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March 29, 2024, 04:53:02 am

Author Topic: VCE Physics Question Thread!  (Read 603318 times)  Share 

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Adequace

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Re: VCE Physics Question Thread!
« Reply #900 on: March 08, 2015, 09:30:43 pm »
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Not sure where to post this but, I'm undertaking physics 1/2 a year earlier this year and want to build a solid foundation for 3/4. I'm aware of that electricity and motion overlap in to 3/4 but I'm not sure what else I can do to prepare myself for 3/4.

I'm looking to get at least 40, or so. What else can I do to prepare better? What test percentages should I be aiming to get? What should I do right now since I have more time?

It might be worth nothing that my confidence for physics is kind of low...I struggled with nuclear physics and will be lucky to get at least 80% on my outcome test when we get our results back.

silverpixeli

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Re: VCE Physics Question Thread!
« Reply #901 on: March 13, 2015, 10:56:26 am »
+1
Not sure where to post this but, I'm undertaking physics 1/2 a year earlier this year and want to build a solid foundation for 3/4. I'm aware of that electricity and motion overlap in to 3/4 but I'm not sure what else I can do to prepare myself for 3/4.

I'm looking to get at least 40, or so. What else can I do to prepare better? What test percentages should I be aiming to get? What should I do right now since I have more time?

It might be worth nothing that my confidence for physics is kind of low...I struggled with nuclear physics and will be lucky to get at least 80% on my outcome test when we get our results back.

luckily nuclear physics is entirely irrelevant to units 3 and 4 (except the occasional non-VCAA practice paper where they assume you know the charge on an alpha particle is +2e but yeah)


forming solid foundations for basic electrical and mechanical concepts should be your goal in year 11 because there's plenty of new stuff that you'll have to wait until year 12 to learn, both content and problem-solving skills.

to help with this, i highly recommend khanacademy's physics playlists, the first 5 of which are relevant to VCE motion and the second last (electricity and magnetism) is relevant to that. This is a resource I regularly go to for a clear and comprehensive explanation of a physics concept! it can't hurt to give it a try :)
[EDIT: I should add, I'm currently watching the thermodynamics playlist because second year thermal physics isn't taught in a very organised way at the university of melbourne, and salman khan has everything sorted. khanacademy is also great for maths and other sciences]

the other thing you can do is work at improving your algebra skills (for VCE physics, algebra is as far as you need to go, no need for calculus). I'm talking about notation, solving for x, fractions skills and mental arithmetic. Being good at maths makes physics problems waaaay more fun than if you're not a maths-y student and you kinda get the physics but you can't follow the working out. Being clear and formal in your working forces you to think methodically and improve your problem solving skills.

By improving your physics intuition and maths skills now you're forming a really really good starting point for year 12 studies!
i mean, you can work ahead if you like, but you'll have plenty of time in year 12 to learn the year 12 content. If you focus on the foundations now you'll be set when things get far busier next year.
also don't worry about test scores, seriously the only thing that matters at this point is your intuition :)
« Last Edit: March 13, 2015, 11:00:39 am by silverpixeli »
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knightrider

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Re: VCE Physics Question Thread!
« Reply #902 on: March 16, 2015, 01:31:27 am »
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How do you get from 16KWh to 5.76*10^7 J

nerdgasm

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Re: VCE Physics Question Thread!
« Reply #903 on: March 16, 2015, 02:00:21 am »
+1
16KWh means "sixteen kilowatt-hours" or "sixteen thousand watt-hours". What precisely is a 'watt-hour'?

Consider that the watt (W) is a unit of power. One formula for power is Power = Energy/Time. Specifically, 1 Watt = 1 Joule/second. This unit is useful when we want to talk about the energy output of something (such as a lightbulb, you may have noticed things like "60W" or "40W" on lightbulb cartons). Now that we know what a Watt (pardon the pun) is, let's think about what a "watt-hour" is. A 'watt-hour' is the amount of energy that a 1W object would produce in an hour. In other words, this would be (1 Joule/sec) * (3600sec) = 3600 Joules. In short, a "Watt-hour" is a unit of energy, and it is equivalent to 3600J. If you like, you can think of this as the result of multiplying the units 'watt' and 'hour' together.

Therefore, 16KWh = 16000Watt-hours = 16000*3600J = 57600000J = 5.76*10^7 J .

knightrider

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Re: VCE Physics Question Thread!
« Reply #904 on: March 16, 2015, 02:11:00 am »
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16KWh means "sixteen kilowatt-hours" or "sixteen thousand watt-hours". What precisely is a 'watt-hour'?

Consider that the watt (W) is a unit of power. One formula for power is Power = Energy/Time. Specifically, 1 Watt = 1 Joule/second. This unit is useful when we want to talk about the energy output of something (such as a lightbulb, you may have noticed things like "60W" or "40W" on lightbulb cartons). Now that we know what a Watt (pardon the pun) is, let's think about what a "watt-hour" is. A 'watt-hour' is the amount of energy that a 1W object would produce in an hour. In other words, this would be (1 Joule/sec) * (3600sec) = 3600 Joules. In short, a "Watt-hour" is a unit of energy, and it is equivalent to 3600J. If you like, you can think of this as the result of multiplying the units 'watt' and 'hour' together.

Therefore, 16KWh = 16000Watt-hours = 16000*3600J = 57600000J = 5.76*10^7 J .

Thanks so much nerdgasm  :)

lolaishappy

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Re: VCE Physics Question Thread!
« Reply #905 on: March 17, 2015, 06:01:51 pm »
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When finding tension in rope or rod of a object going in circular motion, why do some questions consider the weight force in finding T when others don't? And what's the difference between vertical and horizontal motion?
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Zealous

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Re: VCE Physics Question Thread!
« Reply #906 on: March 17, 2015, 06:35:01 pm »
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When finding tension in rope or rod of a object going in circular motion, why do some questions consider the weight force in finding T when others don't? And what's the difference between vertical and horizontal motion?

This image below might help:



When we're looking at an object in circular motion we've got to consider all the forces acting on it - there's going to be gravity always acting downwards and a tension force which needs to apply an inwards force to the object so that it has a centripetal acceleration. At the bottom of the circle, the tension in the rope has to be able to overcome the force of gravity on the object, and then have the required extra force for the centripetal acceleration, that's why we have at the bottom. At the top of the circle, gravity is working in the direction required for centripetal acceleration, so the tension is decreased as dictated by . If you're looking at the tension when the object is on the edges, we don't need to consider the weight force as it is acting perpendicular to the tension in the rope so there's no component of the weight force working against the tension.

Horizontal circular motion questions usually look like this:



The main difference is that the forces just work in different directions, and there are sometimes angles involved, but you can resolve them using simple trigonometry.
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lolaishappy

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Re: VCE Physics Question Thread!
« Reply #907 on: March 17, 2015, 06:41:50 pm »
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This image below might help:

(Image removed from quote.)

When we're looking at an object in circular motion we've got to consider all the forces acting on it - there's going to be gravity always acting downwards and a tension force which needs to apply an inwards force to the object so that it has a centripetal acceleration. At the bottom of the circle, the tension in the rope has to be able to overcome the force of gravity on the object, and then have the required extra force for the centripetal acceleration, that's why we have at the bottom. At the top of the circle, gravity is working in the direction required for centripetal acceleration, so the tension is decreased as dictated by . If you're looking at the tension when the object is on the edges, we don't need to consider the weight force as it is acting perpendicular to the tension in the rope so there's no component of the weight force working against the tension.

Horizontal circular motion questions usually look like this:

(Image removed from quote.)

The main difference is that the forces just work in different directions, and there are sometimes angles involved, but you can resolve them using simple trigonometry.
Thanks zealous, what happens in between the points of the top diagram?
« Last Edit: March 17, 2015, 06:44:22 pm by lolaishappy »
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Kel9901

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Re: VCE Physics Question Thread!
« Reply #908 on: March 17, 2015, 08:01:51 pm »
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Thanks zealous, what happens in between the points of the top diagram?

From the study design
Quote
apply Newton's second law to circular motion in a vertical plane; consider forces at the highest and lowest positions only

so don't worry about it
s=change in displacement for physics
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lolaishappy

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Re: VCE Physics Question Thread!
« Reply #909 on: March 17, 2015, 08:04:26 pm »
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From the study design
so don't worry about it
Thank you kel9901!
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lolaishappy

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Re: VCE Physics Question Thread!
« Reply #910 on: March 22, 2015, 09:29:51 pm »
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How do we distinguish between v and u, and which to put in a equation in projectile motion?
In general I'm struggling to know which equations to apply to different questions :^(
« Last Edit: March 22, 2015, 09:51:02 pm by lolaishappy »
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Floatzel98

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Re: VCE Physics Question Thread!
« Reply #911 on: March 22, 2015, 11:01:25 pm »
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How do we distinguish between v and u, and which to put in a equation in projectile motion?
In general I'm struggling to know which equations to apply to different questions :^(
The simplest thing i do to figure out which equation I should use is to write out all the values for u,v,x,t,a that have been given in the question and also find the one that i need to solve for. Match the equation on the information given. If you aren't already given a diagram, try to draw one and try to draw the values on the diagram to get a sense of where each part applies. You are always given the velocity at the max height and the acceleration (vertical component) so if you split the parabola that makes up the projectile motion into 2, you will always have a final velocity if you use the first half. Then depending on what the questions asks, you can solve from there. You can always solve for initial velocity  if a height or time is given from this point.

Sorry if i am not too clear or anything. It's my first time actually trying to help someone on here. Usually i just get help haha. I can try to explain it a bit more if you need me to or if no one else tries to help.
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Cosec

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Re: VCE Physics Question Thread!
« Reply #912 on: March 23, 2015, 06:41:23 pm »
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Can somebody explain why when connected in series, a resistor and a thermistor, with the output voltage being measured across the fixed resistor which is placed second after the thermistor, that when the temp increases, so does the resistance of the thermistor and thus the output voltage increases. Im kind of confused with it all.

Kel9901

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Re: VCE Physics Question Thread!
« Reply #913 on: March 23, 2015, 09:13:59 pm »
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Can somebody explain why when connected in series, a resistor and a thermistor, with the output voltage being measured across the fixed resistor which is placed second after the thermistor, that when the temp increases, so does the resistance of the thermistor and thus the output voltage increases. Im kind of confused with it all.

I'm pretty sure that when the temperature increases, the resistance of thermistors decrease. Let R be the resistance of the normal resistor and R(th) that of the thermistor. Vout=Vin*R/R(th), so when R(th) decreases, the denominator only decreases and hence the voltage drop across the normal resistor increases

edit: fuck latex
« Last Edit: March 24, 2015, 09:16:24 am by Kel9901 »
s=change in displacement for physics
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lolaishappy

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Re: VCE Physics Question Thread!
« Reply #914 on: March 23, 2015, 10:23:31 pm »
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The simplest thing i do to figure out which equation I should use is to write out all the values for u,v,x,t,a that have been given in the question and also find the one that i need to solve for. Match the equation on the information given. If you aren't already given a diagram, try to draw one and try to draw the values on the diagram to get a sense of where each part applies. You are always given the velocity at the max height and the acceleration (vertical component) so if you split the parabola that makes up the projectile motion into 2, you will always have a final velocity if you use the first half. Then depending on what the questions asks, you can solve from there. You can always solve for initial velocity  if a height or time is given from this point.

Sorry if i am not too clear or anything. It's my first time actually trying to help someone on here. Usually i just get help haha. I can try to explain it a bit more if you need me to or if no one else tries to help.
Thanks!
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