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Edexcel Unit 4: Physics on the Move 6PH04 (11th June 2015)

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Reply 560

h.h.h.h.h

So what would it be for an A roughly?

Reply 561

PhysicsFiend

Isn't time period for the roundabout question 3.03, not 3.08? :

mw^2r=0.35mg
w^2r=0.35g
w=sqrt(0.35g/r)
2pi/T=sqrt(0.35g/r)
T=2pi*sqrt(r/0.35g)
T=3.03 (3.s.f.)

Reply 562

LukeBarnett

Original post by PhysicsFiend

Isn't time period for the roundabout question 3.03, not 3.08? :

mw^2r=0.35mg
w^2r=0.35g
w=sqrt(0.35g/r)
2pi/T=sqrt(0.35g/r)
T=2pi*sqrt(r/0.35g)
T=3.03 (3.s.f.)

I got 3.03s as well

Reply 563

Matteddolls

Do we know how much each question was worth? I've got a couple of them down, not sure about Qs 11,12 and 18 though! Help Please!

Reply 564

Freddy-Francis

Original post by Dancatpro

wait what, how do you know your coursework mark?!

our school didnt tell us ours :frown:

Lol. xD Our teacher told us the moderated mark

Original post by cerlohee

A B or a C. But the actual grade you'll get depends on what you got last year as well as how you do in unit 5!!

i need around 100 UMS for an A in unit 5 :/ I will be happy with a B as uni wants a B. But A will be good :smile:

Reply 565

Dancatpro

Original post by Freddy-Francis

Lol. xD Our teacher told us the moderated mark

i need around 100 UMS for an A in unit 5 :/ I will be happy with a B as uni wants a B. But A will be good :smile:

You can do it bro!

Reply 566

PhysicsFiend

11a=1 mark, 11b = 2 marks

Reply 567

Freddy-Francis

Original post by Dancatpro

You can do it bro!

Thanks buddy :tongue:

Reply 568

Kurly10

Original post by BP_Tranquility

For the capacitor q, I said that since W=½QV , if the charge halves then the energy stored is divided by 4. So the energy must halve before the charge halves . is that correct?

Posted from TSR Mobile

nope sorry mate, voltage aint constant, you have to substitute voltage for Q/C to get W=1/2 Q^2/c

Reply 569

Freddy-Francis

Original post by PhysicsFiend

11a=1 mark, 11b = 2 marks

for people who said current is upwards. will get 1 mark in that section if the magnetic field is in the right direction. As in field must be out of the page.

Reply 570

PhysicsFiend

Original post by Freddy-Francis

for people who said current is upwards. will get 1 mark in that section if the magnetic field is in the right direction. As in field must be out of the page.

I sadly did this exact thing, silly error :/ I'm assuming at least 1 ecf mark. Also, since 11b was 2 marks, I have a feeling they wanted something about it being perpendicular (to the current), perhaps?

(edited 8 years ago)

Reply 571

Edac

My numerical answers:

3.03s (roundabout question)
15V
2.48e-5
7.08s for charge or something to decrease to 0.2
4.1A (emf induced)
0.035N (tension)
41 degrees
18000Hz

Reply 572

LukeBarnett

Original post by Edac

My numerical answers:

3.03s (roundabout question)
15V
2.48e-5
7.08s for charge or something to decrease to 0.2
4.1A (emf induced)
0.035N (tension)
41 degrees
18000Hz

Agreeing with all of them! :smile:

(Apart from I got 1.5V but miscalculated so the 15V is right)

Reply 573

horatiohorn

hey guys for the kid on the roundabout question, i did mv^2/r= 20x0.35. so 20v^2/0.8=7. then rearrange to get 20 v^2=5.6 so v=0.53.
since V=d/t and d=2pi r shouldnt t=d/v work to give 9.48seconds??!?!?!?!?!?

Reply 574

LukeBarnett

Original post by horatiohorn

The frictional force was given by the weight of the child, so the frictional force was 0.35x20x9.81

Reply 575

horatiohorn

Original post by LukeBarnett

The frictional force was given by the weight of the child, so the frictional force was 0.35x20x9.81

oh....fml im an idiot. goodbye uni. cheers mate

Reply 576

LukeBarnett

Original post by horatiohorn

oh....fml im an idiot. goodbye uni.

I did that too at first had to double take, sure you'll still get method marks, don't worry about it too much :smile:

Reply 577

BP_Tranquility

Grade boundary predictions?

Reply 578

h.h.h.h.h

Original post by Edac

My numerical answers:

3.03s (roundabout question)
15V
2.48e-5
7.08s for charge or something to decrease to 0.2
4.1A (emf induced)
0.035N (tension)
41 degrees
18000Hz

2.48e-5?

Reply 579

Edac