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Jun:28.... PHY6.. how was it??

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Reply 20

---Master A---

6 i think.....anywayz wat did every1 write down for the energy transfer between the battery and lamp?

Reply 21

Kabeer

CRTs scan from left to right :eek:

, whaa ! this wasnt in any of the physics modules syllabus, that is a bit unfair how do they expect us to know it!

Unless of course we watched a tv program caled the history of television 8 years ago :wink:

Reply 22

S1M

rastaman

how many marks was the inelastic scattering question?

5 iirc

that paper was quite easy overall IMO... im confident that ill get my A overall now :biggrin:

scattering one was tough... and so was why does shadow on tv shift right... but rest was very easy imo...

Reply 23

Kabeer

---Master A---

6 i think.....anywayz wat did every1 write down for the energy transfer between the battery and lamp?

Chemical potential--->Electric----->heat and light

Reply 24

Bezza

I thought it was generally ok, there were just a couple of dodgy bits. I haven't studied deep inelastic scattering for ages so couldn't remember the details although I knew roughly what the differences were. What did everyone put for the difference between a beam of He ions from a Van de Graaff generator and a beam of alpha particles for a radioactive source?

With the co-axial cable I could have done with 2 right hands!! :rolleyes:

I was using the right and grip rule but with both hands so I ended up thinking that inside the insulator the magnetic fields from the sheath and the core would cancel out, rather than reinforce each other.

For the radioactive power I got 10.9 years (I think) to drop to 15W - anyone else get this?

Reply 25

Camford

Kabeer

Also the question about how that power is 20W :eek:

, i just played arond with the numbers and the closest thing i got to 20 was 31, so i just wrote that.
The next part i gueused as 9 years (3/4 of total power therefore half of half life).

I got 7.47 years. I hope that's acceptable using powers of 2 rather than e^whatever it was.

Kabeer

The last quesiton was terrible, especially with the cable :confused:

anyone know the answer to that one?

I think the electric field is a radio field. The field outside the cable is in one direction; the field inside the cable in another direction. Don't know whether the copper wire is +ve or -ve, they probably would take a mark off me for that one. Then again, how was I to know which is positive.

For the magnetic field inside the cable I said something about the fields close to the wire and the sheath being in opposite directions. and there will be a region where there's no apparent fields. The field outside the cable is main influnced by the current in the sheath.

God I hate that paper.

Reply 26

S1M

Camford

I got 7.47 years.

yeah i got that too :smile:

so did one of my mates :smile:

Reply 27

S1M

Kabeer

Unless of course we watched a tv program caled the history of television 8 years ago :wink:

lol

Reply 28

Kabeer

how did yu get 7... years?

Reply 29

Camford

Kabeer

CRTs scan from left to right :eek:

Apparently it has got something to do with the lawn. :eek:

I think it's something to with the cathode ray osciliscope being scanning from left to right. But I agree it is an unfair question.

Reply 30

Bezza

S1M

yeah i got that too :smile:

so did one of my mates :smile:

Hmm. How did you all do that question?

Reply 31

Hovick

i got the time to be 9 years, i assumed the power droped to 10W in one halflife, so 15W would be half the half life?

Reply 32

S1M

well i cant remember it all so i cant replicate my answer here...

but the steps were
work out power = energy per second
work out lambda
and use energy released per decay with lambda somehow
then work out half life (convert to seconds...)
and i did a few more things to piece it all together but i cant remember...

Reply 33

zoidberg

Bezza

Hmm. How did you all do that question?

you could use the energy per decay to find out how many atoms you'd need to be decaying at that instant, and then find out how many atoms would need to be present at that time. then used the decay constant to find out how long it would take to decay to that amount of atoms, got in the region 7.3 to 7.7 cant remember exactly. i think i did a long winded method

Reply 34

Camford

Bezza

Hmm. How did you all do that question?

Which should come out to be 7.47 or I would lose some marks.

Reply 35

S1M

Hovick

i got the time to be 9 years, i assumed the power droped to 10W in one halflife, so 15W would be half the half life?

no that doesnt work as the decay is not constant or in a linear relationship...

Reply 36

zoidberg

btw what did anyone put for the analogy of A=(lamda)N ? i had something stupid, mind went blank

Reply 37

Camford

zoidberg

You could use the power of 1/2 method or the P=(P0)e^(lambda *t1/2)

Reply 38

Camford

zoidberg

btw what did anyone put for the analogy of A=(lamda)N ? i had something stupid, mind went blank

I=Q/(RC), I is dQ/dt, i.e the rate of discharge.

A is dN/dt, i.e. the rate of decay.

Reply 39

S1M

zoidberg

btw what did anyone put for the analogy of A=(lamda)N ? i had something stupid, mind went blank

yup that one with V=(1/C) x Q i think
edit... actually i think i put what the above pst says.... for some reason i forget it all... well anyway i should be focusing on next exam now... p4...