October 21, 2019, 11:12:45 pm

0 Members and 1 Guest are viewing this topic.

#### Homer

• Victorian
• Forum Obsessive
• Posts: 431
• Respect: +10
##### Re: VCE Physics Question Thread!
« Reply #30 on: May 02, 2013, 09:36:09 pm »
0
Thanks
Bachelor of Laws/Engineering

2013 ATAR: 98.65

Specialist Maths [53.06] Maths Methods [48.83] Physics [48.22]

Donuts. Is there anything they can't do?

#### jono88

• Victorian
• Trendsetter
• Posts: 104
• Respect: 0
##### Re: VCE Physics Question Thread!
« Reply #31 on: May 03, 2013, 05:47:40 pm »
0
What is the difference between tensile strength and yield strength?

#### Phy124

• Honorary Moderator
• Part of the furniture
• Posts: 1355
• Respect: +462
##### Re: VCE Physics Question Thread!
« Reply #32 on: May 03, 2013, 07:31:31 pm »
+1
Ultimate tensile strength is the maximum stress a material can withstand before failing under tension.

Yield strength is the maximum stress which a material can withstand before it begins to deform plastically, rendering it unable to return to its original state.
« Last Edit: May 03, 2013, 07:33:07 pm by 2/cos(c) »
2011
Mathematical Methods | Physics | Chemistry | English | Business Management

2012-2017
Bachelor of Science in Applied Mathematics and Bachelor of Civil Engineering (Honours) @ Monash University

Current
Transport Modeller @ Arup

#### Homer

• Victorian
• Forum Obsessive
• Posts: 431
• Respect: +10
##### Re: VCE Physics Question Thread!
« Reply #33 on: May 13, 2013, 06:56:42 pm »
+1
How to?

EDIT: I actually figured out a way to do this now, but not sure if right or not

$\frac{2}{m^2}\times 16mm^2$

$\frac{2}{(1000mm)^2}\times 16mm^2$

$\frac{32mm^2}{(1000mm)^2}$

$\frac{32mm^2}{1000000mm^2}$

$=\frac{32}{1000000}$

$32\mu W$
« Last Edit: May 13, 2013, 07:12:28 pm by Homer »
Bachelor of Laws/Engineering

2013 ATAR: 98.65

Specialist Maths [53.06] Maths Methods [48.83] Physics [48.22]

Donuts. Is there anything they can't do?

#### jono88

• Victorian
• Trendsetter
• Posts: 104
• Respect: 0
##### Re: VCE Physics Question Thread!
« Reply #34 on: May 13, 2013, 08:06:24 pm »
0
I was wondering if anyone has had there EPI yet for physics. What topics are involved?

#### Professor Polonsky

• Victorian
• Part of the furniture
• Posts: 1154
• Respect: +93
##### Re: VCE Physics Question Thread!
« Reply #35 on: May 15, 2013, 08:17:33 pm »
0
Can anyone explain why the net force (if there is no side friction) for a banked then is $mg\times\tan\theta$?
« Last Edit: May 16, 2013, 12:46:34 am by Polonius »

• Victorian
• Fresh Poster
• Posts: 3
• Respect: 0
##### Re: VCE Physics Question Thread!
« Reply #36 on: May 15, 2013, 10:50:24 pm »
0
Anyone mind helping me out with my circuit question (topic link below)

circuit question

#### Robert123

• Victorian
• Forum Obsessive
• Posts: 201
• Respect: +5
• School: Kyabram P-12 College
##### Re: VCE Physics Question Thread!
« Reply #37 on: May 22, 2013, 04:45:08 pm »
0
Question about the 'gain' in amplifiers
First off, can you have a negative gain? Does that occur when the gradient is negative?
Secondly, my interpretation of the gain is the factor in which the input voltage is multiplied to get the output voltage. However, in Qs 3a in Test 2 on electronic & photonics part in the atar notes book, it doesn't work because the gradient is negative.
Also, my original understanding with circuits, if you have two resistors in a circuit, the resistor with the highest resistance will have the highest power output. But this doesn't seem the case with resistors in parallel. Could someplease please confirm this and try to explain it a bit.
Thanks

#### EspoirTron

• Posts: 598
• Respect: +82
##### Re: VCE Physics Question Thread!
« Reply #38 on: May 22, 2013, 09:50:53 pm »
0
Question about the 'gain' in amplifiers
First off, can you have a negative gain? Does that occur when the gradient is negative?
Secondly, my interpretation of the gain is the factor in which the input voltage is multiplied to get the output voltage. However, in Qs 3a in Test 2 on electronic & photonics part in the atar notes book, it doesn't work because the gradient is negative.
Also, my original understanding with circuits, if you have two resistors in a circuit, the resistor with the highest resistance will have the highest power output. But this doesn't seem the case with resistors in parallel. Could someplease please confirm this and try to explain it a bit.
Thanks

Remember that the formula for gain is constituted by 'change in' vout and vin. The reason you may get a 'negative' gain is that the amplifier is inverting and since the gradient of the graph is negative, you will get a negative answer. To my knowledge gain is only concerned with a magnitude; therefore, while using it to calculate vout it you simply multiply by the gain value.
To address your second question. The reason why in series that a resistor with a higher resistance will output more power is because that it will have a higher potential across it, if you recall in a series circuit each component has the same amount of current flowing through it; hence, to maintain the ratio of v2/v1 = r2/r1 the resistor with the high resistance must have a higher potential across it and consequently a higher power output. The reason why this isn't the case for resistors in parallel is that in parallel the resistors will have the same potential but varying current based on their resistance. Therefore, I believe the lower the resistance the higher the current; thus, the resistor with the lowest resistance will have the highest power output (in the parallel component).
I hope that helped you out!
2012-2013: VCE
2014-2016: Bachelor of Biomedicine at Monash University

#### sin0001

• Victorian
• Forum Obsessive
• Posts: 487
• Respect: +1
##### Re: VCE Physics Question Thread!
« Reply #39 on: May 22, 2013, 11:25:07 pm »
0
Just a made-up example for help:
E.g. Suppose there's a circuit involving 2 resistors (2 and 6 ohms) in parallel, the battery supplies, let's say, 8 V. Current would be 1.5 A, the 2 ohm resistor would receive 3/4 of the total current, so 9/8 A. Whereas the 6 ohm resistor would get 1/4 of the total current, so 3/8 A.
P(2 ohm)= I^2 x R = (9/8)^2 x 2 = 2.53 W
P(6 ohm) = (3/8)^2 x 6 = 0.84 W
So it can be seen that the resistor with the lowest resistance, in a parallel circuit, would have the highest power output.
Hope this helps!
ATAR: 99.00
Monash Commerce Scholars

#### ~T

• Victorian
• Trendsetter
• Posts: 197
• Respect: +4
• School: St Patrick's College
##### Re: VCE Physics Question Thread!
« Reply #40 on: May 27, 2013, 04:51:31 pm »
0
^^ just getting resistance and current mixed up...

$\frac{1}{R_{T}} = \frac{1}{2} + \frac{1}{6} = \frac{2}{3}$

therefore $R_{T} = \frac{3}{2}$, not $I$

Current is thus given  $I_{T} = \frac{V}{R_{T}} = \frac{8}{1.5} = 5.33A$

The point still stands of course, but none of that calculation is even required.
The smaller resistor will receive $I_{2} = \frac{V_{T}}{R} = \frac{8}{2} = 4A$
The larger resistor will receive $I_{2} = \frac{V_{T}}{R} = \frac{8}{6} = 1.33A$

As $V$ is the same for both resistors (parallel circuit), the current will be greater for less resistance as $I=\frac{V}{R}$ and then the power will be greater because $P=VI$
ATAR: 99.95
Specialist 50 | Methods 50 | Physics 50 | Further 49 | Literature 48 | Music Style/Composition 41

2014 - 2016: Bachelor of Science (Chancellor's Scholars' Program) at The University of Melbourne

I will be tutoring in Melbourne this year. Methods, Specialist, and Physics. PM me if you are interested

#### Homer

• Victorian
• Forum Obsessive
• Posts: 431
• Respect: +10
##### Re: VCE Physics Question Thread!
« Reply #41 on: May 27, 2013, 07:54:53 pm »
0
with question 6 in the vcaa 2011 exam 1, how do we know that the switch it attached across the resistor or thermistor?
Bachelor of Laws/Engineering

2013 ATAR: 98.65

Specialist Maths [53.06] Maths Methods [48.83] Physics [48.22]

Donuts. Is there anything they can't do?

#### FlorianK

• Victorian
• Posts: 928
• Respect: +63
##### Re: VCE Physics Question Thread!
« Reply #42 on: May 27, 2013, 08:38:44 pm »
0
When the temperatur increases the resistance goes down and the voltage across the thermistor goes down, hence the voltage across the resistor goes up. If we would have a 500 Ohm resistor and the 1500Ohm Thermistor (at 20°) and the circuit switch across the thermistor then the circuit would switch when the temperature falls below 20° and not when it goes above 20°

#### tote.moore

• Victorian
• Posts: 18
• Respect: 0
• School: St. Joseph's College
##### Re: VCE Physics Question Thread!
« Reply #43 on: May 28, 2013, 06:24:46 pm »
0
from memory the question was...

If a certain amount of Power is transmitted through power lines into a house, why does the voltage of an appliance go up when another appliance is turned off?

2013 Goals(RAW):
Methods 36+
Spec 32+
Physics 35+
English 36+
Further 45+

1st. Pref: Bachelor of Engineering (Civil and Infrastructure)/Bachelor of Business (Management) - 93.20 atar

#### Robert123

• Victorian
• Forum Obsessive
• Posts: 201
• Respect: +5
• School: Kyabram P-12 College
##### Re: VCE Physics Question Thread!
« Reply #44 on: May 29, 2013, 03:05:22 pm »
0
Just a few question related to electronics & photonics.
What is RMS? How is it calculated? How much do we have to know about it for electronics & photonics?
Why do photodiodes have to be in reverse bias?
If a photodiode is in series with a resisitor, what would happen if the light intensity is doubled?
What are the 'reasons' for limits in an amplifier?
Any help will be greatly appreciated