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Edexcel Physics Unit 2 "Physics at work" June 2013 Watch

  • View Poll Results: The last question - Does resistance increase or decrease?
    It increases ( using V=IR or some other method)
    70.73%
    It decreases using the 'lattice vibrations' theory
    29.27%

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    This thread is now completely off topic.

    Can someone explain the entire concept of standing waves? Really dont understand it?
    Eg: how they are formed, what properties they have. stuff like that.

    Thanks in advance.
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    (Original post by JoshThomas)
    Does anyone understand the theory behind max power occurs when load resistance equals internal resistance

    Posted from TSR Mobile
    Well you can show it by calculus

    Spoiler:
    Show



     





V=E-Ir



VI=I(E-Ir)



P=VI=EI-I^2R



therefore \dfrac{dP}{dI}=E-2Ir





At max P,



 \dfrac{dP}{dI} =0



0=E-2Ir



2Ir=E



I= \dfrac{E}{2r} \ast



BUT:



V=E-Ir and V=IR



V=E-Ir=IR



E=I(R+r)



I= \dfrac{E}{R+r} \bullet



at max P,





\ast=\bullet  







Therefore 2r=R+r



R=r

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    (Original post by shaw16)
    This thread is now completely off topic.

    Can someone explain the entire concept of standing waves? Really dont understand it?
    Eg: how they are formed, what properties they have. stuff like that.

    Thanks in advance.
    Standing waves are formed when an incident wave is superposed (NOT superimosed) with a reflected wave traveling in the opposite direction. When these two waves 'superpose', it just means the vector displacement at a point is the sum of the individual displacement of both waves. So if both are crests at one point, you'll get a big crest (antinode). If one is a crest and one is a trough, you get complete cancellation (node). So a standing wave is just a pattern of nodes and anti-nodes, with no net transfer of energy. Amplitudes are maximum at anti-nodes and zero at nodes.

    Hope this helped
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    (Original post by shaw16)
    This thread is now completely off topic.

    Can someone explain the entire concept of standing waves? Really dont understand it?
    Eg: how they are formed, what properties they have. stuff like that.

    Thanks in advance.

    call me on skype right now
    webteacher1
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    Need help with June 2011 Q)15 part a)ii

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    The answer for this in marking scheme is this.

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    Please explain me the theory behind this.Why z is at the bottom of the potential divider.


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    (Original post by blacknightking)
    Need help with June 2011 Q)15 part a)ii

    Name:  ImageUploadedByStudent Room1369656873.834261.jpg
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    The answer for this in marking scheme is this.

    Name:  ImageUploadedByStudent Room1369656905.754342.jpg
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Size:  27.1 KB

    Please explain me the theory behind this.Why z is at the bottom of the potential divider.


    Posted from TSR Mobile
    The voltmeter reading will be zero when IR = 0. For IR to be 0, R must be zero. When z is at the bottom, the resistor is effectively disconnected from the circuit, so theoretically there is no resistance. Therefore the voltmeter reading will be 0 because IR = 0 (V=IR)
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    (Original post by GCSE-help)
    The voltmeter reading will be zero when IR = 0. For IR to be 0, R must be zero. When z is at the bottom, the resistor is effectively disconnected from the circuit, so theoretically there is no resistance. Therefore the voltmeter reading will be 0 because IR = 0 (V=IR)
    "When z is at the bottom, the resistor is effectively disconnected from the circuit" , I couldn't get it .Please can you explain further.


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    (Original post by blacknightking)
    "When z is at the bottom, the resistor is effectively disconnected from the circuit" , I couldn't get it .Please can you explain further.


    Posted from TSR Mobile
    The resistor is no longer part of the circuit, because you're attaching the wire right at the bottom, so the current doesn't go through the resistor any more, it goes straight through the wire.

    Sorry if i'm not explaining well, but its the best I can do
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    (Original post by GCSE-help)
    The resistor is no longer part of the circuit, because you're attaching the wire right at the bottom, so the current doesn't go through the resistor any more, it goes straight through the wire.

    Sorry if i'm not explaining well, but its the best I can do
    Thanx a lot.i understood now.



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    Help needed with Jan 2012 questions 2 and 7

    Here's 2

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    The answer is A according to mark scheme.

    Here's 7

    Name:  ImageUploadedByStudent Room1369670983.190906.jpg
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    The answer is A according to mark scheme.


    Please someone explain the theories behind these two questions.


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    (Original post by blacknightking)
    Help needed with Jan 2012 questions 2 and 7

    Here's 2

    Name:  ImageUploadedByStudent Room1369670871.494657.jpg
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    The answer is A according to mark scheme.

    Here's 7

    Name:  ImageUploadedByStudent Room1369670983.190906.jpg
Views: 371
Size:  59.8 KB

    The answer is A according to mark scheme.


    Please someone explain the theories behind these two questions.


    Posted from TSR Mobile
    I can explain the second one, but first has threw me, so if anyone can help with that, okay so its been proposed that the universe is expanding, so in this case the star is moving away from the earth, so in the opposite direction of the direction of the light emitted, so they are stretched out and have longer wavelength and lower frequency when they reach earth, its a good example of the doppler effect, the frequency emitted or wavelength doesn't change, the only thing that does change is the distance travelled by each successive wave, and since waves are continuous as the star moves it is constantly emitting this light but at the same time its moving away causing a shift towards the red( longer wavelength)

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    (Original post by blacknightking)
    Help needed with Jan 2012 questions 2 and 7

    Here's 2

    Name:  ImageUploadedByStudent Room1369670871.494657.jpg
Views: 373
Size:  56.9 KB

    The answer is A according to mark scheme.

    Here's 7

    Name:  ImageUploadedByStudent Room1369670983.190906.jpg
Views: 371
Size:  59.8 KB

    The answer is A according to mark scheme.


    Please someone explain the theories behind these two questions.


    Posted from TSR Mobile
    I thought the first one was D cause red shift is when the star is moving away and therefore have longer wavelength. No idea how they got A :'(

    *Blue shift is when the star is coming towards us and there is shorter wavelength.
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    (Original post by blacknightking)
    Help needed with Jan 2012 questions 2 and 7

    Here's 2

    Name:  ImageUploadedByStudent Room1369670871.494657.jpg
Views: 373
Size:  56.9 KB

    The answer is A according to mark scheme.

    Here's 7

    Name:  ImageUploadedByStudent Room1369670983.190906.jpg
Views: 371
Size:  59.8 KB

    The answer is A according to mark scheme.


    Please someone explain the theories behind these two questions.


    Posted from TSR Mobile
    Blue light has a higher frequency than red light. In fact, red is the lowest frequency visible light. Therefore, for the frequency to shift towards red it must have decreased.

    We know from c=fλ, where c is a constant value for the speed of light in a vacuum, that if the frequency decreases, subsequently the wavelength, λ, must increase as a result.

    The only option that matches this is A; the distance travelled by each wavefront increases, which means the wavelength increases. The reason for this increase is that the galaxy is travelling at a faster velocity than our galaxy, thus red shift occurs- the light "shift" towards the red end of the spectrum.

    I believe somebody already took care of the second question.

    Hope I helped
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    (Original post by Freddy-Francis)
    I thought the first one was D cause red shift is when the star is moving away and therefore have longer wavelength. No idea how they got A :'(

    *Blue shift is when the star is coming towards us and there is shorter wavelength.
    The star isn't emitting a longer wavelength. If we were to observe the wavelength emitted from the star it would look different to the wavelength we observe here. You're right that it's red shift, but D implies that the star is doing something different. The idea is that it is the difference in velocities of the two galaxies that causes the red shift, not any change in the emission of light that either galaxy makes.

    Instead, because the galaxy is moving away from us at an increased rate the distance that light has to travel increases each time. As a result, the wavelength we measure increases, thus the frequency must decrease in accordance with c=fλ. It's the type of question that people understand the theory for, but get wrong because they've worded it horribly on purpose!

    Does that make sense?
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    When showing the wave behaviour of electrons by diffraction using graphite atoms why is it a good idea to shoot the electrons one by one at the graphite ...?:confused:
    Is it because the electrons would then have no chance of colliding with each other and or repelling each other and "diffract" even while its a particle only?
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    (Original post by studentigcse)
    well you can show it by calculus

    Spoiler:
    Show



     





v=e-ir



vi=i(e-ir)



p=vi=ei-i^2r



therefore \dfrac{dp}{di}=e-2ir





at max p,



 \dfrac{dp}{di} =0



0=e-2ir



2ir=e



i= \dfrac{e}{2r} \ast



but:



V=e-ir and v=ir



v=e-ir=ir



e=i(r+r)



i= \dfrac{e}{r+r} \bullet



at max p,





\ast=\bullet  







therefore 2r=r+r



r=r

    wow! Thanks it makes sense now!.:d
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    Did anyone explain this:'( im so stuck the answer says AName:  1369684050061.jpg
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    (Original post by JoshThomas)
    Did anyone explain this:'( im so stuck the answer says AName:  1369684050061.jpg
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    Posted from TSR Mobile
    Think I explained this to someone else earlier... anyway,

    I = \dfrac{V}{R}

    We're given that V is constant, so the variables are R and I. So the graph of I against R would be a reciprocal graph.
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    http://www.edexcel.com/migrationdocu...e_20090115.pdf
    january 2009, 16a) and b)
    I don't understand why in a) the resistance of A is less than the resistance of B?
    I don't get b) at all :c.
    Thankyou :c
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    (Original post by pureandmodest)
    http://www.edexcel.com/migrationdocu...e_20090115.pdf
    january 2009, 16a) and b)
    I don't understand why in a) the resistance of A is less than the resistance of B?
    I don't get b) at all :c.
    Thankyou :c
    If you think of it like this, brightness of bulb depends on power, power is given by i^2 r, if I is the same for both as in series therefore the brighter bulb must have more resistance, and part b) the same potential difference occurs in both branches the only difference is current and resistance of the lamps, as current takes route of least resistance, so the one with least resistance will shine brighter as p=vi, and as v is equal, a larger I, will cause a larger power, more brighter, from part a, its established that a has less resistance, so A will shine brighter in paralell, hope this helps

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