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Edexcel Physics Unit 2 "Physics at work" June 2013

lol
better way:
1) define voltage
2) define pdf
3) define emf

haha yea it looks weird ..i just noticed

Well if you ever want to define a physical quantity you can form an equation and make the denominator equal to one. (If you want to define a unit then make both the numerator and denominator one)

voltage = \dfrac{energy}{charge}

So voltage is energy transferred per unit charge or per coulomb of charge. (And one volt is one joule transferred per coulomb charge).

This does not specify if it is inputting or outputting electrical energy. So we have emf.

Emf of a battery or cell is the total work done by the cell against both the internal and external circuits per coulomb of charge that passes through it. It is different to the terminal pd learnt at GCSE which does not recognise internal resistance so terminal PD is the work done by the cell in the external circuit only.

In other words emf is the total electrical energy provided to each coulomb of charge by a power supple. This energy is then used to do work in the internal circuit (against internal resistance) and in the external resistance. Lost volts us the work done in the internal circuit against internal resistance.

So emf=lost volt+terminal pd
Terminal PD= emf - lost volt

V=E-Ir

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Reply 161

RoyalBlue7

19

Original post by CharlieTT

I think I have discovered the angriest Physics lecturer ever.
[video="youtube;luXW6AFSloA"]http://www.youtube.com/watch?v=luXW6AFSloA&list=PLEEAF7F80E205E50E[/video]

Haha, he has a PhD though :awesome:

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Reply 162

RoyalBlue7

19

Original post by KBenzema

i have a big doubt on part (a)(iii)
why do i get a different answer depending on the formula i use. im baffled.
P=IV : (0.75)*3=2.25w
P=I^2R : (0.75)^2*3.6 = 2.025w
P=V^2/R : (3)^2/3.6 = 2.5w

The first and third equation (involving V as 3v) is wrong. V is not 3 because the internal resistance uses some of the emf. :smile:

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Reply 163

RoyalBlue7

19

Original post by krisshP

Can somebody help me with section a question 7 on paper January 2012 unit 2 please? I don't get it at all

V=IR
I=V/R

There's constant pd so surely as R increases, I decreases linearly, giving B as the answer?

Thanks

Its not B.
Its not a linear decrease. It does not intersect any of the two axes. Its inversely proportional.

Its like an a y=1/x graph.

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Reply 164

RoyalBlue7

19

Original post by krisshP

In January 2012 paper unit 2 q18a, they said core - cladding. Does this means that light travels from core into cladding, or light travels from cladding? HOW do you know?

Thanks

It does not matter.
The critical angle will always be in the more denser medium or the medium with a higher refractive index or a lower speed.

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Reply 165

Kitnimohabbathai

oopps sorry about the capitals...for that question I did it the other way round and got 0.67 but the answer was 1.49...so is the refractive index always greater than 1? is there any rule? it said core-cladding so I did core over cladding but it was the other way round?

Reply 166

RoyalBlue7

19

Original post by Kitnimohabbathai

oopps sorry about the capitals...for that question I did it the other way round and got 0.67 but the answer was 1.49...so is the refractive index always greater than 1? is there any rule? it said core-cladding so I did core over cladding but it was the other way round?

Use the formula given to do it like this...

Spoiler

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(edited 10 years ago)

Reply 167

SnowStorm94

Original post by JoshThomas

Did anyone explain this:'( im so stuck the answer says A

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This one is easy... What you have is a variable resistor in variable resistors Current is indirectly prop to resistance.... Think it's that when you increase current you cause a heating effect which causes resistance to increase... The voltage is constant so when resistance increase something had to go down/decrease that is the current :smile:

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Reply 168

Kitnimohabbathai

Original post by StUdEnTIGCSE

Use the formula given to do it like this...

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what formula? the only formula is sini/sinr....what formula do you use when you have velocities?

Reply 169

SnowStorm94

Original post by KBenzema

Hi,

Why is it D and not C?

Well when you move away from something what is increasing between you and that person? :smile:

distance... Frequency stays same

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Reply 170

RoyalBlue7

19

Original post by Kitnimohabbathai

what formula? the only formula is sini/sinr....what formula do you use when you have velocities?

The formula given ...this

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1369931970643.jpg29.7KB

Reply 171

krisshP

14

Original post by StUdEnTIGCSE

It does not matter.
The critical angle will always be in the more denser medium or the medium with a higher refractive index or a lower speed.

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For core to cladding ie more dense to less dense
U=1.96/2.03
SinC=1/U
C=sin^-1(1/U)
C=sin^-1(1/(1.96/2.03))
C=math error on calculator! :frown:

WTF

I thought you said it does not matter?

Btw for calculations do you use accurate values e.g. fractions or rounded values? If you use rounded you give slightly different answer as I used 2.03/1.96 rounded when doing cladding to core and got something like 71... is this okay?

Reply 172

RoyalBlue7

19

Original post by krisshP

For core to cladding ie more dense to less dense
U=1.96/2.03
SinC=1/U
C=sin^-1(1/U)
C=sin^-1(1/(1.96/2.03))
C=math error on calculator! :frown:

WTF

I thought you said it does not matter?

Btw for calculations do you use accurate values e.g. fractions or rounded values? If you use rounded you give slightly different answer as I used 2.03/1.96 rounded when doing cladding to core and got something like 71... is this okay?

Ah, it does matter (I mean it doesn't matter how they have worded it :dontknow:

). Core-cladding or cladding-core boundary its the same thing but :sigh:

Better check the marking scheme. :smile:

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Reply 173

B-Stacks

13

Original post by krisshP

For core to cladding ie more dense to less dense
U=1.96/2.03
SinC=1/U
C=sin^-1(1/U)
C=sin^-1(1/(1.96/2.03))
C=math error on calculator! :frown:

WTF

I thought you said it does not matter?

Btw for calculations do you use accurate values e.g. fractions or rounded values? If you use rounded you give slightly different answer as I used 2.03/1.96 rounded when doing cladding to core and got something like 71... is this okay?

For those type questions I always feel its best to know the full snells law. Then you can't go wrong.

n1sini=n2sinr
n1v1=n2v2

Where the '1's' are the speed/refractive index in medium one and '2's' the speed/refractive index in medium two.

Reply 174

Baa

Hi, in the spec it says to "recognise and use the relationship between phase difference and path difference".

I don't really get this... what is the relationship between them. Can someone please explain?
Thank you! :smile:

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Reply 175

mickeycharb

Original post by blacknightking

Help needed with jan 2009 Q12) part b

ImageUploadedByStudent Room1369323801.617115.jpg

I don't know how to do part b.please explain.

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Have done this before a while ago, I think the water is moving vertically down at point Y

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Reply 176

Zoeyyy

I have a question. The wind is blowing across the end of a pipe 1.5 m and sets up a standing wave. The pipe is closed at one end. What is the lowest frequency of sound that will be produced?

since its the lowest frequency I used 2 x 1.5 = 3 cm for the wavelength
and then 330/3 = frequency
But the wavelength is supposed to be 6 cm. What am I doing wrong? O.O

Reply 177

RoyalBlue7

19

Original post by Baa

Hi, in the spec it says to "recognise and use the relationship between phase difference and path difference".

I don't really get this... what is the relationship between them. Can someone please explain?
Thank you! :smile:

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Well phase difference is the fraction of difference of oscillation of one oscillator to another. It measured as an angle in radians or in degrees. It actually sees how the sine curve of one wave has been translated to get another along the x axis. A translation or phase difference of 0,2π,4π
...2nπ (multiples of 2π) gives the same curve back again so its thought to be in phase, while a phase angle difference of π,3π....2n+1π gives a sine curve that is totally opposite so its out of phase.
You can work with degrees also.

Path difference is the difference in the distance travelled between two waves from their source. It can be measured as distance or in wavelength. If the difference shows a multiple of one wavelength they are in phase while a odd multiple of half wavelengths(0.5lamda,1.5lambda) shows an exactly out of phase points in the two wave. The significance of this can be seen in the principle of superposition of waves, diffraction and interference patterns etc

The relationship between these two variables are simple. A phase angle difference of 2π or 360° corresponds to 1 lambda path difference so use this ratio method

Spoiler

(edited 10 years ago)

Reply 178

RoyalBlue7

19

Original post by Zoeyyy

I have a question. The wind is blowing across the end of a pipe 1.5 m and sets up a standing wave. The pipe is closed at one end. What is the lowest frequency of sound that will be produced?

since its the lowest frequency I used 2 x 1.5 = 3 cm for the wavelength
and then 330/3 = frequency
But the wavelength is supposed to be 6 cm. What am I doing wrong? O.O

Now its difference to producing standing waves in a string. The harmonics are produced for odd multiples of half wavelength. Like this

uploadfromtaptalk1369979390183.jpg8.0KB

Reply 179

DanielDaniels

14

Original post by StUdEnTIGCSE

Well if you ever want to define a physical quantity you can form an equation and make the denominator equal to one. (If you want to define a unit then make both the numerator and denominator one)

voltage = \dfrac{energy}{charge}

So voltage is energy transferred per unit charge or per coulomb of charge. (And one volt is one joule transferred per coulomb charge).

This does not specify if it is inputting or outputting electrical energy. So we have emf.

Emf of a battery or cell is the total work done by the cell against both the internal and external circuits per coulomb of charge that passes through it. It is different to the terminal pd learnt at GCSE which does not recognise internal resistance so terminal PD is the work done by the cell in the external circuit only.

In other words emf is the total electrical energy provided to each coulomb of charge by a power supple. This energy is then used to do work in the internal circuit (against internal resistance) and in the external resistance. Lost volts us the work done in the internal circuit against internal resistance.

So emf=lost volt+terminal pd
Terminal PD= emf - lost volt

V=E-Ir

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Thank you sooo much really!
one more question (from CIE) but who cares ...it's still electricity!
http://vvcap.net/db/rhNqORlDMtAxemgMkbWe.htp http://vvcap.net/db/DlCJFeu-fbrM6OA7lLmZ.htp
the second last one when it says : closed open open .... i thought it's 3 ?

(edited 10 years ago)

Edexcel Physics Unit 2 "Physics at work" June 2013

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