[paul272]

http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN11.PDF

May someone please help me with questions 16 and 21

[callmenighthawk]

(Original post by paul272)
http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN11.PDF

May someone please help me with questions 16 and 21

can sort of do 16 by using logic but

45 r = 50 r - 1.5
5r = 50 x 1.5
5r = 75

r = 15

21

force on a moving charge is = Bqv, so I think it's A.

ahh piss, it wants the time, so it wants velocity, so the answer is C because the force is the same for both (same charge) but f = mv^2 / r says the velocity will be slower for the bigger mass, so the time period is bigger

[don'tTRIP.]

Could someone please explain to me why Q4 is -z?

http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN10.PDF

This does involve the left-hand rule, right?

[callmenighthawk]

(Original post by don'tTRIP.)
Could someone please explain to me why Q4 is -z?

http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN10.PDF

This does involve the left-hand rule, right?

the electron is travelling clockwise, the force must always act to the centre, and the middle finger must be in the opposite direction to the direction of the electron, so you can work it out using the LHR

[callmenighthawk]

someone explain 14 please i got the right calculation but don't understand why it's minus

http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN11.PDF

[don'tTRIP.]

(Original post by callmenighthawk)
someone explain 14 please i got the right calculation but don't understand why it's minus

http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN11.PDF

The formula for electric potential energy is (Q1)(Q2)/4pi(epsilonNOUGHT)(r)

So, just sub in your values the charge on an electron is -1.6x10^-19, whilst the charge on a proton is +1.6x10^-19.

Sub in and you should -2.3 x 10^-18

[don'tTRIP.]

Could someone give me an explanation on why racing tracks are banked? I know all of the basics, but I don't how to explain it.

[callmenighthawk]

(Original post by don'tTRIP.)
Could someone give me an explanation on why racing tracks are banked? I know all of the basics, but I don't how to explain it.

you won't need to know that, i'm not sure if I even know that but I wouldn't be worried

mind giving me the basics anyway ?

[FrightBright]

(Original post by callmenighthawk)
you won't need to know that, i'm not sure if I even know that but I wouldn't be worried

mind giving me the basics anyway ?

The main purpose of banking the track is to reduce friction on the tyres from centripetal forces. When you bank the track the contact force can contribute to the centripetal force.

So if we say contact force is N, and its banked (theta) from the horizontal, then Ncos(theta) + Frictional forces = mv^2.

If it wasnt banked then the frictional forces would have to be greater, hence worn tyres (for example). I think it helps you go round bends faster not sure though.

[radiator0505]

Could someone go over how to do proportion calculations?
The entire process baffled me the first time round and I've outright forgotten now.

[internet tough guy]

Can someone explain how a mass spectrometer works please?

It says that the beams enter at the same velocity beause they go through the 'velocity selector' - wth is this?

[radiator0505]

(Original post by internet tough guy)
Can someone explain how a mass spectrometer works please?

It says that the beams enter at the same velocity beause they go through the 'velocity selector' - wth is this?

Are they even on the spec? My teacher never mentioned them.
This is an absolute shot in the dark here but maybe the velocity selector is a uniform magnetic field perpendicular to the velocity of the beams?
Then because F=BQv the faster travelling beams will be deflected further and won't enter the spectrometer?

[callmenighthawk]

Mass spectrometer is definitely not in the physics course

[internet tough guy]

(Original post by radiator0505)
Are they even on the spec? My teacher never mentioned them.
This is an absolute shot in the dark here but maybe the velocity selector is a uniform magnetic field perpendicular to the velocity of the beams?
Then because F=BQv the faster travelling beams will be deflected further and won't enter the spectrometer?

its in one of the 'how science work' textbox in the book, if you just have a look at it, its on somewhere in chapter 7 I think

[internet tough guy]

(Original post by callmenighthawk)
Mass spectrometer is definitely not in the physics course

how do you know for sure?

I mean, they have it on the official textbook, why did they bother having such a detailed page on it , argh wasted about 30mins on trying to understand it.

[Troll the Trolls]

they can ask you questions on it that dont require u to know how it works so u dont need to learn anything about it, its on the as chemistry sylabus though

[schizopear]

(Original post by internet tough guy)
Can someone explain how a mass spectrometer works please?

It says that the beams enter at the same velocity beause they go through the 'velocity selector' - wth is this?

ions are accelerated to a uniform speed using an electric field to ensure that differences in deflection are due to differences in mass rather than momentum

[internet tough guy]

(Original post by schizopear)
ions are accelerated to a uniform speed using an electric field to ensure that differences in deflection are due to differences in mass rather than momentum

I thought it says somewhere in the textbook that electric fields don't change the velocity of the particle or something along the lines of this:

''no work is done by the magnetic field on the particle as the force always acts at right angles to the velocity of the particle . Its direction of motion is changed by the force but not the its speed. The kinetic energy of the particle is unchanged by the magnetic field''

thats on page 113 on nelson thornes aqa textbook

[internet tough guy]

On page 127 of the nelson thornes textbook, why is it that in figure 3, it shows the coil parallel with the field lines with emf at a maximum?

is it just me, or is it contradictory, just on the same page, they mention that ''the induced emf is zero when the sides of the coil move parallel to the field lines''

[callmenighthawk]

(Original post by internet tough guy)
On page 127 of the nelson thornes textbook, why is it that in figure 3, it shows the coil parallel with the field lines with emf at a maximum?

is it just me, or is it contradictory, just on the same page, they mention that ''the induced emf is zero when the sides of the coil move parallel to the field lines''

that threw me off too, but know that the emf is at a maximum when the coil is cutting MOST of the field, as emf = deltaBAN/t , so when the face of the coil is parallel to the field. i.e the change in area or b field usually, as its' difficult to quickly wrap wires on to coils quickly!

i think that sentence meant instead of the coil turning clockwise/anticlockwise perpendicular to the field, it would be turning anticlockwise/clockwise in the same plane, so there would be no change in area cut even though the the area cut is a maximum. a little bit confusing and should really have a diagram considering it's not the easiest topic at all (my worst topic infact)