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# OCR PHYSICS B G491~ 19th May 2015 AM ~ AS Physics watch

1. (Original post by ThatPerson2)
0.018m is 1.8cm - think of the size of a bulb and the filament inside of it seems likely. I got 0.018m and rounded it to 0.02 (I generally do 2dp or 3sf) do you think this will count against me? I had 0.018m in the working.
It might count against you, but I doubt it: it depends on the examiner and mark scheme.
2. (Original post by ThatPerson2)
0.018m is 1.8cm - think of the size of a bulb and the filament inside of it seems likely. I got 0.018m and rounded it to 0.02 (I generally do 2dp or 3sf) do you think this will count against me? I had 0.018m in the working.
They usually like the answers to be to at least 2 significant figures
3. (Original post by Alex .G.)
Let's start the mark scheme?
Yh go on !

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4. Q1 Units

Coulomb - As
Conductance - AV^-1

Q2

3x10^8/1.7

Q3

3.2 or 4.8 ?

Q4

Q5

Q6

Q7

Q8

Q9
5. (Original post by jpetersgill)
Yh go on !

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Ok for
Qu1: Power - AV
Conductance: AV-1
Charge: As

Question 2 anyone? Also D, C, B for graphs as C had largest area using 1/2b*h for tensile strength
6. (Original post by Alex .G.)
Q1 Units

Coulomb - As
Conductance - AV^-1

Q2

3x10^8/1.7

Q3

3.2 or 4.8 ?

Q4

Q5

Q6

Q7

Q8

Q9
The other unit was power - IV

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7. (Original post by HennersPD)
Ok for
Qu1: Power - AV
Conductance: AV-1
Charge: As

Question 2 anyone? Also D, C, B for graphs as C had largest area using 1/2b*h for tensile strength
Surely tensile strength is just A because it had the highest stress?

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8. (Original post by Alex .G.)
Q1 Units

Coulomb - As
Conductance - AV^-1

Q2

3x10^8/1.7

Q3

3.2 or 4.8 ?

Q4

Q5

Q6

Q7

Q8

Q9
Q3 is 4.8 as it was 4-(-0.8) = 4+0.8 = 4.8D
9. (Original post by Alex .G.)
Q1 Units

Coulomb - As
Conductance - AV^-1

Q2

3x10^8/1.7

Q3

3.2 or 4.8 ?

Q4

Q5

Q6

Q7

Q8

Q9
Q3 was like input * 40/280 I think?

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10. (Original post by HennersPD)
Q3 is 4.8 as it was 4-(-0.8) = 4+0.8 = 4.8D
Q3 was a potential divider I'm pretty sure.. But the lens one was 3.2D

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11. Gerrow's Unofficial mark scheme.

Add to it yourselves - I Can't remember the paper!

Q1 Units

Power AV
Coulomb - As
Conductance - AV^-1

Q2

3x10^8/1.7

Q3

3.2 or 4.8 ?

Q4

Q5

Q6

Q7

Q8

Q9
12. (Original post by HennersPD)
Q3 is 4.8 as it was 4-(-0.8) = 4+0.8 = 4.8D
Q3 = 3.2 D as 4.0 D is total curvature needed to be added to the incoming wavefronts and the lens adds 3.2 D and the curvature added by the eye is 0.8 D
13. Q4. 40/280 * 4.5V (not sure about the emf of the cell)
14. (Original post by Alex .G.)
Gerrow's Unofficial mark scheme.

Add to it yourselves - I Can't remember the paper!

Q1 Units

Power AV
Coulomb - As
Conductance - AV^-1

Q2

3x10^8/1.7

Q3

3.2 or 4.8 ?

Q4

Q5

Q6

Q7

Q8

Q9
Let me just say what I remember

The lens one:
i) 1/1.25 = 0.8D
ii) power = 3.2D

The bits one:
i) show that Q, 2^4 = 16
ii) 500 x 300 x 4 / 8 = 75,000
iii) 75,000 x 90 x 5 = 33,750,000

The Scanning Tunnelling Microscope one:
i) sensitivity = 580 pA nm^-1
ii) diagram; one bump then two bumps next to each other
iii) explanation: it'd scan the whole surface and produce a graph of all of it, wherever a "bump" on the graph was, an atom was

The Sounds one:
i) It's logarithmic because as you go to the right it increases with a constant factor of 10
ii) frequency of highest amplitude in orchestra = 350 Hz (I think)
iii) bandwidth = 9900 Hz

The Potential Divider one:
4.5 x 40/280 = 0.64 V
15. (Original post by Zain.chishty)
Q4. 40/280 * 4.5V (not sure about the emf of the cell)
Wouldnt it ve 40\280+40#4.5
16. (Original post by Zain.chishty)
Q3 = 3.2 D as 4.0 D is total curvature needed to be added to the incoming wavefronts and the lens adds 3.2 D and the curvature added by the eye is 0.8 D
But since u was negative so 1/f-0.8=4 so you add 0.8 = 4.8 surely?
17. (Original post by HennersPD)
But since u was negative so 1/f-0.8=4 so you add 0.8 = 4.8 surely?
Nah, -0.8 and -4 were both negative because they were on the same side of the lens in her eye
18. (Original post by LJflowrie)
It probably was to be honest, hopefully i might get one method mark? I was rushing the first page because our invigilators hadn't given us formulae booklets before they started the exam so i didn't have one for the first 5 minutes which was really annoying...
So did you get 4.8D for the power as u was negative?
19. (Original post by StrangeBanana)
Let me just say what I remember

The lens one:
i) 1/1.25 = 0.8D
ii) power = 3.2D

The bits one:
i) show that Q, 2^4 = 16
ii) 500 x 300 x 4 / 8 = 75,000
iii) 75,000 x 90 x 5 = 33,750,000

The Scanning Tunnelling Microscope one:
i) sensitivity = 580 pA nm^-1
ii) diagram; one bump then two bumps next to each other
iii) explanation: it'd scan the whole surface and produce a graph of all of it, wherever a "bump" on the graph was, an atom was

The Sounds one:
i) It's logarithmic because as you go to the right it increases with a constant factor of 10
ii) frequency of highest amplitude in orchestra = 350 Hz (I think)
iii) bandwidth = 9900 Hz

The Potential Divider one:
4.5 x 40/280 = 0.64 V
For the bandwidth it was just largest - smallest?
20. Gerrow's Unofficial mark scheme.

Add to it yourselves - I Can't remember the paper!

Q1 Units

Power AV
Coulomb - As
Conductance - AV^-1

Q2

3x10^8/1.7

Q3

1/1.25 = - 0.8D
ii) power = -0.8 + (ans) = 4. So =4.8D

Q4It's logarithmic because as you go to the right it increases with aconstant factor of 10
ii) frequency of highest amplitude in orchestra =350 Hz (I think)
iii) bandwidth = 9900 Hz

(probably accepts 9500-10500)

Q5 Voltage Resistance Power
small voltage 3.14
6 - 20.05

The Potential Divider one:
4.5 x 40/280 = 0.64 V

Q6 (ish)The bits one:
i) show that Q, 2^4 = 16
ii) 500 x 300 x 4 / 8 = 75,000
iii) 75,000 x 90 x 5 = 33,750,000

Q7

Q8

Q9
The Scanning Tunnelling Microscope one:
i) sensitivity = 260 pA nm^-1
ii) diagram; one bump then two bumps next to eachother
iii) explanation: it'd scan the whole surface andproduce a graph of all of it, wherever a "bump" on the graph was, anatom was

The Sounds one:

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