Post by: Mao on June 11, 2008, 05:53:47 pm
they are just suggestions :)
VCE Physics Unit 3 2008
Motion in one and two dimensions
Question 1 (2 marks)
What is the acceleration of the two ships when the tugboat and the ship are travelling at 2.0 ms-1
from graph: resistance
Question 2 (2 marks)
After a time, the tugboat and ship are travelling at constant speed
constant speed => Fnet=0
hence resistance
simple circular motion...~[m=2.4 kg, v=2.0 ms-1, r=1.6m]
Question 3 (2 marks)
What is the tension in the string?
Question 4 (2 marks)
direction of the resultant force is P [towards the centre]
Simple projectile motion: v=30,
Question 5 (2 marks)
What is the maximum height the ball reaches?
16.22m
Question 6 (2 marks)
resultant force is R [downwards]
now an advertising board is placed 72m from the batsman:
Question 7 (3 marks)
what height above the groud will the ball strike the advertising board?
solving for horizontal velocity gives a time
subbing this time into verticle motion
simple collision, train=20E3 kg, v=8ms-1, trucks=20E3 kg
Question 8 (2 marks)
What is the speed of the coupled locomotive + trucks?
simple proportionality gives 2ms-1
but you'd have to use conservation of momentum to get marks
Question 9 (3 marks)
Impulse TO the locomotive BY the trucks
obviously in a westward direction, 1.2E5 kgms-1
Question 10 (3 marks)
INELASTIC :D
Question 11 (2 marks)
During the collision, compare sizes of the magnitudes of average forces FL and FT
points that I think will give marks:
- Newton's 3rd Law, action/reaction pair
- :. FL = FT
Simple spring system, k=10, m=0.2kg, stretched to 60cm=0.6m, original 40cm=0.4m
Question 12 (2 marks)
Stationary Elastic Potential Energy:
EPE = 0.2J
Question 13 (2 marks)
What is the graph of KINETIC energy as a function of height?
D - upside down parabola
this is a good question that'll trip a lot of people up. there's two ways of thinking about it:
1. the ball changes direction at top and bottom [oscillating], therefore kinetic energy at top and bottom must be 0 [instantaneously stationary], hence only D satisfy this [at both top and bottom it is 0]
2. total energy is constant. GPE increases as a function of height [basically answers the next question].
- if we set GPE to be 0 at the bottom, KE=0 initially [it changes direction => v=0], => EPE=TE initially.
- EPE decreases quadratically [though not from turning point], GPE increases linearly, KE is TE [constant] - GPE [linear] - EPE [quadratic], which will give a quadratic expression
so what i'm trying to say is, D is right =]
Question 14 (2 marks)
What is GPE vs height?
compared to last question this one was a no-brainer. A. [linear increasing]
Eccliptical orbit of a comet around the sun
Question 15 (2 marks)
Describe how speed and total energy change from X [closest point] to Y [furthest point]
Total energy remains constant
Speed decreases as GPE increases hence KE decreases
This page is blank
Simple Gravity
Satellite = 930 kg
Mars = 6.42E23 kg
Radius of Orbit [thank god not above the surface] = 3.83E6 m
Question 16 (2 marks)
Calculate the gravitational FORCE.
not field strength :P
2.71E3 N
Question 17 (3 marks)
Calculate period of orbit
Keplar's law or consequential answer from 16
7.20E3 s
Electronics and Photonics
Simple LED [2.5v] circuit [supply = 8v] with limiting resistor [300 ohms]
Question 1 (2 marks)
current = 18.3 mA
TRANSISTOR AMPLIFIER!!!
Rc = 1k ohm
R1 = 2k ohm
R2 = 1k ohm
point P at 4 way connection
point Q at collector connection
Question 2 (2 marks)
DC voltage at point P
voltage divider gives 2V
current through Rc = 3 mA
Question 3 (2 marks)
voltage between Q and earth
[basically the DC voltage]
3V - crucial step: Vout = Vcc - V-Rc
Question 4 (2 marks)
power dissipated through Rc
P=I2R = 9E-3 W
Vout/Vin transfer characteristics:
[cut off at -60mV,3V] [saturation at 60mV, -3V] [quienscent point 0,0]
Question 5 (2 marks)
voltage gain = 50
Question 6 (3 marks)
Explain the shape of the graph, including why the graph has a negative slope, and why it has horizontal sections at [blah]
points that would be worth marks [i hope]
- as ib increase, ic increase and Vc increase => Vout decrease => inverting [negative gradient] (1 mark)
- cut off/saturation occur because ib is too low/high, resulting in the transistor acting as an open/closed switch with maximum/minimum potential across it. Since Vout is parallel to the transistor, the potential across it will also be maximum/minimum and cannot rise/decrease any further, hence the signal is clipped as reflected by the horizontal sections.
Question 7 (3 marks)
Vin = sinosudial wave, period = 1ms, max = 100mv, min = -100mV
marks allocated:
- correct frequency [unchanged]
- maximum at
- clipping occurs both top and bottom
- inverted
now we are given two transistors in one circuit!
Question 8 (3 marks)
explain in terms of biasing why a capacitor is required between X [collector of first transistor] and Y [4 way connection of second transistor]
marks allocated:
- to avoid distortion, the base [not input] must be biased to the quienscent point [middle of linear operating range], hence point Y must be biased correctly
- voltage at point X has a DC and AC component. the DC component is counterfunctional to the biasing of the second transistor
- a capacitor is installed to decouple the signal, removing the DC component to ensure correct biasing hence optimum amplification
simple thermistor
Question 9 (1 mark)
R at 20 degrees = 1000 ohms
simple thermistor voltage divider in a cooling circuit [they actually got it right this time!] with variable Resistor [vout connected across this], and 12V supply
Question 10 (3 marks)
set R to maintain 5 degrees [Vout is required to be 4]
Rt=4000 ohms
by simple ratio -> R = 2000 ohms
Question 11 (3 marks)
The cool room is not cold enough, should R be increased or decreased?
marks allocated:
- increase resistance
- as T decrease, Rt increase and VRt increase, VR decrease
- to increase VR to turn on the cooling element, R need to be increased
Detailed Studies - Structures and Materials
sorry to those who are not doing S&M, but i have chem exam tomorrow and I realy cbf. :P
these will get done tomorrow after I type up the chem exam.
btw everything is (2 marks) from now, so i wont bother
a two-way stress/strain graph that'll have freaked people out, compression is in quadrant 3.
Question 1
Young's modulus, D = 5.3E10 Nm-2
Question 2
B -> brittle
Question 3
max force given area = 1.50m^2 -> A, 1.2E8 N
Question 4
strain energy per unit volume = B, 1.2E5 J/m^3
Question 5
for the strain energy of this particular block, volume = 30 [20 m * 1.5 m^2], hence C, 30
Question 6
at strain = 500E-4,
Question 7
C, stone is stiffer in tension than compression
(http://obsolete-chaos.wikispaces.com/space/showimage/structures_bridge_pic.gif)
Question 8
"Which of the following best describes the stress in beam YZ?"
A tension
B compression
C neutral
D depends on load
XY is under obvious tension, A
Question 9
"Which of the following best describes the stress in beam XY?"
A tension
B compression
C neutral
D depends on load
the stress experienced by YZ is:
Quote
B, compression
this one took a while
both ZX and XY are under tension. at Z, a force is pulling it towards X, and Y, a force is pulling it towards X also. the horizontal components of these two forces can be resolved and shown that they pull towards the same centre of YZ, hence it is under compression.
it makes sense to think this way, as concrete is weak under tension, and structures have been designed to make the beam under compression to increase its lifetime and safety.
^^^ that's wrong. the actual answer should be [i think it should be, according to itute and a few simulations i've found on the internet] A, tension. as for explanation, i dare not try :P
Question 10
material P is stronger and stiffer than Q, but not as tough. option A.
Code: [Select]
_X______________________Y___________Z
| | | |
| | | |
|<-------8m------------->|<---4m---->|
Beam is 4000 kgQuestion 11
with no one standing at the lookout, what is the force of the beam on the support at X?
using rotational equilibrium, [and remembering 4000kg = 40000N], we arrive at C, 10000N
Question 12
what is the maximum load at Z without the beam tilting?
at that point, the force on X will be 0, and the torque at Z balances the weight of the beam. hence B, 2000kg
Question 13
steel reinforcement of the above beam:
D: ==========================
this is because between XY the beam sags [tension on the bottom], and bends down at Z [tension on top]
Post by: enwiabe on June 11, 2008, 05:55:06 pm
Post by: equinox on June 11, 2008, 06:02:12 pm
Plz :P
Post by: Mao on June 11, 2008, 06:02:52 pm
Mao, you are a champion! Now scan the exam paper and upload it :Pi would if i have a scanner :P
Post by: equinox on June 11, 2008, 06:05:21 pm
"Question 8
XY is under obvious tension, A
Question 9
the stress experienced by YZ is:
B, compression
this one took a while
both ZX and XY are under tension. at Z, a force is pulling it towards X, and Y, a force is pulling it towards X also. the horizontal components of these two forces can be resolved and shown that they pull towards the same centre of YZ, hence it is under compression.
it makes sense to think this way, as concrete is weak under tension, and structures have been designed to make the beam under compression to increase its lifetime and safety."
ALot of people reckon it was tension for both...and itute seems to agree with them...
Post by: Mao on June 11, 2008, 06:10:00 pm
but it makes sense that the answer is that way. if it was tension, [which is what i originally chose], then what is stretching it? this way we can model it with action/reaction, but that just really doesnt seem to work....
Post by: ganges on June 11, 2008, 06:12:48 pm
By they way the structures questions where u got 19.99 is wrong its 20.01 cuz its under tension.
Post by: Antonyant on June 11, 2008, 06:14:19 pm
Post by: fredrick on June 11, 2008, 06:15:13 pm
Post by: equinox on June 11, 2008, 06:15:50 pm
Lol who does this mao guy thinks he is suggesting the solutions by himself, go home man chill out!
By they way the structures questions where u got 19.99 is wrong its 20.01 cuz its under tension.
HUH? isnt it under compression... the roof is sitting on it or somethin lik that....
Post by: Mao on June 11, 2008, 06:16:44 pm
how do you remember all that????????????????i have the exam :P
Post by: Captain on June 11, 2008, 06:17:55 pm
how do you remember all that????????????????i have the exam :P
I was concidering doing solutions, however yours pretty much sum mine up.
I've posted my Further electronics solutions in another thread, feel free to add them here.
Post by: Pop on June 11, 2008, 06:17:59 pm
Post by: fredrick on June 11, 2008, 06:19:26 pm
Post by: iamdan08 on June 11, 2008, 06:19:45 pm
Post by: fredrick on June 11, 2008, 06:22:35 pm
Post by: dcc on June 11, 2008, 06:24:14 pm
Post by: Mao on June 11, 2008, 06:25:06 pm
this was supposed to be embedded in the first post.
but i was in a rush :P
Post by: equinox on June 11, 2008, 06:25:49 pm
"Question 4 (2 marks)
power dissipated through Rc
P=VI^2 = 9E-3 W"
isnt Power =VI not VI^2
or did i just screw up another question
Post by: Mao on June 11, 2008, 06:27:37 pm
Another query...sorry im just stressing out ... :(i swear i meant
"Question 4 (2 marks)
power dissipated through Rc
P=VI^2 = 9E-3 W"
isnt Power =VI not VI^2
or did i just screw up another question
Post by: fredrick on June 11, 2008, 06:28:49 pm
i got 9E-3 i think
Post by: dancing_jesus on June 11, 2008, 06:37:57 pm
These are my answers, and are not to be considered as the right answer.
they are just suggestions :)
VCE Physics Unit 3 2008
Motion in one and two dimensions
Question 1 (2 marks)
What is the acceleration of the two ships when the tugboat and the ship are travelling at 2.0 ms-1
from graph: resistanceN
Question 2 (2 marks)
After a time, the tugboat and ship are travelling at constant speed
constant speed => Fnet=0
hence resistance
simple circular motion...~[m=2.4 kg, v=2.0 ms-1, r=1.6m]
Question 3 (2 marks)
What is the tension in the string?
Question 4 (2 marks)
direction of the resultant force is P [towards the centre]
Simple projectile motion: v=30,=36.9, VCELand ignores air friction
Question 5 (2 marks)
What is the maximum height the ball reaches?
16.22m
Question 6 (2 marks)
resultant force is R [downwards]
now an advertising board is placed 72m from the batsman:
Question 7 (3 marks)
what height above the groud will the ball strike the advertising board?
solving for horizontal velocity gives a time
subbing this time into verticle motiongives 9 meters [VCAA is too nice :P ]
simple collision, train=20E3 kg, v=8ms-1, trucks=20E3 kg
Question 8 (2 marks)
What is the speed of the coupled locomotive + trucks?
simple proportionality gives 2ms-1
but you'd have to use conservation of momentum to get marks
Question 9 (3 marks)
Impulse TO the locomotive BY the trucks
obviously in a westward direction, 1.2E5 kgms-1
Question 10 (3 marks)
INELASTIC :D
Question 11 (2 marks)
During the collision, compare sizes of the magnitudes of average forces FL and FT
points that I think will give marks:
- Newton's 3rd Law, action/reaction pair
- :. FL = FT
Simple spring system, k=10, m=0.2kg, stretched to 60cm=0.6m, original 40cm=0.4m
Question 12 (2 marks)
Stationary Elastic Potential Energy:
EPE = 0.2J
Question 13 (2 marks)
What is the graph of KINETIC energy as a function of height?
D - upside down parabola
this is a good question that'll trip a lot of people up. there's two ways of thinking about it:
1. the ball changes direction at top and bottom [oscillating], therefore kinetic energy at top and bottom must be 0 [instantaneously stationary], hence only D satisfy this [at both top and bottom it is 0]
2. total energy is constant. GPE increases as a function of height [basically answers the next question].
- if we set GPE to be 0 at the bottom, KE=0 initially [it changes direction => v=0], => EPE=TE initially.
- EPE decreases quadratically [though not from turning point], GPE increases linearly, KE is TE [constant] - GPE [linear] - EPE [quadratic], which will give a quadratic expression
so what i'm trying to say is, D is right =]
Question 14 (2 marks)
What is GPE vs height?
compared to last question this one was a no-brainer. A. [linear increasing]
Eccliptical orbit of a comet around the sun
Question 15 (2 marks)
Describe how speed and total energy change from X [closest point] to Y [furthest point]
Total energy remains constant
Speed decreases as GPE increases hence KE decreases
This page is blank
Simple Gravity
Satellite = 930 kg
Mars = 6.42E23 kg
Radius of Orbit [thank god not above the surface] = 3.83E6 m
Question 16 (2 marks)
Calculate the gravitational FORCE.
not field strength :P
2.71E3 N
Question 17 (3 marks)
Calculate period of orbit
Keplar's law or consequential answer from 16
7.20E3 s
Electronics and Photonics
Simple LED [2.5v] circuit [supply = 8v] with limiting resistor [300 ohms]
Question 1 (2 marks)
current = 18.3 mA
TRANSISTOR AMPLIFIER!!!
Rc = 1k ohm
R1 = 2k ohm
R2 = 1k ohm
point P at 4 way connection
point Q at collector connection
Question 2 (2 marks)
DC voltage at point P
voltage divider gives 2V
current through Rc = 3 mA
Question 3 (2 marks)
voltage between Q and earth
[basically the DC voltage]
3V - crucial step: Vout = Vcc - V-Rc
Question 4 (2 marks)
power dissipated through Rc
P=I2R = 9E-3 W
Vout/Vin transfer characteristics:
[cut off at -60mV,3V] [saturation at 60mV, -3V] [quienscent point 0,0]
Question 5 (2 marks)
voltage gain = 50
Question 6 (3 marks)
Explain the shape of the graph, including why the graph has a negative slope, and why it has horizontal sections at [blah]
points that would be worth marks [i hope]
- as ib increase, ic increase and Vc increase => Vout decrease => inverting [negative gradient] (1 mark)
- cut off/saturation occur because ib is too low/high, resulting in the transistor acting as an open/closed switch with maximum/minimum potential across it. Since Vout is parallel to the transistor, the potential across it will also be maximum/minimum and cannot rise/decrease any further, hence the signal is clipped as reflected by the horizontal sections.
Question 7 (3 marks)
Vin = sinosudial wave, period = 1ms, max = 100mv, min = -100mV
marks allocated:
- correct frequency [unchanged]
- maximum at3
- clipping occurs both top and bottom
- inverted
now we are given two transistors in one circuit!
Question 8 (3 marks)
explain in terms of biasing why a capacitor is required between X [collector of first transistor] and Y [4 way connection of second transistor]
marks allocated:
- to avoid distortion, the base [not input] must be biased to the quienscent point [middle of linear operating range], hence point Y must be biased correctly
- voltage at point X has a DC and AC component. the DC component is counterfunctional to the biasing of the second transistor
- a capacitor is installed to decouple the signal, removing the DC component to ensure correct biasing hence optimum amplification
simple thermistor
Question 9 (1 mark)
R at 20 degrees = 1000 ohms
simple thermistor voltage divider in a cooling circuit [they actually got it right this time!] with variable Resistor [vout connected across this], and 12V supply
Question 10 (3 marks)
set R to maintain 5 degrees [Vout is required to be 4]
Rt=4000 ohms
by simple ratio -> R = 2000 ohms
Question 11 (3 marks)
The cool room is not cold enough, should R be increased or decreased?
marks allocated:
- increase resistance
- as T decrease, Rt increase and VRt increase, VR decrease
- to increase VR to turn on the cooling element, R need to be increased
Detailed Studies - Structures and Materials
sorry to those who are not doing S&M, but i have chem exam tomorrow and I realy cbf. :P
these will get done tomorrow after I type up the chem exam.
btw everything is (2 marks) from now, so i wont bother
a two-way stress/strain graph that'll have freaked people out, compression is in quadrant 3.
Question 1
Young's modulus, D = 5.3E10 Nm-2
Question 2
B -> brittle
Question 3
max force given area = 1.50m^2 -> A, 1.2E8 N
Question 4
strain energy per unit volume = B, 1.2E5 J/m^3
Question 5
for the strain energy of this particular block, volume = 30 [20 m * 1.5 m^2], hence C, 30
Question 6
at strain = 500E-4,= 0.01m, hence C, 19.99m
Question 7
C, stone is stiffer in tension than compression
(http://obsolete-chaos.wikispaces.com/space/showimage/structures_bridge_pic.gif)
Question 8
XY is under obvious tension, A
Question 9
the stress experienced by YZ is:
B, compression
this one took a while
both ZX and XY are under tension. at Z, a force is pulling it towards X, and Y, a force is pulling it towards X also. the horizontal components of these two forces can be resolved and shown that they pull towards the same centre of YZ, hence it is under compression.
it makes sense to think this way, as concrete is weak under tension, and structures have been designed to make the beam under compression to increase its lifetime and safety.
Question 10
material P is stronger and stiffer than Q, but not as tough. option A.Code: [Select]_X______________________Y___________ZBeam is 4000 kg
| | | |
| | | |
|<-------8m------------->|<---4m---->|
Question 11
with no one standing at the lookout, what is the force of the beam on the support at X?
using rotational equilibrium, [and remembering 4000kg = 40000N], we arrive at C, 10000N
Question 12
what is the maximum load at Z without the beam tilting?
at that point, the force on X will be 0, and the torque at Z balances the weight of the beam. hence B, 2000kg
Question 13
steel reinforcement of the above beam:
D: ==========================
this is because between XY the beam sags [tension on the bottom], and bends down at Z [tension on top]
OH FUCK I GOT THAT ONE WRONG
Post by: chid on June 11, 2008, 06:41:47 pm
http://itute.com/physicsline/phys_unit3_summary_sheets.pdf
Post by: Mao on June 11, 2008, 06:43:52 pm
and chid: that's not what I got tho? i said it was compression....
Post by: chid on June 11, 2008, 06:46:08 pm
Post by: Mao on June 11, 2008, 06:48:52 pm
it was tension first and then compression...right? As the right answer?
itute summary sheet says tension then tension.
we are talking about YZ here :P
Post by: chid on June 11, 2008, 06:50:17 pm
Post by: 4xrp on June 11, 2008, 06:58:50 pm
Post by: chid on June 11, 2008, 07:12:01 pm
Post by: 4xrp on June 11, 2008, 07:15:55 pm
Post by: anonymous12 on June 11, 2008, 07:33:23 pm
This is a pre-transposed equation for T that I found.
T= ((4*pi^2*R^3)/g)^(1/2)
Use R as 3.83*10^6
g=2.91
then I got T=2.7*10^10
am I wrong?
Post by: Mao on June 11, 2008, 07:40:10 pm
are you sure it wasn't XZ. That's what I thought all this time.
that was the first one
the second one was on YZ, where I was wrong [i am pretty sure i'm wrong.... i hope i double-wrong myself tho :P ]
Post by: 4xrp on June 11, 2008, 08:32:36 pm
For question 17 can you do.
This is a pre-transposed equation for T that I found.
T= ((4*pi^2*R^3)/g)^(1/2)
Use R as 3.83*10^6
g=2.91
then I got T=2.7*10^10
am I wrong?
How can you do that? You don't have g do you...?
I used T=[(4pi^2R^3)/(GM)]^(1/2)
Thats seems to get the same answer as Keplers law.
Post by: Mao on June 11, 2008, 08:37:11 pm
For question 17 can you do.
This is a pre-transposed equation for T that I found.
T= ((4*pi^2*R^3)/g)^(1/2)
Use R as 3.83*10^6
g=2.91
then I got T=2.7*10^10
am I wrong?
that is wrong.
your formula is suggesting that:
that'll create
what your formula should have read is
[ps btw 2000 posts, woot, lol]
Post by: 4xrp on June 11, 2008, 08:52:57 pm
For question 17 can you do.
This is a pre-transposed equation for T that I found.
T= ((4*pi^2*R^3)/g)^(1/2)
Use R as 3.83*10^6
g=2.91
then I got T=2.7*10^10
am I wrong?
that is wrong.
your formula is suggesting that:, where
that'll create, which is really a bad transposing from keplar's law.
what your formula should have read is
[ps btw 2000 posts, woot, lol]
So was my way right? I'm assuming it was because I got a similar answer...
Post by: Mao on June 11, 2008, 08:58:55 pm
Post by: 4xrp on June 11, 2008, 09:04:30 pm
Post by: anonymous12 on June 11, 2008, 09:21:10 pm
Post by: Mao on June 11, 2008, 09:30:41 pm
Fuck TSFX man. That's the formula they gave me. Shit!
you will find that they all do the same:
http://vcenotes.com/forum/index.php/topic,2645.0.html
Post by: 4xrp on June 11, 2008, 09:35:40 pm
Fuck TSFX man. That's the formula they gave me. Shit!
I went to TSFX as well and got that same sheet...That formula isn't on it. You must have it confused with the one I used.
Post by: dooopy on June 11, 2008, 10:10:02 pm
Post by: porsca911 on June 11, 2008, 10:17:37 pm
Post by: dcc on June 11, 2008, 10:24:22 pm
Post by: dooopy on June 11, 2008, 10:35:34 pm
Post by: Pop on June 11, 2008, 10:39:53 pm
Post by: nickalaz on June 11, 2008, 10:42:55 pm
I believe I got 100% after checking my answers with other variable elite students.
I have doing 83 hours in WoW that week and I still won you all
Regardations,
3D | dohteM
Post by: Russelljohn on June 11, 2008, 10:49:18 pm
Post by: Pop on June 11, 2008, 10:54:39 pm
That was being the easiest exam for physics for yetazn gold farmer?
I believe I got 100% after checking my answers with other variable elite students.
I have doing 83 hours in WoW that week and I still won you all
Regardations,
3D | dohteM
Post by: AppleXY on June 11, 2008, 10:57:25 pm
That was being the easiest exam for physics for yet
I believe I got 100% after checking my answers with other variable elite students.
I have doing 83 hours in WoW that week and I still won you all
Regardations,
3D | dohteM
Gtfo. Your arrogance isn't welcome here.
Post by: dcc on June 11, 2008, 10:58:19 pm
That was the easiest exam for physics yet.
I believe I got 100% after comparing my answers with those of other various elite students.
I have played WoW for 83 hours this week and I still defeated you all comprehensively.
Regards,
3D | dohteM
Post by: 4xrp on June 11, 2008, 11:01:59 pm
Post by: 4xrp on June 11, 2008, 11:03:55 pm
That was the easiest exam for physics yet.
I believe I got 100% after comparing my answers with those of other various elite students.
I have played WoW for 83 hours this week and I still defeated you all comprehensively.
Regards,
3D | dohteM
Oh. And that is a massive win. Respect.
Post by: AppleXY on June 11, 2008, 11:06:10 pm
Post by: Pop on June 11, 2008, 11:08:27 pm
Post by: 4xrp on June 11, 2008, 11:11:48 pm
I totally want to be that guy! Doesn't everyone else here idolize someone who can spend that much time away from society! Awesome!
Post by: enwiabe on June 11, 2008, 11:18:18 pm
Post by: ozzaJ on June 11, 2008, 11:26:51 pm
That was being the easiest exam for physics for yet
I believe I got 100% after checking my answers with other variable elite students.
I have doing 83 hours in WoW that week and I still won you all
Regardations,
3D | dohteM
nickalaz-
1. Learn to write a sentence
2. 4xrp you are spot on
Thanks for the solutions Mao!
Post by: onlyfknhuman on June 12, 2008, 08:34:27 am
instead of the mass of the ship, i used the mass of the tugboat
because the driving force they gave us was in tension wasnt it? i remmbered in past exams to find the tension you had to consider the object thats being pulled not the object pulling and the corresponding mass/friction towards it.
so if u you found the acelartion of the boat it should be the same as the ship?
i got 5 questions wrong, dumb mistakes
Post by: ell on June 12, 2008, 09:31:38 am
http://www.itute.com/physicsline/ph_vcaa_exam1_sol_2008.pdf
Post by: ed_saifa on June 12, 2008, 09:57:20 am
itute solutions are up for anyone interested:It says the file is damaged :(
http://www.itute.com/physicsline/ph_vcaa_exam1_sol_2008.pdf
Post by: orangez on June 12, 2008, 10:01:15 am
itute solutions are up for anyone interested:It says the file is damaged :(
http://www.itute.com/physicsline/ph_vcaa_exam1_sol_2008.pdf
Serious? I downloaded it.
Post by: bigtick on June 12, 2008, 10:03:47 am
Post by: deano on June 12, 2008, 02:37:40 pm
Post by: orsel on June 12, 2008, 03:41:47 pm
Assuming equilibrium:
- Reaction force from ground at point Z must be upwards (ignoring horizontal component of friction).
- Only way to balance this upwards force is a downwards compressive force from beam ZX (ZY has no vertical component)
- Hence ZX = compression
- Only way to balance forces at point X is if XY is under tension (since we already know ZX is compressed and pushing upwards)
- Back to point Z, ZX still has a horizontal component to the 'left'. A force to the 'right' must be supplied. Thus ZY is under tension so that it pulls to the 'right'.
btw I reckon they made this year too easy...I'm a bit concerned about whether they'll accept my explain-type answers; other than that full marks lol.
Post by: Mao on June 12, 2008, 04:27:01 pm
Structures and materials q9 aka 'da bridge':
Assuming equilibrium:
- Reaction force from ground at point Z must be upwards (ignoring horizontal component of friction).
- Only way to balance this upwards force is a downwards compressive force from beam ZX (ZY has no vertical component)
- Hence ZX = compression
- Only way to balance forces at point X is if XY is under tension (since we already know ZX is compressed and pushing upwards)
- Back to point Z, ZX still has a horizontal component to the 'left'. A force to the 'right' must be supplied. Thus ZY is under tension so that it pulls to the 'right'.
btw I reckon they made this year too easy...I'm a bit concerned about whether they'll accept my explain-type answers; other than that full marks lol.
that's a very good explanation :) thank you.
I was resolving forces at the point Y. XY under tension made me think that ZY need to be balanced with a tensile force as well :P
oh well :)
Post by: pinchies on June 12, 2008, 04:41:06 pm
Post by: Mao on June 12, 2008, 05:47:55 pm
Thanks Mao! =D (I don't think I got *any* wrong!) I 'll just need to see the multi though... that dodgy capacitor one...
well done!
[dcc: $10 on this guy duxing physics]
Post by: zzdfa on June 12, 2008, 06:43:36 pm
Post by: chid on June 12, 2008, 06:51:06 pm
what were the two questions asked on the exam (in order) again (and was choice A tension for both questions)
Thanks, I'm just a little forgetful/confused
Post by: 4xrp on June 12, 2008, 07:04:40 pm
Post by: jay. on June 12, 2008, 07:06:22 pm
Is there anywhere we can get a copy of the exam online? Or is anyone who has it able to put it up?
that would be mighty helpful..
Post by: costargh on June 12, 2008, 07:09:36 pm
Post by: iamdan08 on June 12, 2008, 07:15:10 pm
Post by: Mao on June 12, 2008, 07:16:26 pm
Don't you guys take your exam papers with you? We are told to take EVERYTHING with us (except obviously the answer book)
we answer in the question booklet in science subjects. [just like math]
Post by: Mao on June 12, 2008, 07:17:26 pm
Captain, (or anyone else who has the exam paper), Q12 Further electronics, was it asking for power dissapation over the 10kO resistor or the 10O resistor? Was there even a 10kO resistor in the circuit?it was the 10 ohms resistor
Post by: Mao on June 12, 2008, 07:18:30 pm
Is there anywhere we can get a copy of the exam online? Or is anyone who has it able to put it up?we would, except neither of us have a scanner...
if you'd like a particular question to be posted up, say so and I'll try my best to replicate [but no guarantees on the diagrams :P ]
Post by: chid on June 12, 2008, 07:23:53 pm
That would be great.
Post by: Mao on June 12, 2008, 07:58:43 pm
Post by: anonymous12 on June 12, 2008, 08:36:32 pm
For question 17 can you do.
This is a pre-transposed equation for T that I found.
T= ((4*pi^2*R^3)/g)^(1/2)
Use R as 3.83*10^6
g=2.91
then I got T=2.7*10^10
am I wrong?
that is wrong.
your formula is suggesting that:
that'll create
what your formula should have read is
[ps btw 2000 posts, woot, lol]
Do I get zero marks due to the incorrect formula I used? Please say no.
I've stated the correct R value, g value and G value.
Post by: zzdfa on June 12, 2008, 09:03:23 pm
Captain, (or anyone else who has the exam paper), Q12 Further electronics, was it asking for power dissapation over the 10kO resistor or the 10O resistor? Was there even a 10kO resistor in the circuit?it was the 10 ohms resistor
Argh. That's 4 marks I've lost from misreading a question. damn multiple choice.
Post by: Skullmaster4 on June 12, 2008, 11:00:55 pm
My answers where actually rather like Mao's
I'm starting to see hope now!!
P.s I'm in the same specialist maths class as Mao..
He kills me :'(
(This is David btw Mao :P )
Post by: dcc on June 13, 2008, 08:03:17 am
Sweet,
My answers where actually rather like Mao's
I'm starting to see hope now!!
P.s I'm in the same specialist maths class as Mao..
He kills me :'(
(This is David btw Mao :P )
lol Mao's spec class only has 2 people, from what I hear :)
Post by: bigtick on June 13, 2008, 08:12:00 am
Post by: Fidgey on June 13, 2008, 02:58:01 pm
Post by: anonymous12 on June 13, 2008, 04:14:41 pm
Post by: Mao on June 13, 2008, 04:30:06 pm
Can someone plz answer my question?
mmm, I cant guarantee :P sorry
and hi David
btw bigtick are you in Narre south or lyndhurst? :P I think i might have an idea who you might be :P
Post by: chid on June 13, 2008, 04:36:30 pm
Can someone plz answer my question?I remember reading somewhere (might have been an assessment report and I think coblin mentioned it one time) that using the derived projectile motion formulas was 'all or nothing.' If you got the answer wrong it was zero but if you got the answer right then you got full marks. I think that applies here as well. I hope for your sake that you get a mark for your working though. (I cant stand losing marks when you don't deserve to lose them)
Post by: Skullmaster4 on June 13, 2008, 05:27:19 pm
Atleast I'm guaranteed second place :D
On a further note he is also in my physics and Chem classes..
I can't escape him :'(
Post by: bigtick on June 15, 2008, 09:07:44 am
well i have a query..http://itute.com/board/viewtopic.php?p=695#695
"Question 8
XY is under obvious tension, A
Question 9
the stress experienced by YZ is:
B, compression
this one took a while
both ZX and XY are under tension. at Z, a force is pulling it towards X, and Y, a force is pulling it towards X also. the horizontal components of these two forces can be resolved and shown that they pull towards the same centre of YZ, hence it is under compression.
it makes sense to think this way, as concrete is weak under tension, and structures have been designed to make the beam under compression to increase its lifetime and safety."
ALot of people reckon it was tension for both...and itute seems to agree with them...