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AQA PA10 Physics A Synoptic (June 16th)

Reply 80

c12andn15

10

I would love to see the mark scheme for this tbh!

Reply 81

golddaytonas

i got 400K or something for the lightning q,

What did everyone draw for the feynman diagram,

i drew neutron to proton

then W-

Then Ve to e-

Otherwise it would have been a beta plus decay which didnt sound right for the question?

Lepton = 1 for e- and Ve, 0 for proton and neutron

and uud to udd for neutron to proton

Reply 82

drumbum99

did u not have to do the quarks so D to U
also anti electron neutrino?

Reply 83

TheTallOne

18

drumbum99

did u not have to do the quarks so D to U
also anti electron neutrino?

That's what I did.

[i.e, the corrections are correct]

Reply 84

ibs137

11

anyone actually did the very last question about the acceleration??

and the elastic collision one??? :confused:

maybe if I had more time.....

Reply 85

persia90

drumbum99

did u not have to do the quarks so D to U
also anti electron neutrino?

I think it was electron and anti neutrino

owyeh lol udd to uud

persia90

I think it was electron and anti neutrino

whoop, that's what i put :biggrin:

Reply 88

drumbum99

persia90

I think it was electron and anti neutrino

haha yer sorry typo =P
EDIT: no it wasnt
the neutrino is an anti electron neutrino
but yer there is an electron made too

Reply 89

username233985

I well and truly ****ed that up..... i made so many stupid mistakes like not reading the absolute temperature thing.... bye bye durham

Reply 90

daveyb91

no its neutron -> proton + electron + electron neutrino
but neutron -> proton in quarks like so d to u
isnt an anti as then the number of anti particles wouldnt balance as there are no anti particles on the right hand side,

the very last question was mind boggling? anyone have a clue?

Reply 91

Antineutrino

Here's my view on the lightning question...

Electric field strength...
E=V/d=(3.2*10^6)/(2*10^3)=1.6kV/m

Average current...
Q=It

I=Q/t=5/(0.1*10^-3)=50,000 (or 50kA)

Energy stored in cloud...
E=QV=5*(3.2*10^6)=16MJ

OR

P=IV=50,000*(3.2*10^6)=160GW
P=E/t
E=Pt=(160*10^9)*(0.1*10^-3)=16MJ

Then use Resisitivity to get 0R1 Ohms.

The next bit was the temperature rise...
Power converted in the conductor was given by P=(I^2)R
That was a tiny fraction of the toal power output of the lightning strike. Which is logical if you think about it. The strike came from 2km up and only 50m of the currents journey was in the conductor. The other 1,950m was in the insulator air. If you imagine a voltage divider where the top resistor(R1) is the resistance of the 1,950m of air and the bottom resistor(R2) is the 0R1 ohm resistance of the conductor cable then the majority of the voltage is dropped across the top resistor(the air). This means as electrical power is given by P=IV and the current is equal in both resistances as they're in series then the most energy is converted in the top resistor, R1 (the air).
P=(I^2)R=(50,000^2)*0.1=250MJ
E=pt=(250*10^6)*(0.1*10^-3)=25kJ

That then translated into a 3K or degree rise.

Reply 92

Chris12002

1

The melting point of lead is only around 330 degrees, i doubt a lightining rod would be made that would melt every time it was struck

Reply 93

persia90

daveyb91

no its neutron -> proton + electron + electron neutrino
but neutron -> proton in quarks like so d to u
isnt an anti as then the number of anti particles wouldnt balance as there are no anti particles on the right hand side,

the very last question was mind boggling? anyone have a clue?

I think the electron always goes with anti neutrino(W- interaction) and positron goes with neutrino(W+ interation) :confused:

Reply 94

drumbum99

daveyb91

no its neutron -> proton + electron + electron neutrino
but neutron -> proton in quarks like so d to u
isnt an anti as then the number of anti particles wouldnt balance as there are no anti particles on the right hand side,

the very last question was mind boggling? anyone have a clue?

im sure its an anti electron neutrino
so that it would cancel out the lepton +1 from the electron?

Reply 95

c12andn15

10

Antineutrino

Here's my view on the lightning question...

Electric field strength...
E=V/d=(3.2*10^6)/(2*10^3)=1.6kV/m

Average current...
Q=It

I=Q/t=5/(0.1*10^-3)=50,000 (or 50kA)

Energy stored in cloud...
E=QV=5*(3.2*10^6)=16MJ

OR

P=IV=50,000*(3.2*10^6)=160GW
P=E/t
E=Pt=(160*10^9)*(0.1*10^-3)=16MJ

Then use Resisitivity to get 0R1 Ohms.

The next bit was the temperature rise...
Power converted in the conductor was given by P=(I^2)R
That was a tiny fraction of the toal power output of the lightning strike. Which is logical if you think about it. The strike came from 2km up and only 50m of the currents journey was in the conductor. The other 1,950m was in the insulator air. If you imagine a voltage divider where the top resistor(R1) is the resistance of the 1,950m of air and the bottom resistor(R2) is the 0R1 ohm resistance of the conductor cable then the majority of the voltage is dropped across the top resistor(the air). This means as electrical power is given by P=IV and the current is equal in both resistances as they're in series then the most energy is converted in the top resistor, R1 (the air).
P=(I^2)R=(50,000^2)*0.1=250MJ
E=pt=(250*10^6)*(0.1*10^-3)=25kJ

That then translated into a 3K or degree rise.

I agree 100% exactly what I got my friend 8-)

Reply 96

c12andn15

10

just looked in mod 1 book

it's n >> p + e- + anti neutrino.

Reply 97

Bobbage

c12andn15

I agree 100% exactly what I got my friend 8-)

Ditto. I got the first crazy large answer, and realised that it'd be a rather crap lightning rod if it actually absorbed the entire energy of the strike instead of conducting it away. Scoured the formula book for something a little more normal.