Edexcel Physics Unit 2 "Physics at work" June 2013

Uni students - Complete our survey >>

Scroll to see replies

Reply 380

blacknightking

Original post by B-Stacks

Yes, if you know the the EMF and two Resistance values of the resistors.

What I mean is , which method awards the full marks.

Posted from TSR Mobile

Reply 381

B-Stacks

Original post by blacknightking

What I mean is , which method awards the full marks.

Posted from TSR Mobile

Either way usually. But they'd expect you to apply your knowledge of potential dividers by using the formula.

Reply 382

blacknightking

Help needed with jan 2011 question 5

ImageUploadedByStudent Room1370343371.093676.jpg

Please show the working.

Posted from TSR Mobile

Reply 383

Jaydude

Can someone help me with question 9, about I=nqva on june 2012 paper. link below: Thanks!

http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202012%20-%20QP/6PH02_01_que_20120525.pdf

Why is the answer 1/2 and not 2? Surely the drift velocity doubles when area has been doubled.
Therefore Speed of A/Speed of B = 2. But its not...help!

Reply 384

CWE

Original post by blacknightking

Help needed with jan 2011 question 5

ImageUploadedByStudent Room1370343371.093676.jpg

Please show the working.

Posted from TSR Mobile

The filament of M breaks. This means that the lamp in p-arallel is effectively removed. Now we have a bulb that was previoussly in parallel in series. When in parralel the resistance across the bulbs as a whole was lower (1/RT=1/R1+1/R2 etc.) but now the bulb is in series the resistance has increased. The total EMF stays constant so the voltage across the whole circuit is still the same. As the resistance has increased, looking at the equation V=IR we can see that the current must have decreased. We know that P=IV and that Voltage is split in a circuit equal to the ratio of the resistances. The lamp now has more resistance compared to before and so is a larger proportion of the total resistance than before , and hence takes more of the voltage. The lamp N is now a smaller proportion of the total resistance than it was before and so takes less P.D. P=IV and a greater V for the first lamp than previously and a smaller V for lamp N shows that the power will increase and decrease respectively.

Reply 385

CWE

Original post by Jaydude

They are in series and therefore they have the same current:

I=nAvq ~ If I is to stay the same whilst A has been doubled, its obvious that to offset this one of the other variables must be halved to maintain the same current. We are talking about A , so the Drift velocity of A will now be 0.5 its original drift velocity and the current has stayed constant. Now the ratio would just be the velocity of A over the velocity of B. 0.5v/1v = 0.5. Half anything over one of something will always be half.

Reply 386

username1190975

Original post by Jaydude

I = nqvA which means that v = I/nqvA
So when x-sectional area is doubled, CURRENT doubles, but the drift velocity doesn't - it halves.
Hope this helps

(edited 10 years ago)

Reply 387

kitty-boo

Can someone help me with question 10, about diffraction on January 2013 paper.:

The effect of diffraction is more noticeable, in everyday life, with sound than with light. This is because
A sound has a much longer wavelength than light
B sound is a longitudinal wave, light is a transverse wave
C sound is a mechanical wave, light is an electromagnetic wave
D sound travels more slowly in air than light does

Apparently the answer is A. But i don't understand. Please help! Thanks!

Reply 388

B-Stacks

Original post by kitty-boo

We know that light has a very small wavelength. In order to see diffraction we would need light to pass a really small gap/object which is approximately the same size as its wavelength. So it's highly unlikely for us to see this.

(edited 10 years ago)

Reply 389

CharlieTT

Original post by kitty-boo

Because of a longer wavelwngth the dustanxe between each wavefront is longer rherefore us easier to see. That question confused my class too dw.

Posted from TSR Mobile

Reply 390

kitty-boo

Original post by CharlieTT

Because of a longer wavelwngth the dustanxe between each wavefront is longer rherefore us easier to see. That question confused my class too dw.

Posted from TSR Mobile

Original post by B-Stacks

Thank you! :smile:

Reply 391

CWE

Original post by kitty-boo

Diffraction is the bending of a wave that occurs when a wave travels through a gap. The amount of diffraction is dependent on the wavelength of the wave relative to the size of the gap. If the wavelength is a similar size to the gap it will undergo diffraction and the light will bend, if you have the same gap but a wave that has a wavelength much smaller then the wave will be much smaller in size relative to the gap and will pass through will less diffraction. A good thing to thing about when visualizing diffraction is light and sound. We can hear around doors because the wavelength of sound is similar in size relative to the gaps and therefore sound bends round the corners allowing it to be heard even when not directly in the line of sight, the wavelength of light on the other hand is very small and is much smaller than the gap, it therefore undergo's very little diffraction and doesn't bend, so you must be in the Line of sight to see it.

Back to the question, the longer wavelength of sound means it is similar in size relative to the gaps encountered in the world and therefore diffracts more, the much smaller wavelength of light means it is often much smaller than gaps it passes through and undergoes no observable diffraction.

Hope that helped.

Reply 392

Jaydude

I cant get the unit 2 paper! Ffs. Its a unit 1 paper that i can only find, wheres unit 2?!? The link was for unit 1 paper :frown:

EDIT: nevermind

(edited 10 years ago)

Reply 393

KBenzema

Original post by cwe

here goes:

As we know, 2 resistors in parallel have less resistance than they would have if they were connected in series. 1/r1 + 1/r2 + 1/r3 = 1/total resistance. When this lamp is removed resistors that were in parallel and hence each had a lower resistance now have a greater resistance. The circuit without the lamp is effectively a series circuit in which current is split amongst components based on the proportion of the total resistance they represent. This component will have a larger resistance, will therefore receive a larger proportion of the voltage and the voltmeter will therefore read a larger value for the potential difference.

thank you! Brilliant! :smile:

Reply 394

Jaydude

Original post by krisshP

Oops sorry
http://www.mediafire.com/?5djntd8dd7ykb

I never actually checked the link, i wanted the paper as a mock. But Im trying to look for it and the link only has unit1 2013 jan papers....Wheres Unit 2!?!? :frown:

EDIT: nevermind , got it. Checked every PDF which were labblled as "6ph01" and got it. Please make titles more clearer next time!

(edited 10 years ago)

Reply 395

KBenzema

Energy loss depends on Current,Resistance or Power? when do we know if teres greater or smaller energy loss?

Reply 396

KBenzema

Energy loss depends on Current,Resistance or Power? when do we know if *theres greater or smaller energy loss? :smile:

Reply 397

CWE

Original post by KBenzema

Energy loss depends on Current,Resistance or Power? when do we know if *theres greater or smaller energy loss? :smile:

I believe it is essentially power. If something was to loose 3j/s it would be the same principal as using 3j/s, if that makes sense. P=IV so the power is dependant on those things, but you have to look at the power to get a decisive answer.