2012 Higher Physics Discussion
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Original post by Blue7195
please do, really intrigued about revised higher to see what was in it 
I'm just trying to upload it now!
aha it's taking a while though
Original post by Maryam A
Are the section B answers up yet? - from an authentic source
Not sure anyone other than bob169 in post #610 but im nearly there - just need Q27biii onwards as scan corrupted - the graph thing whats the question asking?
Am I authentic though - who is!
Original post by tomctutor
Not sure anyone other than bob169 in post #610 but im nearly there - just need Q27biii onwards as scan corrupted - the graph thing whats the question asking?
Am I authentic though - who is!
Am I authentic though - who is!
ok that's great. anyone who's good at physics really!
Original post by n00b/
b iii) A) Calculate the maximum potential difference between X and Y during this time. (2)
B) Calculate the maximum length of the fabric during this time. (3)
B) Calculate the maximum length of the fabric during this time. (3)
ta
Original post by Jamietaylor1995
I said that the xenon engine gives a greater force because since they are accelerated at the same speed and F=ma, the force will be greater because the Xeon ions have a larger mass. I think that's basically the same as you? I think it's right
Xenon engine is the correct answer. Think it's more to do with the conservation of momentum than it is to do with ma though. (Although they all tied together somewhere anyway).
My names Emily
For 21) I got 19.21km cause using pic was 15.7 but Pythagoras gave me that, then 38.7 degrees from B toA doing tan 12/15.
B) 33
Ii) 0.366m/s
No one else has this, ahhhh
For 21) I got 19.21km cause using pic was 15.7 but Pythagoras gave me that, then 38.7 degrees from B toA doing tan 12/15.
B) 33
Ii) 0.366m/s
No one else has this, ahhhh
Hey here is my solutions if any use, not sure if they are all correct so give me a shout if not. Hope it went well for everybody today.
1.E
2.A
3.C
4.C
5.C
6.D
7.B
8.D
9.B
10.C
11.B
12.E
13.D
14.E
15.D
16.A
17.A
18.D
19.B
20.B
21.a)i)15.6km 64degrees south of east
a)ii) v=s/t
= 15.6/1.25
=12.5km/hr
b)i) 15.6km at 64degrees south of east
b)ii) t=d/s
=33/22
=1.5 hours
v=s/t
=15.6/1.5
=10.4km/hr
22a)i) D=st
=3.06x20
=61.2m
ii)s=ut+1/2at^2
=15x1.53+1/2x(-9.8)x1.53^2
=22.95-11.47
=11.5m
b) The ball is more likely to hit the tree as the maximum height will occur sooner so less of a chance for the ball to clear the tree.
23.a)i) W=Qv
=1.6x10^-19x1.22x10^3
=1.95^-16J
ii) Ek=W
1/2mv^2=QV
V^2=1.95^-16/1.09x10^-25
V^2=1.80x109
V=42296m/s
b) Ft=mv
v=ft/m
v=0.07x60/750
change in v=5.6x10^-3m/s
c)Xenon as f=ma so if acceleration is kept constant then the force exerted will be larger as mass is larger
24.a)k=p/t
=0.347 roughly for most so as there is a constant then temperature is directly proprtional to pressure
b) As the temperature is increased the particles have greater kinetic energy. This increase the speed of the particles and cause more frequent and more forceful collisions with the walls of the container. As the force exerted on the walls increase and the volume remains constant then the pressure increases.
c) So that all the gas inside the beaker has the same temperature.
25.a)i) I=v/r
=12/8
=1.5A
ii) E=IR+Ir
Ir=E-IR
Ir=12-1.5x6
Ir=3V
iii)P=IV
=1.5x9
=13.5W
b) As E=IR=Ir then if there is a greater power over the lamp then IR must have increased. As E is constant and the current is the same throughout the circuit then r must have decreased.
26.a) increases as a curve before levelling off at 12 v
b)R=V/I
= 12/2x10^-3
= 6000ohms
c)i)As V=IR and as voltage and resistance is the same so is the current.
ii) Smaller as Q=It and if the time taken is smaller to discharge then the charge is smaller. As C=Q/V as the charge is smaller and the voltage constant then the capacitance is smaller
27.a)R1/R2=R3/R4
R1/40=240/80
80R1=9600
R1=120ohms
b)i) Differential mode
ii)gain=Rf/R1
=560/100
=5.6
ii)A) Vo=(V2-V1)xgain
V2-V1=1.93V
B) V2=2.25
2.25-V1=1.93
-V1=-0.32
V1=0.32V
V1=(R1/R1+R2)xVs
4.5+0.32=(120/120+R2)x9
120/120+R2=0.535
64.267+0.535R2=120
0.535R2=55.73
R2=104.17Ohms
I took it to be 104Ohms for graph which came to be 66mm of fabric.
28.a)n=sinX/sin2
1.33=sinX/sin36
sinX=0.781...
X=51.4degrees
b)i)As there is a refracted ray at 90 degrees to the normal as well as a reflected ray
ii)sinc=1/n
sinc=1/1.33
c=48.8degrees
c) sketch of total internal reflection occuring.
29.a) nlambda=dsinA
3x633x10-9=dsin35.3
d=1.899x10-6/sin35.3
d=3.29x10-6M
b)lines per m=1/3.29x10^-6
=304295lines per M
c)102% of 3x10^5=30600
As the figure calculated in part b) is less than this figure then it satisfies the manufacturers claim.
30.a) decreases
b)i)photoconductive mode.
ii) The current increases as more photons acting on the plate and this creates more electron hole pairs available for conduction and thus the current increases.
c)Irradiance is directly proportianal to current
so I1d1^2=I2d2^2
3x10-6x1.2^2=I2x0.8^2
0.64I2=4.32x10-6
I2=6.75x10-6A
31.a) E=Dm
=500x10-6x0.04
=2x10-5J
b) H=Hdotxt
=2x5x10-3
=10x10-3sv
H=Dwr
wr=H/D
wr=10x10-3/500x10-6
=20
so radiation is alpha.
1.E
2.A
3.C
4.C
5.C
6.D
7.B
8.D
9.B
10.C
11.B
12.E
13.D
14.E
15.D
16.A
17.A
18.D
19.B
20.B
21.a)i)15.6km 64degrees south of east
a)ii) v=s/t
= 15.6/1.25
=12.5km/hr
b)i) 15.6km at 64degrees south of east
b)ii) t=d/s
=33/22
=1.5 hours
v=s/t
=15.6/1.5
=10.4km/hr
22a)i) D=st
=3.06x20
=61.2m
ii)s=ut+1/2at^2
=15x1.53+1/2x(-9.8)x1.53^2
=22.95-11.47
=11.5m
b) The ball is more likely to hit the tree as the maximum height will occur sooner so less of a chance for the ball to clear the tree.
23.a)i) W=Qv
=1.6x10^-19x1.22x10^3
=1.95^-16J
ii) Ek=W
1/2mv^2=QV
V^2=1.95^-16/1.09x10^-25
V^2=1.80x109
V=42296m/s
b) Ft=mv
v=ft/m
v=0.07x60/750
change in v=5.6x10^-3m/s
c)Xenon as f=ma so if acceleration is kept constant then the force exerted will be larger as mass is larger
24.a)k=p/t
=0.347 roughly for most so as there is a constant then temperature is directly proprtional to pressure
b) As the temperature is increased the particles have greater kinetic energy. This increase the speed of the particles and cause more frequent and more forceful collisions with the walls of the container. As the force exerted on the walls increase and the volume remains constant then the pressure increases.
c) So that all the gas inside the beaker has the same temperature.
25.a)i) I=v/r
=12/8
=1.5A
ii) E=IR+Ir
Ir=E-IR
Ir=12-1.5x6
Ir=3V
iii)P=IV
=1.5x9
=13.5W
b) As E=IR=Ir then if there is a greater power over the lamp then IR must have increased. As E is constant and the current is the same throughout the circuit then r must have decreased.
26.a) increases as a curve before levelling off at 12 v
b)R=V/I
= 12/2x10^-3
= 6000ohms
c)i)As V=IR and as voltage and resistance is the same so is the current.
ii) Smaller as Q=It and if the time taken is smaller to discharge then the charge is smaller. As C=Q/V as the charge is smaller and the voltage constant then the capacitance is smaller
27.a)R1/R2=R3/R4
R1/40=240/80
80R1=9600
R1=120ohms
b)i) Differential mode
ii)gain=Rf/R1
=560/100
=5.6
ii)A) Vo=(V2-V1)xgain
V2-V1=1.93V
B) V2=2.25
2.25-V1=1.93
-V1=-0.32
V1=0.32V
V1=(R1/R1+R2)xVs
4.5+0.32=(120/120+R2)x9
120/120+R2=0.535
64.267+0.535R2=120
0.535R2=55.73
R2=104.17Ohms
I took it to be 104Ohms for graph which came to be 66mm of fabric.
28.a)n=sinX/sin2
1.33=sinX/sin36
sinX=0.781...
X=51.4degrees
b)i)As there is a refracted ray at 90 degrees to the normal as well as a reflected ray
ii)sinc=1/n
sinc=1/1.33
c=48.8degrees
c) sketch of total internal reflection occuring.
29.a) nlambda=dsinA
3x633x10-9=dsin35.3
d=1.899x10-6/sin35.3
d=3.29x10-6M
b)lines per m=1/3.29x10^-6
=304295lines per M
c)102% of 3x10^5=30600
As the figure calculated in part b) is less than this figure then it satisfies the manufacturers claim.
30.a) decreases
b)i)photoconductive mode.
ii) The current increases as more photons acting on the plate and this creates more electron hole pairs available for conduction and thus the current increases.
c)Irradiance is directly proportianal to current
so I1d1^2=I2d2^2
3x10-6x1.2^2=I2x0.8^2
0.64I2=4.32x10-6
I2=6.75x10-6A
31.a) E=Dm
=500x10-6x0.04
=2x10-5J
b) H=Hdotxt
=2x5x10-3
=10x10-3sv
H=Dwr
wr=H/D
wr=10x10-3/500x10-6
=20
so radiation is alpha.
(edited 11 years ago)
Original post by Webbster91
Hey here is my solutions if any use, not sure if they are all correct so give me a shout if not. Hope it went well for everybody today.
1.E
2.A
3.C
4.C
5.C
6.D
7.B
8.D
9.B
10.C
11.B
12.E
13.D
14.E
15.D
16.A
17.A
18.D
19.B
20.B
21.a)i)15.6km 64degrees south of east
a)ii) v=s/t
= 15.6/1.25
=12.5km/hr
b)i) 15.6km at 64degrees south of east
b)ii) t=d/s
=33/22
=1.5 hours
v=s/t
=15.6/1.5
=10.4km/hr
22a)i) D=st
=3.06x20
=61.2m
ii)s=ut+1/2at^2
=15x1.53+1/2x(-9.8)x1.53^2
=22.95-11.47
=11.5m
b) The ball is more likely to hit the tree as the maximum height will occur sooner so less of a chance for the ball to clear the tree.
23.a)i) W=Qv
=1.6x10^-19x1.22x10^3
=1.95^-16J
ii) Ek=W
1/2mv^2=QV
V^2=1.95^-16/1.09x10^-25
V^2=1.80x109
V=42296m/s
b) Ft=mv
v=ft/m
v=0.07x60/750
change in v=5.6x10^-3m/s
c)Xenon as f=ma so if acceleration is kept constant then the force exerted will be larger as mass is larger
24.a)k=p/t
=1.347 roughly for most so as there is a constant then temperature is directly proprtional to pressure
b) As the temperature is increased the particles have greater kinetic energy. This increase the speed of the particles and cause more frequent and more forceful collisions with the walls of the container. As the force exerted on the walls increase and the volume remains constant then the pressure increases.
c) So that all the gas inside the beaker has the same temperature.
25.a)i) I=v/r
=12/8
=1.5A
ii) E=IR+Ir
Ir=E-IR
Ir=12-1.5x6
Ir=3V
iii)P=IV
=1.5x9
=13.5W
b) As E=IR=Ir then if there is a greater power over the lamp then IR must have increased. As E is constant and the current is the same throughout the circuit then r must have decreased.
26.a) increases as a curve before levelling off at 12 v
b)R=V/I
= 12/2x10^-3
= 6000ohms
c)i)As V=IR and as voltage and resistance is the same so is the current.
ii) Smaller as Q=It and if the time taken is smaller to discharge then the charge is smaller. As C=Q/V as the charge is smaller and the voltage constant then the capacitance is smaller
27.a)R1/R2=R3/R4
R1/40=240/80
80R1=9600
R1=120ohms
b)i) Differential mode
ii)gain=Rf/R1
=560/100
=5.6
ii)A) Vo=(V2-V1)xgain
V2-V1=1.93V
B) V2=2.25
2.25-V1=1.93
-V1=-0.32
V1=0.32V
V1=(R1/R1+R2)xVs
4.5+0.32=(120/120+R2)x9
120/120+R2=0.535
64.267+0.535R2=120
0.535R2=55.73
R2=104.17Ohms
I took it to be 104Ohms for graph which came to be 66mm of fabric.
28.a)n=sinX/sin2
1.33=sinX/sin36
sinX=0.781...
X=51.4degrees
b)i)As there is a refracted ray as well as a reflected ray
ii)sinc=1/n
sinc=1/1.33
c=48.8degrees
c) sketch of total internal reflection occuring.
29.a) nlambda=dsinA
3x633x10-9=dsin35.3
d=1.899x10-6/sin35.3
d=3.29x10-6M
b)lines per m=1/3.29x10^-6
=304295lines per M
c)102% of 3x10^5=30600
As the figure calculated in part b) is less than this figure then it satisfies the manufacturers claim.
30.a) decreases
b)i)photoconductive mode.
ii) The current increases as more photons acting on the plate and this creates more electron hole pairs available for conduction and thus the current increases.
c)Irradiance is directly proportianal to current
so I1d1^2=I2d2^2
3x10-6x1.2^2=I2x0.8^2
0.64I2=4.32x10-6
I2=6.75x10-6A
31.a) E=Dm
=500x10-6x0.04
=2x10-4J
b) H=Hdotxt
=2x5x10-3
=10x10-3sv
H=Dwr
wr=H/D
wr=10x10-3/500x10-6
=20
so radiation is alpha.
1.E
2.A
3.C
4.C
5.C
6.D
7.B
8.D
9.B
10.C
11.B
12.E
13.D
14.E
15.D
16.A
17.A
18.D
19.B
20.B
21.a)i)15.6km 64degrees south of east
a)ii) v=s/t
= 15.6/1.25
=12.5km/hr
b)i) 15.6km at 64degrees south of east
b)ii) t=d/s
=33/22
=1.5 hours
v=s/t
=15.6/1.5
=10.4km/hr
22a)i) D=st
=3.06x20
=61.2m
ii)s=ut+1/2at^2
=15x1.53+1/2x(-9.8)x1.53^2
=22.95-11.47
=11.5m
b) The ball is more likely to hit the tree as the maximum height will occur sooner so less of a chance for the ball to clear the tree.
23.a)i) W=Qv
=1.6x10^-19x1.22x10^3
=1.95^-16J
ii) Ek=W
1/2mv^2=QV
V^2=1.95^-16/1.09x10^-25
V^2=1.80x109
V=42296m/s
b) Ft=mv
v=ft/m
v=0.07x60/750
change in v=5.6x10^-3m/s
c)Xenon as f=ma so if acceleration is kept constant then the force exerted will be larger as mass is larger
24.a)k=p/t
=1.347 roughly for most so as there is a constant then temperature is directly proprtional to pressure
b) As the temperature is increased the particles have greater kinetic energy. This increase the speed of the particles and cause more frequent and more forceful collisions with the walls of the container. As the force exerted on the walls increase and the volume remains constant then the pressure increases.
c) So that all the gas inside the beaker has the same temperature.
25.a)i) I=v/r
=12/8
=1.5A
ii) E=IR+Ir
Ir=E-IR
Ir=12-1.5x6
Ir=3V
iii)P=IV
=1.5x9
=13.5W
b) As E=IR=Ir then if there is a greater power over the lamp then IR must have increased. As E is constant and the current is the same throughout the circuit then r must have decreased.
26.a) increases as a curve before levelling off at 12 v
b)R=V/I
= 12/2x10^-3
= 6000ohms
c)i)As V=IR and as voltage and resistance is the same so is the current.
ii) Smaller as Q=It and if the time taken is smaller to discharge then the charge is smaller. As C=Q/V as the charge is smaller and the voltage constant then the capacitance is smaller
27.a)R1/R2=R3/R4
R1/40=240/80
80R1=9600
R1=120ohms
b)i) Differential mode
ii)gain=Rf/R1
=560/100
=5.6
ii)A) Vo=(V2-V1)xgain
V2-V1=1.93V
B) V2=2.25
2.25-V1=1.93
-V1=-0.32
V1=0.32V
V1=(R1/R1+R2)xVs
4.5+0.32=(120/120+R2)x9
120/120+R2=0.535
64.267+0.535R2=120
0.535R2=55.73
R2=104.17Ohms
I took it to be 104Ohms for graph which came to be 66mm of fabric.
28.a)n=sinX/sin2
1.33=sinX/sin36
sinX=0.781...
X=51.4degrees
b)i)As there is a refracted ray as well as a reflected ray
ii)sinc=1/n
sinc=1/1.33
c=48.8degrees
c) sketch of total internal reflection occuring.
29.a) nlambda=dsinA
3x633x10-9=dsin35.3
d=1.899x10-6/sin35.3
d=3.29x10-6M
b)lines per m=1/3.29x10^-6
=304295lines per M
c)102% of 3x10^5=30600
As the figure calculated in part b) is less than this figure then it satisfies the manufacturers claim.
30.a) decreases
b)i)photoconductive mode.
ii) The current increases as more photons acting on the plate and this creates more electron hole pairs available for conduction and thus the current increases.
c)Irradiance is directly proportianal to current
so I1d1^2=I2d2^2
3x10-6x1.2^2=I2x0.8^2
0.64I2=4.32x10-6
I2=6.75x10-6A
31.a) E=Dm
=500x10-6x0.04
=2x10-4J
b) H=Hdotxt
=2x5x10-3
=10x10-3sv
H=Dwr
wr=H/D
wr=10x10-3/500x10-6
=20
so radiation is alpha.
Hi. Thanks for posting your solutions! great help
i agree with everything except:
isn't P/T = 0.347
it is the critical angle because the refracted ray is 90 degreees to the normal, having both reflected and refracted rays is not unique to the the critical angle, i think
Original post by Maryam A
Hi. Thanks for posting your solutions! great help
i agree with everything except:
isn't P/T = 0.347
it is the critical angle because the refracted ray is 90 degreees to the normal, having both reflected and refracted rays is not unique to the the critical angle, i think
i agree with everything except:
isn't P/T = 0.347
it is the critical angle because the refracted ray is 90 degreees to the normal, having both reflected and refracted rays is not unique to the the critical angle, i think
yes you are correct i will change them now. I think i said it was 90 degrees in exam, or at least im hoping so...
Thank you for posting the solutions, I need a B to get into uni and I have just marked my paper and got 66% will this be enough for a B?
Original post by AmandaD163
Thank you for posting the solutions, I need a B to get into uni and I have just marked my paper and got 66% will this be enough for a B?
Yes, easily.
for the first few questions what if you changed into metres and ms^-1? Do people agree with the answers: 21. ii) 3.5 m/s (also for each of my velocities I said at 154.5 degrees, will I lose any marks for this?)
21. ii) 3.5 ms^-1 at 154.5
b) 15.7 km at 154.5
ii)2.9 ms^-1
21. ii) 3.5 ms^-1 at 154.5
b) 15.7 km at 154.5
ii)2.9 ms^-1
Original post by Tycho
Yes, easily.
Great thanks!
Original post by Tycho
Comeon, tomctutor post the answers already! I don't think my nerves can take any more suspense! 
lol
lolis someone posting the answers?
For the critical angle question, if you say because the ray refracts along the surface of the water is not ok?
Quick Reply
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