It might count against you, but I doubt it: it depends on the examiner and mark scheme.(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.
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OCR PHYSICS B G491~ 19th May 2015 AM ~ AS Physics watch
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- 19-05-2015 17:48
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milliethemoo
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- 19-05-2015 17:54
(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. -
jpetersgill
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- 19-05-2015 17:57
(Original post by Alex .G.)
Let's start the mark scheme?
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- 19-05-2015 18:02
Q1 Units
Coulomb - As
Conductance - AV^-1
Q2
3x10^8/1.7
Q3
3.2 or 4.8 ?
Q4
Q5
Q6
Q7
Q8
Q9 -
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- 19-05-2015 18:03
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 -
jpetersgill
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- 19-05-2015 18:03
(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
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- 19-05-2015 18:04
(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
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- 19-05-2015 18:05
(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 -
jpetersgill
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- 19-05-2015 18:07
(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
Posted from TSR Mobile -
jpetersgill
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- 19-05-2015 18:08
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- 19-05-2015 18:08
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 -
Zain.chishty
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- 19-05-2015 18:13
Q4. 40/280 * 4.5V (not sure about the emf of the cell)
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StrangeBanana
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- 19-05-2015 18:16
(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
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 VLast edited by StrangeBanana; 19-05-2015 at 18:17. -
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- 19-05-2015 18:16
(Original post by Zain.chishty)
Q4. 40/280 * 4.5V (not sure about the emf of the cell) -
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- 19-05-2015 18:18
(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 -
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- 19-05-2015 18:21
(Original post by HennersPD)
But since u was negative so 1/f-0.8=4 so you add 0.8 = 4.8 surely? -
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- 19-05-2015 18:24
(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... -
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- 19-05-2015 18:30
(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 -
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- 19-05-2015 18:31
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:
�Last edited by Alex .G.; 19-05-2015 at 18:35.
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