hey David! You are on the right track for 2.2 - You just need to do a bit more work to get your voltage ($V$). That 120V isn't directly applied to the bar, the bar is in parallel with that second resistor. You need to do a bit of circuit analysis - Draw a version of the circuit with the bar replaced with a 10 ohm resistor, and try and figure out the voltage across it (Hint - The two parallel resistors can be reduced to one resistor using a formula, which makes the analysis easier!). After that, you do use the current you get from Ohm's Law like you did:

$ma=BIL\implies a=\frac{BIL}{m}$

Only other thing is I think the 2.6N is supposed to be 2.6kg

2.3 is actually really tricky! Your Part (a) looks about right - To do Part (b) onwards, I'm fairly sure you need to make an assumption that the radius of curvature of that curved motion is large enough that you can assume the magnetic force is always directed to the left. Then, it becomes a projectile motion question - With the horizontal acceleration taking the place of gravity, meaning the horizontal direction actually becomes the one with acceleration!! Turn the image 90 degrees to the left, and you'll see what I mean

Once it leaves the magnetic field, it's moving at a constant speed (another assumption, but an appropriate one), and it becomes $v=\frac{d}{t}$ and a bit of vector arithmetic! I'll write out a proper solution if you need it, but it might need to wait until after lecture weekend! Or better yet, come chat to me about it in the Physics lecture tomorrow tough question!!

#### david.wang28

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##### Re: HSC Physics Question Thread
« Reply #3597 on: January 11, 2019, 10:35:14 pm »
+1
Hey David! You are on the right track for 2.2 - You just need to do a bit more work to get your voltage ($V$). That 120V isn't directly applied to the bar, the bar is in parallel with that second resistor. You need to do a bit of circuit analysis - Draw a version of the circuit with the bar replaced with a 10 ohm resistor, and try and figure out the voltage across it (Hint - The two parallel resistors can be reduced to one resistor using a formula, which makes the analysis easier!). After that, you do use the current you get from Ohm's Law like you did:

$ma=BIL\implies a=\frac{BIL}{m}$

Only other thing is I think the 2.6N is supposed to be 2.6kg

2.3 is actually really tricky! Your Part (a) looks about right - To do Part (b) onwards, I'm fairly sure you need to make an assumption that the radius of curvature of that curved motion is large enough that you can assume the magnetic force is always directed to the left. Then, it becomes a projectile motion question - With the horizontal acceleration taking the place of gravity, meaning the horizontal direction actually becomes the one with acceleration!! Turn the image 90 degrees to the left, and you'll see what I mean

Once it leaves the magnetic field, it's moving at a constant speed (another assumption, but an appropriate one), and it becomes $v=\frac{d}{t}$ and a bit of vector arithmetic! I'll write out a proper solution if you need it, but it might need to wait until after lecture weekend! Or better yet, come chat to me about it in the Physics lecture tomorrow tough question!!
Shame I can't come to the lectures tomorrow- I'm very busy with my homework. I would have definitely registered to listen to your top-quality lecture. Anyways, thanks for the help Jamon!
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#### jamonwindeyer

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##### Re: HSC Physics Question Thread
« Reply #3598 on: January 12, 2019, 02:23:09 pm »
+5

#### kahliacopland

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##### Re: HSC Physics Question Thread
« Reply #3599 on: January 14, 2019, 02:18:41 pm »
0
Hi, I'm having trouble with this circular motion question-

"A proton is undergoing circular motion in a magnetic field with an angular speed of 1500 rad/s. If the net force acting on the proton is 3.8x10^-22N, what is the radius of its path?"

Could someone help me with this?? thanks