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#### xZero

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##### Re: Newton's Laws and probably more questions
« Reply #30 on: November 03, 2011, 04:50:46 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?

I think this diagram answers your question, the observed depth Y is less than the actual depth X but this doesnt happen when you're looking straight down the pool from above.
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#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #31 on: November 03, 2011, 04:57:29 pm »
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yep, thanks for putting up with these questions lol

ah that makes a bit more sense. so, an ocean wave - that would be an elastic medium? i think in the book it said none of the molecules that crash on the beach would have been at the source of the wave so there's no transfer of matter

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About the units, that' was a question from the alternative energy topic (something like that). I don't remember it exactly but basically there's a solar panel and you had to work out the rate of energy absorbed per sqm…

So 10J per square metre per second is 10J/m^2/s and when you simplify it mathematically , you get 10J/m^2 s which reads 10J per square metre second which don't make sense. I guess basically what I wanted to know if it's still considered J per metre squared per second or does metre squared second mean the same thing?

@xZero - what about when you're looking from directly above? i can't find anything in my textbook about that but in the handouts, it says  a pool appears shallower when looking from directly above (no refraction)

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##### Re: Newton's Laws and probably more questions
« Reply #32 on: November 03, 2011, 05:09:26 pm »
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Quote
Yeah, I found that as well.

Quote
So 10J per square metre per second is 10J/m^2/s and when you simplify it mathematically , you get 10J/m^2 s
$m^2/s = m^2*s^{-1}$ not $m^2*s$

Quote
@xZero - what about when you're looking from directly above? i can't find anything in my textbook about that but in the handouts, it says  a pool appears shallower when looking from directly above (no refraction)
Snell's law was it? If you plug in the angle 0 for incident angle, what is the refracted angle? It's 0 is it not. (I hope I used the correct formula).

http://en.wikipedia.org/wiki/Snell's_law

I used the one off Wikipedia, if I remember correctly, angle 1 is incident, angle 2 is refracted. n2/n1 = 1.33 for air into water (rounded off air to 1). If I did mix up my angles, either way, they will both be zero.

#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #33 on: November 04, 2011, 12:05:25 pm »
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So you treat the $m^2/s$ as a fraction and not $m^2$ and and $10J$ together? This is what I did:

$\frac{10J}{m^2} \times\frac{1}{s}$

Would everything together be

$10J m^-^2s$

?

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##### Re: Newton's Laws and probably more questions
« Reply #34 on: November 04, 2011, 04:39:12 pm »
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So you treat the $m^2/s$ as a fraction and not $m^2$ and and $10J$ together? This is what I did:

$\frac{10J}{m^2} \times\frac{1}{s}$

Would everything together be

$10J m^-^2s$

?
Oh yes, you're (almost) correct, I see my mistake.

I assumed: $\frac{J}{(\frac{m^2}{s})}$, but since it's Rate of "energy absorbed per sqm", it would be:
$\frac{(\frac{J}{m^2})}{s}$ which simplifies to: $\frac{J}{m^2 s}$ (this is what you had)
This further simplifies to:
${J m^{-2} s^{-1}}$ (Joules per sq. metre per second)

Note that $\frac{1}{s} = s^{-1}$ (I'm guessing you simply forgot the type up the -1 at the end in that post)

A bit of beating around the bush, but I think I can confirm that $\frac{\frac{J}{m^2}}{s} = \frac{J}{m^2 s}$

#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #35 on: November 04, 2011, 05:38:19 pm »
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That was based on your last post :p

so does that read 'joules per metre squared per second', or 'joules per metre squared second' (what didn't make sense to me, unless it means the same thing...i want to make sure i got this lol)

#### xZero

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##### Re: Newton's Laws and probably more questions
« Reply #36 on: November 04, 2011, 05:53:10 pm »
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they are different, joules per metre squared per second is $Js/m^2$ and joules per metre squared second is $J/m^2s$
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##### Re: Newton's Laws and probably more questions
« Reply #37 on: November 04, 2011, 05:57:00 pm »
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they are different, joules per metre squared per second is $Js/m^2$ and joules per metre squared second is $J/m^2s$
Hmm, well that makes a lot more sense.

I am not entirely sure what I've been doing wrong though (somewhere in my mathematics probably, I'm guessing probably with my initial assumptions).

#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #38 on: November 04, 2011, 07:10:52 pm »
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they are different, joules per metre squared per second is $Js/m^2$ and joules per metre squared second is $J/m^2s$

So $\frac{J}{1}$ over $\frac{m^2}{s}$...how did you know?

#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #39 on: November 10, 2011, 05:32:57 pm »
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How come it isn't J/m^2 and s/1?

And I just noticed you mention kg/s^3 laseredd, that makes no sense to me haha (i think i know how the calculator got that though)

If a jet fires a missile, it pushes back against it right (if not released first), but what if the missile was fired at a low acceleration so it gradually increased it's velocity to match the aircraft's before being released, would this act as a boost for the jet? There would be no push back, yes?

EDIT: How would one go abut deriving the work formula?
« Last Edit: November 10, 2011, 05:36:39 pm by #1procrastinator »

#### xZero

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##### Re: Newton's Laws and probably more questions
« Reply #40 on: November 10, 2011, 05:43:21 pm »
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they are different, joules per metre squared per second is $Js/m^2$ and joules per metre squared second is $J/m^2s$

So $\frac{J}{1}$ over $\frac{m^2}{s}$...how did you know?
replace per with /
joules per metre squared per second = joules/m^2/s = Js/m^2

joules per metre squared second = j/m^2s

If a jet fires a missile, it pushes back against it right (if not released first), but what if the missile was fired at a low acceleration so it gradually increased it's velocity to match the aircraft's before being released, would this act as a boost for the jet? There would be no push back, yes?
jet usually release missile before it accelerates, letting it accelerate before release is just wasting fuel, prob run out of juice before you can fire it
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#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #41 on: November 10, 2011, 05:48:00 pm »
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joules/m^2/s , i don't get why m^2 and second are together in a fraction and not joules and m^2

hypothetically :p

it's the acceleration that pushes it back though, right? if it was increasing velocity slowly, there would be no (or very little) push back?

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