Advanced Higher Physics 2017-18 Watch

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Labrador99
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Hi everyone :wavey:

Now that the projects will have been collected by the SQA, I though we might need a place to discuss revision and the exam, which will be on 8th May :eek:

How is everyone getting on with exam prep? Which areas do you find the toughest?


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(Original post by Ethan100)
Hey!

I am doing my project on Young's Modulus and I was wondering if any others here are doing the same?

How is everyone finding the project and what topics are you doing also?
(Original post by Canerom)
I am doing mine on Young's Modulus as well, how are you finding it? Mine is due in 2 days :/
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Ethan100
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Hello everyone,

Hope revision is going well. I'm struggling to understand unit three so that's what I've been focusing on.
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uzi24
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Hi everyone!

I can't for the life of me figure this out. Can anyone help?

Two in-phase speakers A and B are emitting a signal of wavelength 1.06 m. A tape recorder is placed on the straight line between the speakers, 2.53 m from speaker A.

Calculate the shortest distance from the tape recorder that speaker B should be placed, to ensure constructive interference where the signal is recorded.

Answer = 0.41 m
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Feynboy
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(Original post by uzi24)
Hi everyone!

I can't for the life of me figure this out. Can anyone help?

Two in-phase speakers A and B are emitting a signal of wavelength 1.06 m. A tape recorder is placed on the straight line between the speakers, 2.53 m from speaker A.

Calculate the shortest distance from the tape recorder that speaker B should be placed, to ensure constructive interference where the signal is recorded.

Answer = 0.41 m
For constructive interference you want the waves to meet at the tape recorder in phase.

From A to tape recorder:
2.53/1.06 = 2.386... wavelengths
So at the tape recorder the wave emitted from A is at 0.386... of a wavelength. i.e. there has been 2.386 waves from A to recorder.

So to be constructive, you want the wave from B to be at 0.386... of a wavelength also. So this could be 0.386, 1.386, 2.386 of a wavelength.
It asks for the shortest so :
0.386..*1.06 = 0.41m
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tomctutor
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(Original post by Labrador99)
Hi everyone :wavey:

Now that the projects will have been collected by the SQA, I though we might need a place to discuss revision and the exam, which will be on 8th May :eek:

How is everyone getting on with exam prep? Which areas do you find the toughest?


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Thanks to Labrador99 for kicking off this thread for the SQA Advanced Higher Physics 2017/2018 exam.

Past papers here:- http://mrmackenzie.co.uk/advanced-higher/

(also relationship and curriculum info there)

Last years forum was at
https://www.thestudentroom.co.uk/sho....php?t=4282746
(also by Labrador99)
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tomctutor
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(Original post by Ethan100)
Hello everyone,

Hope revision is going well. I'm struggling to understand unit three so that's what I've been focusing on.
CfE has no unit number references, you mean the third unit you done at your school!

Just fire up any problems your having here please be clear about paper and Q.No's as there was multiple papers in a single year past- please specify revised, newer or older curriculum- even better a snapshot image of the question pasted into the thread!:crossedf:
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Ethan100
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Can someone explain how to do questions that deal with uncertainties? We haven't been taught any of it at all.
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tomctutor
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(Original post by Ethan100)
Can someone explain how to do questions that deal with uncertainties? We haven't been taught any of it at all.
Thats a big question - its like saying can you tell me how to answer Advanced Higher Questions?:rolleyes:

I refer you to pages 9 through 21 on the course notes regarding Uncertainties pointed to in this - link.

e.g. Ohms Law? V=IR,  I = 0.25(\pm0.02)A,  R=4.0(\pm0.2)\Omega gives
\frac{\Delta I}{I}=\frac{0.02}{0.25}=0.080
\frac{\Delta R}{R}=\frac{0.2}{4.0}=0.050

\frac{\Delta V}{V}=\sqrt{(.080^2+0.050^2)} = 0.094
\Delta V = 0.094\times V=0.094\times1.00=0.094volt

V=1.0(\pm0.09) volt would do.
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A12322
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Name:  Screenshot_315.png
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Size:  268.9 KB Hi, can anyone help with these? Would be much appreciated!
Attached files
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tomctutor
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(Original post by A12322)
Hi, can anyone help with these? Would be much appreciated!
Name:  NC_AHPhysics2018_electricfields.jpg
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tomctutor
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(Original post by A12322)
Hi, can anyone help with these? Would be much appreciated!
Name:  NC_AHPhysics2018_electricfieldsq5.jpg
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hydroxide
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(Original post by tomctutor)
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Is Ep=Fr a formula we need to know?
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tomctutor
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(Original post by hydroxide)
Is Ep=Fr a formula we need to know?
Gain in Potential energy is equal to the work done,

 W=\int_{\inf}^{r'} \ F \ .dr

(the force is Coulomb force)
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A12322
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(Original post by tomctutor)
Gain in Potential energy is equal to the work done = force x distance,

(the force is Coulomb force)
Many thanks for this! Could you possibly give a list of derivations we must know?

Also, when working out electric field strength, or force due to a field, is the negative sign ever included?
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tomctutor
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(Original post by A12322)
Many thanks for this! Could you possibly give a list of derivations we must know?

Also, when working out electric field strength, or force due to a field, is the negative sign ever included?
I've just amended the previous post to highlight the definition
but it simply works out to Fr in magnitude,

don't worry about negative signs as long as you are aware that the potential from +Q is positive and the field is outwords. (opposite for -Q).

An electric dipole is +Q and -Q pair and obviously the force is one of attraction.

Electric Potential energy U= Vq (which is a scalar) and
Electric Field intensity(strength) E = F/q (which is a vector same direction as force) for a small charge "q" placed in an electric potential V volt region.

You dont need to know the fact that the field, as a function E(r):= -\nabla V(r)
or the potential gradient.

Should add you would be expected to derive the relationship E=V/d for the situation where there is a uniform field between to plates where there exists a potential difference V volt and separation d:

this is easy since work done is gain in potential energy of charge q E_w=Fd= U=Vq , (since F is uniform we need not worry about calculus here),
so \frac{F}{q}=\frac{V}{d}, i.e. the field intensity E=\frac{V}{d}
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Ethan100
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Does anybody have any last minute tips that we may forget in the exam?
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tomctutor
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Im signing off now so just to say good luck tomorrow,
Someone please upload quality images of paper if you can (after the exam concludes- ask the invigilator)

remember - relationship formulae- substitution- work out- answer correct number of sig. figs and SI units.

It would be nice if some of you would spend some time learning to use Latex scripting here (TSR link)!:cry:
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username3509752
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Good luck everyone. I'll be honest I'm not looking forward to tomorrow because unit 3 was kinda rushed by our teacher.
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HenryZ
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How many sigfig is considered correct?
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Labrador99
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All the best for tomorrow everyone- hope it goes well Make sure you get some rest tonight and just do your best in the morning
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