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AQA A Physics Unit 4 24th Jan 2012

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just did the rest of the multiple choice paper (january 2011) and got questions 2, 21 and 22 wrong... can anyone explain those for me? thank you!

particularly frustrated by question 2 as it doesn't look that hard but i keep getting the wrong answer...

Original post by LOLford
u guys are amazing thanx! (and i thought some1 would need my help on the other questions ^^)


i'd like some help with the above questions if you don't mind :smile:
Reply 181
Original post by schizopear
just did the rest of the multiple choice paper (january 2011) and got questions 2, 21 and 22 wrong... can anyone explain those for me? thank you!

particularly frustrated by question 2 as it doesn't look that hard but i keep getting the wrong answer...



i'd like some help with the above questions if you don't mind :smile:


2- omega = theta/ time
convert 30 degrees to radians and put that in to get omega which is 25PI/3
then v = r omega, times your omega by the radius (the diameter is given so you need to half) then you end up with 0.5 PI :smile:
21 - flux linkage equation, (BANcos theta) using 50 degrees as the angle then do the same but use 0 as the angle and do the answer from 0- answer from 50
you end up with 2.1x10^-3 as required
22- 60 * 150 * 10 gives area the plane crosses then multiply this by the magnetic field strength to give the value of flux cut which equals 0.9 or 9*10^-1

hope that helped and the explanations were alright :biggrin:
Original post by Sam-8
2- omega = theta/ time
convert 30 degrees to radians and put that in to get omega which is 25PI/3
then v = r omega, times your omega by the radius (the diameter is given so you need to half) then you end up with 0.5 PI :smile:
21 - flux linkage equation, (BANcos theta) using 50 degrees as the angle then do the same but use 0 as the angle and do the answer from 0- answer from 50
you end up with 2.1x10^-3 as required
22- 60 * 150 * 10 gives area the plane crosses then multiply this by the magnetic field strength to give the value of flux cut which equals 0.9 or 9*10^-1

hope that helped and the explanations were alright :biggrin:


thanks :smile:

i knew i was being completely thick with the first one - i just forgot to halve the diameter to get the radius :redface:
Reply 183
Original post by schizopear
thanks :smile:

i knew i was being completely thick with the first one - i just forgot to halve the diameter to get the radius :redface:


Haha I know what you mean, its when you know your doing it right, but for some reason the answer on the calculator isn't the one in the book! :frown:
Does anybody know what synoptic stuff (from AS) we have to learn? Because I thought it was just simple stuff like F=kx and density, but on Jan 2011 past paper there was a question on standing waves/fundamental vibrations :s Now I'm worried because I can't possibly learn the whole AS stuff (since they don't tell you what of the AS you need to know) by next tuesday, but neither can I afford to drop the few marks if they throw in a synoptic question like that again
Original post by Sam-8
Haha I know what you mean, its when you know your doing it right, but for some reason the answer on the calculator isn't the one in the book! :frown:


The NT book have a lot of errors in it! :smile:
Original post by Bombosh Man
Does anybody know what synoptic stuff (from AS) we have to learn? Because I thought it was just simple stuff like F=kx and density, but on Jan 2011 past paper there was a question on standing waves/fundamental vibrations :s Now I'm worried because I can't possibly learn the whole AS stuff (since they don't tell you what of the AS you need to know) by next tuesday, but neither can I afford to drop the few marks if they throw in a synoptic question like that again


Which question was that?
Could someone please help me with Q22 the Jan 2011 paper. I'm not sure which equations are relevant.

http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA4-1-W-QP-JAN11.PDF

Thank you!
Original post by jones_wise
Could someone please help me with Q22 the Jan 2011 paper. I'm not sure which equations are relevant.

http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA4-1-W-QP-JAN11.PDF

Thank you!


see five or six posts above yours... i asked the same question earlier!
Original post by jones_wise
Could someone please help me with Q22 the Jan 2011 paper. I'm not sure which equations are relevant.

http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA4-1-W-QP-JAN11.PDF

Thank you!


Find the area covered in 10s by finding area in 1s (which is 60m*150m/s = 9000m^2/s) and then multiply that by 10s - 9000m^2/s*10s = 90000m^2
Then find flux by doing BA = 1*10^-5 * 90000 = 9*10^-1 which is answer choice D.
Someone help me with Q7 on the specimen paper?
http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA4A-W-SQP-07.PDF
Original post by schizopear
see five or six posts above yours... i asked the same question earlier!

Oops, I didn't spot that, haha. Thanks!

Original post by handsome7654
Find the area covered in 10s by finding area in 1s (which is 60m*150m/s = 9000m^2/s) and then multiply that by 10s - 9000m^2/s*10s = 90000m^2
Then find flux by doing BA = 1*10^-5 * 90000 = 9*10^-1 which is answer choice D.
Thank you!
Original post by handsome7654
Someone help me with Q7 on the specimen paper?
http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA4A-W-SQP-07.PDF


The answer is A (confirmed on mark scheme).

Work=Force×distance×cos(θ) Work = Force \times distance \times cos(\theta)

Where θ=\theta = angle between force and velocity (movement).

Here, the angle is 90° - the force is towards the centre of the circle, but the velocity is at right angles to this.

Therefore, no work is done. :smile:
(edited 12 years ago)
Original post by Xero Xenith
The answer is A (confirmed on mark scheme).

Work=Force×distance×cos(θ) Work = Force \times distance \times cos(\theta)

Where θ=\theta = angle between force and velocity (movement).

Here, the angle is 90° - the force is towards the centre of the circle, but the velocity is at right angles to this.

Therefore, no work is done. :smile:


Thankyou very much.. I'v been thinking it was answer choice B but obv I was wrong! lol :biggrin:
Reply 194
does anyone have the june 2011 paper urgently needed for revision
Reply 195
A mass on the end of a string is whirled round in a horizontal circle at increasing speed until
the string breaks. The subsequent path taken by the mass is
A a straight line along a radius of the circle.
B a horizontal circle.
C a parabola in a horizontal plane.
D a parabola in a vertical plane.

sum1 explain what a bloody parabola is, and the answer, please
Reply 196
oh and when do we use right hand and left hand, im so lost
Original post by paul272
A mass on the end of a string is whirled round in a horizontal circle at increasing speed until
the string breaks. The subsequent path taken by the mass is
A a straight line along a radius of the circle.
B a horizontal circle.
C a parabola in a horizontal plane.
D a parabola in a vertical plane.

sum1 explain what a bloody parabola is, and the answer, please


A parabola is an upside down U kinda shape. Imagine when you throw a ball - it goes up then down. That path is a parabola.

I think the answer is a parabola in a vertical plane - it will act like a projectile, though it'll be projected horizontally. I think it's still a parabola though :smile:

Original post by paul272
oh and when do we use right hand and left hand, im so lost


Left hand rule is for when a wire which has a current ALREADY IN IT has a magnetic field put round it. F=BIL - there's an L in there - Left hand!

Right hand rule is for the opposite situation - when a conductor is moving through a magnetic field, it will get an induced emf (and a current if it's connected up in a circuit).

Original post by swert
does anyone have the june 2011 paper urgently needed for revision


See attached; also includes markscheme (not posted here before I believe).
Gonna start revising today. :colone:
Can someone help me with question 4b iv on jan 2010?
http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA4-2-W-QP-JAN10.PDF

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