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Orchid XP v8 wrote:
>
> Still, you'd think it would be possible to somehow take plants living
> today and transform their tissues into some kind of highly flammable
> hydrocardon with reasonably efficiency. This would seem the obvious
> sustainable solution.
>
They do something of the sort - Biodiesel
But that's what politicians are blaming on the food shortages and price
increases. Go figure....
Tom
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> They do something of the sort - Biodiesel
> But that's what politicians are blaming on the food shortages and price
> increases. Go figure....
And for screwing up engines.
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How about this classical physics problem:
You drop a rock into a water well. Then you hear the splash. Exactly
10 seconds pass between you letting go of the rock and when you hear the
splash. How deep is the well?
We can assume that air friction is negligible, that g = 9.81 m/s^2,
and that the speed of sound is 340 m/s.
Don't just give the answer, also write your whole reasoning.
--
- Warp
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Warp wrote:
> How about this classical physics problem:
>
> You drop a rock into a water well. Then you hear the splash. Exactly
> 10 seconds pass between you letting go of the rock and when you hear the
> splash. How deep is the well?
>
> We can assume that air friction is negligible, that g = 9.81 m/s^2,
> and that the speed of sound is 340 m/s.
>
> Don't just give the answer, also write your whole reasoning.
Hmm, takes me back to school.
10 = t(sound) + t(rockfall)
t(sound) = x/340 where x is the well depth
x = 1/2 * g * t(rockfall)^2
so
t(rockfall) = sqrt(2x/g)
substituting back in,
10 = x/340 + sqrt(2x/g)
sub y = sqrt(x), rearrange
(1/340)*y^2 + sqrt(2/g)*y - 10 = 0
solve quadratic for positive y, square y to get x
I get x = 393.25m, sounds about right but my arithmetic might be wonky.
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Warp wrote:
> How about this classical physics problem:
>
> You drop a rock into a water well. Then you hear the splash. Exactly
> 10 seconds pass between you letting go of the rock and when you hear the
> splash. How deep is the well?
>
> We can assume that air friction is negligible, that g = 9.81 m/s^2,
> and that the speed of sound is 340 m/s.
>
> Don't just give the answer, also write your whole reasoning.
Hmm, the object has an initial velocity v0 = 0 m/s.
Assuming zero air resistence, the object as infinite terminal velocity
and accelerates uniformly at 9.81 m/s^2. This yields
vt = 9.81 t
The integral of this would be the object's position:
pt = 9.81/2 t^2 = 4.905 t^2
We hear the splash at t=10. Assuming that sound is initiated at time t=K
and it propogates towards the observer at 340 m/s, we have
K = 10 - d/340
where d is the distance to the bottom of the well.
So, we know that d must obey
K = 10 - d/340 and 4.905 K^2 = d
Rearranging, we have
K - 10 = -d/340
340 (K - 10) = -d
340 K - 3,400 = -d
-340 K + 3,400 = d
So d = -340 K + 3,400 = 4.905 K^2.
At this point, I have insufficient knowledge of algebra to deduce a
solution, although obviously one exists.
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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> So d = -340 K + 3,400 = 4.905 K^2.
>
> At this point, I have insufficient knowledge of algebra to deduce a
> solution, although obviously one exists.
You should learn how to calculate the solution to equations like this, they
come up everywhere:
http://en.wikipedia.org/wiki/Quadratic_equation
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>> So d = -340 K + 3,400 = 4.905 K^2.
>>
>> At this point, I have insufficient knowledge of algebra to deduce a
>> solution, although obviously one exists.
>
> You should learn how to calculate the solution to equations like this,
> they come up everywhere:
>
> http://en.wikipedia.org/wiki/Quadratic_equation
Solving a single quadratic equation is trivial. Solving a pair of linear
equations is trivial. But I have no idea how to solve a pair of
equations where one is linear and the other isn't.
[Actually, I can still remember sitting down and deriving the quadratic
formula from first principles. It's quite instructive! Whoever it was
that figured out that A*B - 4*(A+B) = A-B is some kind of genius if you
ask me... The next day I tried to do the same with a cubic, but... uh...
failed.]
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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>>> So d = -340 K + 3,400 = 4.905 K^2.
>>>
>>> At this point, I have insufficient knowledge of algebra to deduce a
>>> solution, although obviously one exists.
...
> Solving a single quadratic equation is trivial.
What do you call that equation you got at the top there?
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scott wrote:
>>>> So d = -340 K + 3,400 = 4.905 K^2.
>>>>
>>>> At this point, I have insufficient knowledge of algebra to deduce a
>>>> solution, although obviously one exists.
>
> ...
>
>> Solving a single quadratic equation is trivial.
>
> What do you call that equation you got at the top there?
A pair of equations welded together?
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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>>>>> So d = -340 K + 3,400 = 4.905 K^2.
> A pair of equations welded together?
A = B = C implies that B = C, no?
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