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#### xZero

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##### Re: Newton's Laws and probably more questions
« Reply #15 on: September 25, 2011, 12:53:03 pm »
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i mean like i can't say an object of mass 2kg moving at 2m/s^2 would exert a force of 4N on whatever it hits, i can only say it's acceleration*mass is equal to the sum of forces acting on it (which brings to mind, weight, that's mass*acceleration is it not?)
you can say that an object of mass 2kg moving at 2ms^-1 has a momemtum of p=mv=4 and will exert a force on impact base on the other object's momemtum and the time interval of the collosion. In short, you use momemtum and impluse to work out the force exerted on objects in a collosion
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#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #16 on: September 25, 2011, 09:33:04 pm »
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^
interesting, i'm not up to momentum yet lol. everyone else is - i'm waay behind

what about weight? you can use mass * g to find that - is that the only exception?

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##### Re: Newton's Laws and probably more questions
« Reply #17 on: September 25, 2011, 09:35:05 pm »
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It's not an exception. 'g' is the acceleration due to gravity.
Weight (or force of gravity on object)
$F_g=ma=mg$

$W=mg$

Same thing really.
« Last Edit: September 25, 2011, 09:36:57 pm by laseredd »

#### Kiro Kompiro

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##### Re: Newton's Laws and probably more questions
« Reply #18 on: September 27, 2011, 02:00:34 pm »
+1
@procrastinator:  I found wording Newton's Laws this way helpful:

first law: for an object to change its speed or direction, an UNBALANCED force must act on it.  to get the unbalanced force and see if it is more than zero, add all the forces up taking into account their DIRECTIONS.

law 2:  The UNBALANCED force equals the body's mass times its acceleration.  The 'F' in F=ma refers to the UNBALANCED force.  To work this out you add up all the forces acting on the object, taking into account their direction.  I always use positive for forces acting to the right and negative for forces acting to the left.  When talking about vertical motion, I default to positive if the force acts upward, and negative if it acts downward.  If an object is thrown up, gravity acts down so I make down negative because it acts in the opposite direction to the initial movement. But if the object is thrown down, then gravity acts in the same direction so it then is positive.

law 3:every action has an equal an opposite reaction.  Deceptively simple.  It usually refers to the forces on objects that are in contact but not moving RELATIVE to each other, on the contact surfaces. eg you are standing on a concrete floor.  The earth's gravity exerts a force on you downward, yet you do not move through the floor because the floor pushes up on you with an equal but opposite force.  To test this, take a sheet of card board, lay it across two chairs, and stand on it.  What happens?  You fall through the cardboard because the upward force of the cardboard is insufficient to balance your weight.

Eg that uses all three laws:  two square blocks are in contact and sit on a frictionless floor.  Their masses are 2kg and 5 kg.  A push of 21 N is applied to the small block, and the blocks move together.  What is their acceleration?

Ans: unbalanced force is 21 N, total mass is 2+5=7 kg, a=F/m= 3 m/sec sq

What is the force of the little block on the big block?

Ans: the big block accelerates at 3 m/sec-sq.  It must have an unbalanced force of 5x3=15 N.  This is the force the little block exerts on the big block.

What is the force of the big block on the little block?

ans: Since there is no relative movement between the blocks, the force the big block exerts on the little block must also be 15 N in the opposite direction.

Does this make sense?

Draw the forces acting on the little block and add them up:

21N push acts on the side of the little block, -15N acts in the opposite direction from the big block, unbalanced force on the little block is 21-15=6 N, mass is 2 kg so a=F/m=3 m/sec-sq, which is what we had at the start.

Similarly, the unbalanced force on the big block comes from the little block which is 15N, m=5 kg, a=F/m=3 m/sec-sq

#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #19 on: October 03, 2011, 12:57:55 pm »
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It's not an exception. 'g' is the acceleration due to gravity.
Weight (or force of gravity on object)
$F_g=ma=mg$

$W=mg$

Same thing really.

so as long as you know there's only one force in one direction you can use m*a to find it

thanks kiro, it's coming to me now albeit slowly

---

in physics, is it always possible to visualise what the equations mean physically? e.g. squared constants/variables, i can't imagine. is it just an abstract thing that describes the physical thing but you can't imagine the individual terms and what they mean physically?

how is coulomb's law derived?

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##### Re: Newton's Laws and probably more questions
« Reply #20 on: October 03, 2011, 01:35:36 pm »
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in physics, is it always possible to visualise what the equations mean physically? e.g. squared constants/variables, i can't imagine. is it just an abstract thing that describes the physical thing but you can't imagine the individual terms and what they mean physically?
Apparently you can't with really visualise string theory, but with the stuff we do in VCE, yeah definitely.

I'll use the example of Coulomb's Law:
Here's the fancy definition: "The magnitude of the Electrostatics force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of charges and inversely proportional to the square of the distances between them."

I haven't found a derivation of it that's really simple for it yet (didn't really care about derivations in units 1&2 lol). I've seen it derived from Maxwell's equation. http://planetmath.org/encyclopedia/DerivationOfCoulombsLawFromGaussLaw.html. I think there was also a method about something to do with electric fields.

$\int_S {E_n dA = } \frac{q}{{\varepsilon _0 }}$ (copied this from some equation database, don't know if it's correct)

Gauss's Law states (in fancy terms): "The electric flux through any closed surface is proportional to the enclosed electric charge."

I have also seen this even fancier version: ""the divergence of the electric field equals charge density divided by ε_0" That fancier one is just the equation spelled out in words.

Also this fancy definition kind of combines the two: "The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity."

So you can see that equations are really just a fancy way of describing something that sounds even fancier. However, this even fancier thing (the words), really just describes an idea that's fairly straight forward a lot of the time.

Flux is sort of a measure of 'flow' through a surface. What we've been doing in Unit 4 is magnetic flux. A magnetic field can be modeled using magnetic field lines. If you take an area (e.g. a rectangular piece of paper), magnetic flux is the measure of how many magnetic field lines pass through the area. So flux can be considered to be the "amount of field in an area".

So in less fancy words, Gauss's Law states that "The amount of electric field in an area" is proportional to the "electric charge in that area".  That's what I understood of it anyway.

Damn, I was supposed to be talking about Coulomb's Law. Flux is a concept that clicked for me when the teacher demonstrated it using a piece of paper. I assume you know what electric charge is. I guess it still shows my approach though, just start off with the fancy definitions you are given and then boil it down to the simpler terms. I have a feeling I didn't answer your question entirely.

Maybe this is kind of relevant (probably not, I just like xkcd): http://xkcd.com/895/. I like the mouse-over text: "Space-time is like some simple and familiar system which is both intuitively understandable and precisely analogous, and if I were Richard Feynman I'd be able to come up with it."

#### xZero

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##### Re: Newton's Laws and probably more questions
« Reply #21 on: October 03, 2011, 01:54:02 pm »
+2
$\int_S {E_n dA = } \frac{q}{{\varepsilon _0 }}$
I'll continue from here, Coulomb's law is for a point charge yeh? so essentially we can think of it as a sphere, the integral above is the surface integral and if you follow the planetmath link, it converts to spherical co-ordinate and use triple integral and what it finds is the amount of electric field going through the surface of the sphere. After integrating it just rearrange it a bit and you'll get Coulomb's law, what does this imply? It implies that for a sphere with uniformly distributed electrons (a conductor for example) it acts as a point charge. (This stuff is way beyond VCE btw)
« Last Edit: October 03, 2011, 02:31:39 pm by xZero »
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#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #22 on: October 03, 2011, 02:10:21 pm »
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so for example, what is c^2 physically? i'm talking about the equations that can't really be derived intuitively, like finding distance is simply speed * time, it's obvious, but some of the other kinematic equations for example, they have to be derived mathematically and not using intuition

i'm just reading your reply right now, a bit loss, but that's a fine looking equation. it's crazy how they come up with those equations. have you heard of abstrusegoose/

that sucks for me cause it bothers me when i'm not getting the whole picture or learning from first principles

#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #23 on: October 04, 2011, 04:19:43 pm »
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this is probably more a math question but i'll ask anyway. if f is proportional to q1q2 and also proportional to 1/r^2, why does multiplying them both give a common constant?

if you find the constants individually and combine them, i notice i always get f^2 (so far)

#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #24 on: October 24, 2011, 08:43:17 pm »
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Is 10J/sqm/s the same as 10J/ms?  metres-second doesn't make much sense to me

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##### Re: Newton's Laws and probably more questions
« Reply #25 on: October 24, 2011, 09:18:18 pm »
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Is 10J/sqm/s the same as 10J/ms?  metres-second doesn't make much sense to me
square metres is m^2. m^2 and m is not the same as you know.

10 Joules per m^2/s (Energy is changing with acceleration?)
10 Joules per metres per second - (Energy is changing with velocity?)
Can't think of the relevant equation for those units.

Wait "ms" - milliseconds? Energy over time? That's power: $P = \frac{\Delta E}{\Delta t}$

#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #26 on: October 25, 2011, 10:01:40 am »
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ah yes, sorry, i forgot the ^2. 10J per square metre per second is what i'm trying to get at

#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #27 on: November 02, 2011, 06:18:20 pm »
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Why does a pool appear shallower if you look at the bottom from directly above? what does the speed of light have to do with this? If you're looking from directly above, wouldn't there only be a change in speed and no change in angle?

#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #28 on: November 03, 2011, 11:56:09 am »
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also, when they (the heinemann book) says a mechanical wave involves the passing of a vibration through an elastic medium, does that just mean any physical medium? it doesn't explicitly define 'elastic medium'

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##### Re: Newton's Laws and probably more questions
« Reply #29 on: November 03, 2011, 04:41:07 pm »
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I'll have an attempt at trying to answer your questions. I don't think any of my answers here are fully satisfactory, but better than nothing hopefully

also, when they (the heinemann book) says a mechanical wave involves the passing of a vibration through an elastic medium, does that just mean any physical medium? it doesn't explicitly define 'elastic medium'
Elastic medium I think is one that can retain it's own form after a disturbance. I think that sort of makes sense with my understanding of a mechanical wave. The energy can be carried by the wave through the medium, but the medium is not permanently affected. A guitar string vibrating is a mechanical wave - it goes back to it's regular form. Sound waves - the air particles are not permanently changed.

"Net transfer of energy without net transfer of matter" if you want the fancy wording.

I don't ever recall coming across the term elastic medium in this context - but doing some googling, there are a few websites that use the terminology. They all enjoy neglecting the actual definition as well - it might be defined with 'elastic' being something to do with energy conservation or something. I think the definition I gave is more for springs and all that.

Why does a pool appear shallower if you look at the bottom from directly above? what does the speed of light have to do with this? If you're looking from directly above, wouldn't there only be a change in speed and no change in angle?
I have forgotten a lot about this :/ but it's do with refraction. The speed of light stuff comes into when you're talking about speed of light in different mediums (i.e. the refractive index?). I don't think that helped much, it's probably just restating what you already know.

I still don't quite know the answer to that question you had about units either. But http://en.wikipedia.org/wiki/Joule gives a few other definitions. Perhaps doing a bit of arranging with those definitions might get you to what you want to find out. Whether or not those two units you asked about earlier are equal - I don't really know.

edit: wrote out the definition from wikipedia out wrong, so I've cut it out. I made a typo and subsequently all the stuff I figured out didn't work. Typing 10J per square metre per second into google tells you that it's equivalent to kg/s though.

This relates to that other question you had, about derivations and such not making sense to you. My entire answer to that is unsatisfactory (and such I've decided to delete most of it) and that's usually the reason why I'll leave one of your questions unanswered. I can say that perhaps that the for the constant accel. equations, graphical derivations might make more sense to you.

I will also show you this post by paulsterio - which I thought that as well being a very nice explanation, is relevant. If you need to, reading his post before may help make more sense of what he is talking about Re: Vincezor's Unit 4 Physics Questions~ [this is one of the things we cover in Unit 4 btw].

He's showing where the formula $F=nILBsin \theta$ (force created with a current in a wire placed in a magnetic field) came from - and also the right hand slap rule. I think you may be familiar with the right hand slap rule. I remember covering it in Year 9 science. His post should show you another example of how a lot of intuition and logic actually is involved in getting to a conclusion that might seem pretty random. I'm not expecting you to understand the mathematics he's talking about (I don't really understand it that well either, I only currently know the general gist of vector cross product etc.) - it's irrelevant to the argument though - just trying to show you an example of how starting from something simple and then a few steps will get you to the conclusion (even if it does seem odd).
If you were interested though, the formula and the sin(theta) part come from the vector cross product. The length of the wire is a scalar quantity, so it doesn't matter, same as the number of wires, n, but the two vector quantities, the current, I, and the field, B, are multipled using a vector cross product (there are two ways of multiplying vectors, the cross product (which you'll do in Specialist) where you get a scalar, and the cross product (which you'll do in First Year Uni Maths), where you get a vector) - and the vector cross product is defined as the Product of the Magnitudes of the two vectors by the sin of the angle between them in the direction perpendicular to both of them.

So the Vector Cross Product explains the formula F = nIBL sin(theta)  as well as the Right-Hand slap rule - finding the direction of the Force (direction perpendicular (orthagonally) to the Current and Field)

That's the maths behind it
« Last Edit: November 03, 2011, 04:44:02 pm by laseredd »