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May 09, 2026, 07:03:14 pm

Author Topic: Bazza's 3/4 Physics question thread  (Read 23847 times)  Share 

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taiga

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Re: Bazza's 3/4 Physics question thread
« Reply #60 on: June 07, 2012, 01:08:12 pm »
+1
I can't recall if these are SAC Questions or not, so I'll leave it here, but do note that schools can come down pretty hard if you are asking SAC questions in a public forum like this. If they're not SAC questions then it's all good.
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WhoTookMyUsername

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Re: Bazza's 3/4 Physics question thread
« Reply #61 on: June 07, 2012, 02:12:50 pm »
0
none of them are sac questions, and we've already handed in electricity sac.
If you're referring to the earlier questions, some are related to SAC but i have definitely not taken any answers and copied them onto my SAC.
Personally i don't think simply asking a question can be regarded as plagiarism if you just use it as a starting point for research and the answer doesn't even appear on your sac, but nevertheless i'll take note for future sacs.
« Last Edit: June 07, 2012, 02:32:29 pm by Bazza16 »

nina_rox

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Re: Bazza's 3/4 Physics question thread
« Reply #62 on: June 07, 2012, 09:10:30 pm »
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I need help with a vcaa questions 2006 exam. Question 8 in the motion section. I'm not sure where they got the vertical speed of 15m/s? Could anyone help?

StumbleBum

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Re: Bazza's 3/4 Physics question thread
« Reply #63 on: June 07, 2012, 10:14:05 pm »
0
The vertical speed of 15m/s comes from the formulae v=u+at
We know that:
u=0 as it initially has no vertical velocity.
a=-10 due to gravity.
t=1.5 which is stated in the question.
therefore once applied to the equation v=-15m/s, but as we only need the magnitude its just 15m/s (but ensure that you understand it is in a downward direction when making a triangle to find the total speed and direction)
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nina_rox

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Re: Bazza's 3/4 Physics question thread
« Reply #64 on: June 07, 2012, 10:59:04 pm »
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The vertical speed of 15m/s comes from the formulae v=u+at
We know that:
u=0 as it initially has no vertical velocity.
a=-10 due to gravity.
t=1.5 which is stated in the question.
therefore once applied to the equation v=-15m/s, but as we only need the magnitude its just 15m/s (but ensure that you understand it is in a downward direction when making a triangle to find the total speed and direction)

Ohh yes of course. Thank you so much! Really appreciated :)

WhoTookMyUsername

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Re: Bazza's 3/4 Physics question thread
« Reply #65 on: June 10, 2012, 04:33:45 pm »
0
If there an infintesimally small section of concrete hanging over the edge is the top still in tension?

+ generally how to determine compression / tension in concrete beam apart from "intuition"

WhoTookMyUsername

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Re: Bazza's 3/4 Physics question thread
« Reply #66 on: June 11, 2012, 10:33:40 am »
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Whats the best definition of modulation and demodulation?

Wingtips

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Re: Bazza's 3/4 Physics question thread
« Reply #67 on: June 11, 2012, 10:42:18 am »
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Whats the best definition of modulation and demodulation?

According to the study design for physics from VCAA:

Quote
describe the transfer of information in analogue form (not including the technical aspects of
modulation and demodulation) using:
– light intensity modulation, i.e. changing the intensity of the carrier wave to replicate the amplitude
variation of the information signal so that the signal may propagate more efficiently
– demodulation, i.e. the separation of the information signal from the carrier wave;

nina_rox

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Re: Bazza's 3/4 Physics question thread
« Reply #68 on: June 11, 2012, 09:14:59 pm »
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Could someone explain to me 2009 vcaa exam question 11? I don't understand why it is increased while for another question very similar 2008 vcaa Q11 it is decreased? Any help is reallly appreciated!

StumbleBum

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Re: Bazza's 3/4 Physics question thread
« Reply #69 on: June 11, 2012, 09:25:41 pm »
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Because when the temperature is lower the resistance is higher, and to maintain the correct voltage drop we have to satisfy the ratio of (R1/R1+R2) so therefore when R1 is increases R2 must be increased to maintain the same resistance ratio and in turn the same voltage drop across the thermistor so the the switch will still get the required voltage.

And for Q11 in 2008 it is increased which can be checked against the assessors report.
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nina_rox

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Re: Bazza's 3/4 Physics question thread
« Reply #70 on: June 11, 2012, 09:36:57 pm »
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Because when the temperature is lower the resistance is higher, and to maintain the correct voltage drop we have to satisfy the ratio of (R1/R1+R2) so therefore when R1 is increases R2 must be increased to maintain the same resistance ratio and in turn the same voltage drop across the thermistor so the the switch will still get the required voltage.

And for Q11 in 2008 it is increased which can be checked against the assessors report.

ohh okay, because I've been taught otherwise. I thought that because the temperature has to be lower the resistance of the thermistor would be higher and therefore a lower resistance would be required (as the current is constant) When is this the case or is the resistor and thermistor also proportionate to each other (one increases other increases)? Thank you so much! :)

StumbleBum

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Re: Bazza's 3/4 Physics question thread
« Reply #71 on: June 11, 2012, 09:41:18 pm »
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They must stay proportionate to each other only when the same voltage drop is required, if you play around with the formula of Vout=(R1/R1+R2)*Vin then to maintain the same Vout if you increase R1 then R2 must also be increased.
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nina_rox

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Re: Bazza's 3/4 Physics question thread
« Reply #72 on: June 11, 2012, 09:47:29 pm »
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They must stay proportionate to each other only when the same voltage drop is required, if you play around with the formula of Vout=(R1/R1+R2)*Vin then to maintain the same Vout if you increase R1 then R2 must also be increased.

I can't believe I haven't realized that until now. Thank you very much! I'll make sure I don't get this one wrong on the exam! haha

StumbleBum

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Re: Bazza's 3/4 Physics question thread
« Reply #73 on: June 11, 2012, 09:49:31 pm »
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I can't believe I haven't realized that until now. Thank you very much! I'll make sure I don't get this one wrong on the exam! haha
Glad to be of assistance  :D
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WhoTookMyUsername

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Re: Bazza's 3/4 Physics question thread
« Reply #74 on: July 10, 2012, 06:47:07 pm »
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UNIT 4

...

Ugh, trying wade through Unit 4 AOS 1 atm but i'm struggling quite a bit

Managed to finally finish chapter 9 after struggling with the concepts initially lol

i'm not sure what's the problem, maybe it's just me, but for some of the questions in the textbook i can find absolutely no information regarding within the textbook itself, and the textbook seems to have very shallow explanations for most things in this AOS

For the below questions (6 + 7), (attached) i have absolutely no idea what is going on here

1) How is there magnetic flux when there's no magnetic field?
2) WS say something about "opposing flux change" but i can't find anything about this in the textbook or online (link?)


With the second Q, how do you determine the direction of the current? Book say's initially clockwise then later anticlockwise, but how can you determine this?

thanks
« Last Edit: July 10, 2012, 08:32:43 pm by Bazza16 »