OCR Physics A G485 - Frontiers of Physics - June 2016

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Your explanation is very clear sir, thank you.
just one thing, for the graph, can i start from Positive 80mV?

I know what you mean, but it starts at 80 mV cause of Lenz's Law, it really is
E = (-) gradient. You can see the gradient at 0 is max and is positive but since E is in the opposite direction it has to be negative.

Reply 81

XY123

Original post by Reda2

I have another one of these lol. They help understand them as I am explaining it. And might as well post it here.

X - Ray

How it is made:
- Electrons are produced in a hot filament and targeted and accelerated towards the anode, a rotating metal with high melting point such as tungsten.
- Most of the KE of the electron is transferred as thermal energy which heats up the target, this can be dealt with by circulating the metal with water inside it and by rotating it.
- Only 1% of the KE of the electrons are transferred to x-ray photons.

X-ray interactions:
When the x-ray passes through matter it's intensity decreases because they are either scattered or stopped by the atoms of the material.

The 3 ways that x-ray interacts with matter, 1. Photoelectric effect, 2. Pair production, 3. Compton effect.

Photoelectric effect: happens when energy of x-ray photo < 0.1 MeV, photon disappears and gives all of its energy to the electron which ejects from the metal.
Pair production: happens when energy of x-ray is between 0.5 and 5.0 MeV. X-ray photon loses some of it's energy to eject electron from the metal, but it loses energy and it's wavelength increases.
Compton effect: happens when energy of x-ray photon is > 1.02 MeV. Photon disappears and produces a electron-positron.

I was going to go about the detection systems, but man I need to get back to revising the rest lol.

You've got Compton effect and pair production the wrong way round I think

Reply 82

Xuancong

Anyone got some question want to help ? please ask me.

Reply 83

L'Evil Wolf

Original post by Xuancong

Anyone got some question want to help ? please ask me.

In a circuit, with a battery or cell w/e is the longer line the positive side, and the negative side the shorter line?

Reply 84

box036

Does anyone know the condition for an overall A* in physics is?

Many thanks!

Reply 85

Tara123459

Does anyone have the june 2015 fields, frontiers and particles paper?

Reply 86

HarshaPeed

Original post by Tara123459

Does anyone have the june 2015 fields, frontiers and particles paper?

Yes plz

Reply 87

HarshaPeed

Original post by box036

Does anyone know the condition for an overall A* in physics is?

Many thanks!

90% of A2 and 80% of all marks in both AS and A2.

Reply 88

box036

Original post by HarshaPeed

90% of A2 and 80% of all marks in both AS and A2.

Is it 90% on UMS or 90% on marks?

Reply 89

PotAuFeu

Original post by box036

Is it 90% on UMS or 90% on marks?

UMS

Reply 90

PotAuFeu

Original post by Tara123459

Does anyone have the june 2015 fields, frontiers and particles paper?

Original post by HarshaPeed

Yes plz

See attached

Original post by L'Evil Wolf

In a circuit, with a battery or cell w/e is the longer line the positive side, and the negative side the shorter line?

That is correct.

G485_MS_June15.pdf309.8KB

G485-01 QP Jun15.pdf682.5KB

Reply 91

QuantumSylar

there's like 50 people do this exam board for physics at A2 loool

Reply 92

Algeriandz

Original post by QuantumSylar

there's like 50 people do this exam board for physics at A2 loool

50????

Reply 93

QuantumSylar

Original post by Algeriandz

50????

There's obviously more but there isn't a trending topic for this subject and exam board and it's day before

Reply 94

Xuancong

Original post by L'Evil Wolf

In a circuit, with a battery or cell w/e is the longer line the positive side, and the negative side the shorter line?

Are you asking about the symbol ? then yes

Reply 95

Xuancong

This question 4,b,iii, from paper June 2015 G485.
Anyone know what is the answer ? and Why ? ( I checked the mark scheme but still don't understand)

Reply 96

Xuancong

Original post by Xuancong

This question 4,b,iii, from paper June 2015 G485.
Anyone know what is the answer ? and Why ? ( I checked the mark scheme but still don't understand)

Reply 97

lai812matthew

Original post by Xuancong

This question 4,b,iii, from paper June 2015 G485.
Anyone know what is the answer ? and Why ? ( I checked the mark scheme but still don't understand)

not enough time to discharge.

Reply 98

Xuancong

Original post by lai812matthew

not enough time to discharge.

The markscheme said period is "comparable" to time constant
What is the work "comparable" really mean ?

Reply 99

lai812matthew

Original post by Xuancong

The markscheme said period is "comparable" to time constant
What is the work "comparable" really mean ?

this is what i have written

frequency is doubled, period is halved. I=Ioe-t/CR, (1/120)/(1.66x10^-3)=5 time constants is required to fully discharge for original frequency. when period is halved, 5/2=2.5 time constants is required to fully discharge. for exponential decrease, there are still a small part of the original current which is still present for 2.5 time constant, (1/e^2.5 =8.21%), unlike 5 time constant, which the remaining amount can be considered as negligible. (1/e⁵ =0.673%). therefore the time is not enough for the system to discharge, the current would not be doubled.