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Physics Unit 2. How was it ???!!!

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did anyone use E=hf for working out the frequency, i multiplied the work function by 1.6*10-19 and divided by planks constant, who else used my method :) and i said its ultraviolet range
 
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i think we are right cause using a simple equation was used in the area of gauge resistance question i mean it doesnt make sense to give us 2 questions depending on equations

That's true! Sounds good. Also, how did you find the polarising questions? I thought they were fairly easy but they all seemed to be asking the same thing lol
 
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did anyone use E=hf for working out the frequency, i multiplied the work function by 1.6*10-19 and divided by planks constant, who else used my method :) and i said its ultraviolet range

I used that and got everything right but then i crossed out uv and wrote x-ray :( I think they MIGHT accept either though because they are fairly close and we're not supposed to remember the EXACT wavelengths. Who knows, maybe MS will be generous
 
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i think we are right cause using a simple equation was used in the area of gauge resistance question i mean it doesnt make sense to give us 2 questions depending on equations
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hopefully we are right, i think the grade boundries are going to 57/59 for an A, unless people did bad on the electricity questions
 
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I used that and got everything right but then i crossed out uv and wrote x-ray :( I think they MIGHT accept either though because they are fairly close and we're not supposed to remember the EXACT wavelengths. Who knows, maybe MS will be generous

my ans was a pure guess, i was referring to the gold leaf experiment where you shine uv light on the surface
 
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Yesss! Do you think we were right though? :S Every one else seems to think resistance increased, which seems like an obvious answer because of less current.

i thought the resistance was going to increase too, but the question said 'what EFFECT does this have on the resistance' that means the current flowing through it is low, because that's how we're sending it. not because of a higher resistance.
i wrote what you did. hopefully it is right. i'm not sure though, just a guess :)
 
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That's true! Sounds good. Also, how did you find the polarising questions? I thought they were fairly easy but they all seemed to be asking the same thing lol

yaa first part i wrote difference between two filter is 90 so that they are perpendicular to eacch other
second part i wrote that its polarized so waves only allowed to vibrate in one directions hence less intensity
and finally i wrote that the light which was supposed to hit the left glass for example might hit the right glass which has a different plane of polarization hence not allowing the light to pass so image appears faint
and the definition i wrote photon is a particle of a light and a packet of energy since it was 2 marks
 
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I think the grade boundaries will be 55 for an A. It's a Unit 2 Physics paper after all, so they can't make it too high. And besides, no one from the exam hall said it was easy, many thought it was 'decent'.
 
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I think the grade boundaries will be 55 for an A. It's a Unit 2 Physics paper after all, so they can't make it too high. And besides, no one from the exam hall said it was easy, many thought it was 'decent'.

55 that seems kinda of low 58-59 is more like it
 
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Yeah maybe you're right. What do you think full UMS will be? I haven't seen the questions yet but I don't think I dropped more than 10 marks maybe. Inshallah

well just add 10 marks to the A boundary so if it was 55 a 120 would be 65 and so on
 
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The drift velocity of the electron was V/4 guys because as length increases the number of charge carriers also increases!
 
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The final question:

As the light output of bulbs is low power output is low.
P=I^2R therefore if P is low and I is decreased and remained constant the Resistance also decreases.

Can't remember what else I put but.
 
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For the Voltmeter one:

As the resistance is low the the current will be high.
As the voltmeter is connected in parallel to the cell if the resistance is low hence current is high therefore current is taken away from the cell resulting in less "lost volts''.

The ammeter is negligee since the ammeter is connected in series regardless of the resistance the current is unaffected throughout the rest of te cell.
 
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