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March 28, 2024, 10:11:51 pm

Author Topic: VCE Physics Question Thread!  (Read 603186 times)  Share 

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Conic

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Re: VCE Physics Question Thread!
« Reply #90 on: July 24, 2013, 07:00:33 pm »
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Oh how I can relate to that! VCE physics is especially vague in explaining Electronics & Photonics.
Didn't the formula for cross product contain a 'cos', where did the 'sin' come from? O.o
Thanks, sorta makes sense
Basically, dot/scalar product is cos and cross product is sin. You use a few of each in VCE physics torque, force (in magnetism), work etc.
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lzxnl

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Re: VCE Physics Question Thread!
« Reply #91 on: July 24, 2013, 07:56:27 pm »
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Basically, dot/scalar product is cos and cross product is sin. You use a few of each in VCE physics torque, force (in magnetism), work etc.

I think you mean "you're meant to use a few of each in VCE physics torque, force, work etc".
The dot product is "too difficult" for VCE physics. Funny. Maths was made primarily for physics, so why are we omitting it like this. Hmph.
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SocialRhubarb

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Re: VCE Physics Question Thread!
« Reply #92 on: July 24, 2013, 08:25:12 pm »
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To be fair, most of my classmates are struggling as is.
Fight me.

Conic

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Re: VCE Physics Question Thread!
« Reply #93 on: July 24, 2013, 08:41:58 pm »
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I think you mean "you're meant to use a few of each in VCE physics torque, force, work etc".
The dot product is "too difficult" for VCE physics. Funny. Maths was made primarily for physics, so why are we omitting it like this. Hmph.
They use them without knowing that they are using them :P
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sin0001

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Re: VCE Physics Question Thread!
« Reply #94 on: July 25, 2013, 11:08:14 pm »
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Hey, can someone please help me with question 2a, I thought the max. EMF was induced after the coil is turned 90 (i.e. 1/4 of a turn), but the worked solutions contain a '2pi'
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lzxnl

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Re: VCE Physics Question Thread!
« Reply #95 on: July 26, 2013, 04:24:27 pm »
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They're looking for the max instantaneous emf induced in the coil, not the average. You'll need a derivative here.
Frequency of 100 Hz => 0.01 s = period
Let's denote the rotation speed by w rad/s. Then the angle made between the magnetic field and the coil changes, so the flux is going to be B*A*cos(wt)
As it's a uniform rotation rate, the coil rotates by 2pi radians in one period. Therefore the rotation rate is 2pi/T = 2pi rad/(0.01s) = 200pi rad/s = w
So now we can differentiate the flux with respect to time. This yields -w*BAsin(wt). The maximum of this is simply wBA, which by Faraday's law is the maximum emf induced at any time.
But from before, w = 200 pi rad/s, so plugging in all the values yields emf = 200 pi rad/s * 8*10^-4 T * 40*10^-4 m^2 = 64000pi * 10^-8 Wb/s = 2.01^-3 V

Someone check my working.
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sin0001

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Re: VCE Physics Question Thread!
« Reply #96 on: July 26, 2013, 05:06:34 pm »
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They're looking for the max instantaneous emf induced in the coil, not the average. You'll need a derivative here.
Frequency of 100 Hz => 0.01 s = period
Let's denote the rotation speed by w rad/s. Then the angle made between the magnetic field and the coil changes, so the flux is going to be B*A*cos(wt)
As it's a uniform rotation rate, the coil rotates by 2pi radians in one period. Therefore the rotation rate is 2pi/T = 2pi rad/(0.01s) = 200pi rad/s = w
So now we can differentiate the flux with respect to time. This yields -w*BAsin(wt). The maximum of this is simply wBA, which by Faraday's law is the maximum emf induced at any time.
But from before, w = 200 pi rad/s, so plugging in all the values yields emf = 200 pi rad/s * 8*10^-4 T * 40*10^-4 m^2 = 64000pi * 10^-8 Wb/s = 2.01^-3 V

Someone check my working.
Thanks :)
It mathematically makes sense but are we allowed to solve it using differentiation? Otherwise, is there some alternative way of solving this (i.e. how would one answer it graphically?)
Also, can you please explain the 'cos(wt)' part of the flux formula?
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lzxnl

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Re: VCE Physics Question Thread!
« Reply #97 on: July 26, 2013, 05:50:31 pm »
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Technically, you can't use differentiation in VCE physics as it's "too difficult". Somehow they decided that setting the mathematics level to year 9 was appropriate for a VCE level subject that is meant to have deep connections to mathematics.
Graphically? It's like asking someone what the maximum rate of change of a sine curve is. Graphically? Good luck with that. I'd be stuck for an answer.
As for the cosine term...I'm not surprised you ask. VCE physics's definition of flux has a serious problem. It only considers the one special case where the magnetic field is perfectly perpendicular to the surface/parallel to the surface normal. Now for a general magnetic field, if you resolve it into components parallel and perpendicular to the surface, the component parallel to the surface will not pass through the surface as they are parallel, so no flux there. Only the perpendicular component contributes to the flux, so if theta is the angle between the perpendicular component of the magnetic field and the field itself, the perpendicular component has magnitude B cos theta. Therefore the flux is BA cos theta.
Now initially, theta = 0 in the diagram; the field lines are perfectly perpendicular to the surface. As it rotates with constant angular velocity w, the angle theta at any time would be given by wt + c where c is a constant. But when t = 0, theta = 0, so c=0. This is where the cos(wt) term comes from.
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sin0001

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Re: VCE Physics Question Thread!
« Reply #98 on: July 26, 2013, 06:47:57 pm »
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I think the only other approach would be to memorize the formula... typical VCE Physics...
I was surprised this question came up in the Heinemann book, coz I dont think the exercises showed us how to find the max. EMF; is it just me or is Electric Power a hard topic (especially since the Heinemann book doesn't help)
Thanks again!
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lolipopper

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Re: VCE Physics Question Thread!
« Reply #99 on: July 27, 2013, 03:07:53 pm »
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I think the only other approach would be to memorize the formula... typical VCE Physics...
I was surprised this question came up in the Heinemann book, coz I dont think the exercises showed us how to find the max. EMF; is it just me or is Electric Power a hard topic (especially since the Heinemann book doesn't help)
Thanks again!

Dont even start with the Hinemann book. It is the worst. Im using an old Jacaranda one to accompany my understanding and a few youtube videos. 
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Shyam995

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Re: VCE Physics Question Thread!
« Reply #100 on: July 31, 2013, 01:06:36 pm »
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Hey Guys

Just been wondering, as i love maths and physics, i thought this question can bring out some curiosity.

As i have been researching, i was actually fascinated about the whole concept of a "second".

As i have been told and have studied, the result:

"the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom."

I have a possible rebuttal to this "too accurate" theory.

If this is a second, does that mean that, at  the big bang, how the different phases of the explosion measure in seconds, is this deemed to be inaccurate?

- sorry for the defective expression, as all of you know already, physics requires a substantial amount of communication and reasoning, as it can be hard to express, especially in writing.
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lolipopper

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Re: VCE Physics Question Thread!
« Reply #101 on: August 02, 2013, 08:37:10 pm »
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question...

A double slit arrangement is illuminated by light that consists of two wavelengths, one of which is known to be 600nm. the interference pattern on a screen shows that the fourth dark fringe for the known wavelength coincides with the fifth bright fringe for the unknown wavelength. What is the unknown wavelength?

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lzxnl

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Re: VCE Physics Question Thread!
« Reply #102 on: August 02, 2013, 09:21:32 pm »
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Formula for dark fringes is d sin theta = (n+1/2) wavelength where the first fringe is at n = 0
So the fourth dark fringe for the known wavelength has a path difference of 9/2*600 nm
This corresponds to the fifth dark fringe, which has a path difference of (11/2)*w
9/2*600 = 11/2 * w
w = 5400/11 nm
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lolipopper

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Re: VCE Physics Question Thread!
« Reply #103 on: August 02, 2013, 09:26:48 pm »
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Formula for dark fringes is d sin theta = (n+1/2) wavelength where the first fringe is at n = 0
So the fourth dark fringe for the known wavelength has a path difference of 9/2*600 nm
This corresponds to the fifth dark fringe, which has a path difference of (11/2)*w
9/2*600 = 11/2 * w
w = 5400/11 nm

are these sort of q's in vce?
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lzxnl

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Re: VCE Physics Question Thread!
« Reply #104 on: August 02, 2013, 09:32:06 pm »
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I haven't bothered with many VCAA questions on interference yet, although I can tell you that UMEP physics has a LOT of questions like these.
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