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March 29, 2024, 04:48:19 am

Author Topic: VCE Physics Question Thread!  (Read 603317 times)  Share 

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silverpixeli

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Re: VCE Physics Question Thread!
« Reply #750 on: November 10, 2014, 04:09:50 pm »
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How should we deal with the negative sign in Faraday's law when computing the average emf induced?

Quite clearly we have emf = -N(change in flux/change in time), which is likely to come out as a negative value - how should we 'convert' this to a positive average emf induced, in terms of mathematically sound working?

If you have a negative change in flux, the negatives will cancel. Otherwise you get a negative number of volts, but it doesn't matter because in VCE you figure out the direction of current analytically with right hand rules and you wont be tested on the specifics of which way is 'positive voltage' because that's not the focus of these questions.
If you do find yourself going through the working and get a negative answer at the end, don't sweat, just leave it or say the magnitude of the voltage is |your answer| = your answer without the negative.
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myanacondadont

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Re: VCE Physics Question Thread!
« Reply #751 on: November 10, 2014, 06:38:45 pm »
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Is there solutions to the sample exam VCAA provided last year? http://www.vcaa.vic.edu.au/Documents/exams/physics/physics-specs-samp-w.pdf

Or am I missing something?

Zealous

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Re: VCE Physics Question Thread!
« Reply #752 on: November 10, 2014, 06:41:12 pm »
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Is there solutions to the sample exam VCAA provided last year? http://www.vcaa.vic.edu.au/Documents/exams/physics/physics-specs-samp-w.pdf

Or am I missing something?

VCAA didn't put out any solutions - interestingly the sample exam is just a compilation of old VCAA questions.

iTute has some solutions here: http://www.itute.com/wp-content/uploads/2013-2016-vcaa-physics-sample-exam-solutions.pdf
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theshunpo

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Re: VCE Physics Question Thread!
« Reply #753 on: November 10, 2014, 08:15:35 pm »
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Is there solutions to the sample exam VCAA provided last year? http://www.vcaa.vic.edu.au/Documents/exams/physics/physics-specs-samp-w.pdf

Or am I missing something?

In addition to Zealous's link i found this while looking through VicPhysics once, hope it helps.
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rui97

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Re: VCE Physics Question Thread!
« Reply #754 on: November 10, 2014, 10:01:09 pm »
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Could someone explain why I cannot find the answer to Q14 as kx=mg?

The answer to Q13 is 19.6 btw
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silverpixeli

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Re: VCE Physics Question Thread!
« Reply #755 on: November 10, 2014, 10:16:43 pm »
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kx=mg applies to when the guy was hanging in equilibrium (net F = 0 = kx up - mg down --> kx = mg), and we don't know what the spring's equilibrium length is (natural length is x=0 and the point where he comes to rest isnt equilibrium because he's oscillating)
so in short, it's a valid formula but we aren't given what x is in that situation

this means you need to use conservation of energy here!
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lzxnl

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Re: VCE Physics Question Thread!
« Reply #756 on: November 10, 2014, 10:33:54 pm »
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Quick check for what to use for springs:
If you're told energies or speeds, use conservation of energy. If you're told something isn't moving, using forces (net force = 0)
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Re: VCE Physics Question Thread!
« Reply #757 on: November 11, 2014, 01:03:48 pm »
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hello can somebody please explain the difference and hoe they work: dynamic and velocity micriphones?

whats the difference between dynamic microphones and dynamic loudspeaker loudspeakers? VCAA separated the two in the study design???

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Re: VCE Physics Question Thread!
« Reply #758 on: November 11, 2014, 02:40:36 pm »
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can someone pwease explain the highlighted bit? what can't it be squarish shape?
VCAA 2011 Exam 2 q11 electric power
http://www.vcaa.vic.edu.au/Pages/vce/studies/physics/exams.aspx

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Re: VCE Physics Question Thread!
« Reply #759 on: November 11, 2014, 03:48:09 pm »
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It's a bit late but oh well. How do I tackle these questions? I found the acceleration and then consider about the forces in question.

In part c I considered the forces acting on box C. I found net force from F=ma and then subtracted friction force and since thats the 'driving' force on box C then Force of B on C must be equal to it?

However I always get 'em wrong :(

Plitzer

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Re: VCE Physics Question Thread!
« Reply #760 on: November 11, 2014, 04:16:04 pm »
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It's a bit late but oh well. How do I tackle these questions? I found the acceleration and then consider about the forces in question.

In part c I considered the forces acting on box C. I found net force from F=ma and then subtracted friction force and since thats the 'driving' force on box C then Force of B on C must be equal to it?

However I always get 'em wrong :(
The net force on Box A is equal to the 140N driving force minus the frictional force of Box A.
Frictional force is 0.2 x ( 10 x 9.8 ) = 19.6N. Therefore the net force on Box A is 120.4N to the right, which is now exerted on to Box B.
Net force on box B = 120.4 - (0.2 x (9.8 x 7)) = 106.68 to the right, which is exerted on to Box C.
i.e. the Force of Box B on Box C is 106.68N to the right, is this correct?
« Last Edit: November 11, 2014, 04:17:45 pm by Plitzer »
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myanacondadont

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Re: VCE Physics Question Thread!
« Reply #761 on: November 11, 2014, 04:24:03 pm »
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The net force on Box A is equal to the 140N driving force minus the frictional force of Box A.
Frictional force is 0.2 x ( 10 x 9.8 ) = 19.6N. Therefore the net force on Box A is 120.4N to the right, which is now exerted on to Box B.
Net force on box B = 120.4 - (0.2 x (9.8 x 7)) = 106.68 to the right, which is exerted on to Box C.
i.e. the Force of Box B on Box C is 106.68N to the right, is this correct?

No :( Sorry. The answer for part c is 26.7N and the answer for part d is 73.4N

myanacondadont

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Re: VCE Physics Question Thread!
« Reply #762 on: November 11, 2014, 04:37:29 pm »
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And one more sorry - For Q7 they times the torque supplied by the pole by 1.5. I thought you times it by 3 since its 3m from the pivot point?

Why is that?

Camo15

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Re: VCE Physics Question Thread!
« Reply #763 on: November 11, 2014, 04:41:58 pm »
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What are some devices that can act as modulators and demodulators, and what can act as a carrier wave?


EDIT: Also, how do you go about finding the rebound Force of a wall that is holding a cantilever as well as the angle of said force?


And one more sorry - For Q7 they times the torque supplied by the pole by 1.5. I thought you times it by 3 since its 3m from the pivot point?

Why is that?

The overall torque clockwise due to the pole is 1.5*100 because that's its horizontal length from the pivot, I believe. It's the same reason why the torque due to the 20kg mass is 3*200, even though the mass is at the end of the 6m rod
« Last Edit: November 11, 2014, 05:18:15 pm by Camo15 »

Zealous

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Re: VCE Physics Question Thread!
« Reply #764 on: November 11, 2014, 05:21:20 pm »
+1
In part c I considered the forces acting on box C. I found net force from F=ma and then subtracted friction force and since thats the 'driving' force on box C then Force of B on C must be equal to it?

However I always get 'em wrong :(

Look at all three blocks as a whole to find the acceleration of the system: m=21kg, F(applied)=140N, friction=42N, acceleration=14/3

Force of B on C, set up an equation for block C individually:



Force of B on A, look at Box A by itself, you've got the push force, force of B on A and the frictional forces:



And one more sorry - For Q7 they times the torque supplied by the pole by 1.5. I thought you times it by 3 since its 3m from the pivot point?

Why is that?

Torque is calculated as distance multiplied by perpendicular component of force, the 100N of weight is not perpendicular to the pole PQ, so we multiply the 100N of weight by the component of the pole which is perpendicular to the weight force which is 1.5m (using a bit of trigonometry).

What are some devices that can act as modulators and demodulators, and what can act as a carrier wave?
Modulators: you're mainly looking for transducers (electro-optic) which convert electrical signals to some physical signal, where their output can be adjusted based on an input signal. So things like an LED (can't think of any others right now).

Demodulators: take the opposite, so opto-electric devices such as LDR's and photodiodes. They will basically take the variations in the signal and convert it back into an electrical signal.

Carrier Waves: VCAA usually uses light waves in their scenarios, or even the brightness of an LED.

EDIT: Also, how do you go about finding the rebound Force of a wall that is holding a cantilever as well as the angle of said force?
Well, if you've got a rope or something at an angle holding up the cantilever, the force from the wall will just ensure the whole system is equilibrium. So find the value of tension in a cable/rope using torque calculations (to oppose the vertical weight force of the cantilever), then look at translational equilibrium - if the system isn't in translational equilibrium then the wall is likely applying a force to the cantilever in some direction.


So in this example, we'd find the tension in the cable such that it opposes the weight of the beam. But the tension in the cable would be made up of a vertical component which opposes the weight of the beam, and a horizontal component (because the tension is at an angle). So the wall (point A) will provide a horizontal force to work against the tension so the whole thing stays in equilibrium.
« Last Edit: November 11, 2014, 05:33:34 pm by Zealous »
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