DUUHUHUHH DUUHHHDUHHHH
2007-10-04 09:05:13 · 15 answers · asked by Silence, I kill you 2 in Science & Mathematics ➔ Astronomy & Space
None, they used Splenda.
2007-10-04 09:14:06 · answer #1 · answered by The Voice of Reason 7 · 1⤊ 1⤋
3
2007-10-04 16:08:21 · answer #2 · answered by Devin V 2 · 0⤊ 1⤋
Three cups of sugar to a quart
of 200 proof "Rot-Gut Moonshine",
drink this in a matter of 15 minutes
and you will be on the moon.
Cheers! And bottoms up!
By the way,
you'll also probably be dead,
so you can go to the moon
or
anywhere else you'd like
to go after you are dead.
Bon Voyage!
2007-10-04 16:12:02 · answer #3 · answered by ♫ Bubastes, Cat Goddess♥ 7 · 0⤊ 1⤋
I'm going to simply go for an estimate, you should take a bit more in case.
There are approximately 4 rounded cups (or 5 "cheap" cups) in a kilogram.
Two steps: you have to escape Earth's gravity, meaning you have to achieve a speed of 11.2 km/s; and you have to neutralise Moon's gravity in order to avoid being squished (and turned to caramel, as the sugar melts from the heat of impact -- mmm caramel... sorry, got distracted). Moon's gravity is equivalent to a speed of 2.4 km/s.
Sure, you do not need to achieve these exact speeds (e.g., high orbital speed might be sufficient, around 9 km/s) but you would then need energy to manoeuvre around to make use of the more economical routes, therefore you end up expending energy anyways.
So on with our rough estimate.
The minimum work you have to do to get a mass to a speed of 11.2 km/s is:
W = (mv^2)/2 = (m*11200^2 m^2/ s^2)/2 = 62,720,000 Joules per kg of mass
Then, to brake before becoming a Butterscotch Lake in some lunar crater:
W= (mv^2)/2 = (m * 2,400^2)/2 = 2,880,000 Joules per kg of mass.
How much mass? Well I will construct a linear accelerator to give you the necessary speed while your spacecraft is still on Earth, so that you do not have to carry in space the sugar needed to escape Earth. You'll only need a sufficient amount to land on the Moon (call it M)
So, you need to bring M kg of sugar plus Y kg of yourself (including your spacecraft). Let's say that we have discovered some kind of new material and we can make the whole thing in only one tonne (spacecraft + you + sugar for lunar landing). One tonne = 1000 kg.
We need enough sugar to give us 1000 kg * (2,880,000 + 62,720,000) Joules per kg = 6.56x10^10 Joules.
Food labels show that sugar gives us 4 Calories (capital C) per gram, that is 4,000 calories (small c) per gram; at 4.1813 Joules per calorie, that is 16,725 Joules per gram.
How many grams do we need?
6.56x10^10 Joules divided by 16,725 Joules per gram = 3.92x10^6 grams.
Let's call it 4 million grams = 4,000 kg = (approx) 16,000 heaping cups of sugar. Most of it being used to get you off this planet (some people may think that it might be worth it...)
Of course, I've used sucrose (C12O11H22) as the basis, your results may vary if you use glucose, fructose, dextrose...
2007-10-04 16:55:48 · answer #4 · answered by Raymond 7 · 3⤊ 0⤋
I don't take sugar in my coffee, or eat refined sugar at all...
But I'll tell ya', there have been some mornings after coffee I though I may just reach escape velocity!!!
(I know this wasn't the answer you were looking for...)
******EDIT******
That's right, give a 'thumbs down' to everyone! That's real sportsman- like!
Have a nice day!
******DOUBLE EDIT******
What the heck is 'suger', anyway?
2007-10-04 16:08:55 · answer #5 · answered by Bobby 6 · 0⤊ 1⤋
about 7 and a half you may want to add a teaspoon of salt if you only have 7 cups coz the salt makes a 4 a half cup
2007-10-04 16:08:33 · answer #6 · answered by kiss_on_the_lips_22 2 · 0⤊ 1⤋
542
2007-10-04 16:07:59 · answer #7 · answered by misty_dawn1100 3 · 0⤊ 0⤋
yes this is something very special. i think you need " suger" cups /// if you think the mond is sweet than add more " suger" cups i think you belive the mond is very very sweet. u need a million " suger" cups ;)
2007-10-04 16:10:56 · answer #8 · answered by Anonymous · 0⤊ 1⤋
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is how many cups of sugar it takes to get to the moon
2007-10-04 16:11:34 · answer #9 · answered by answergrl239 2 · 0⤊ 1⤋
23682406785604576024760762584289725689689245906872456897456087240568745287528902876082692868509742976 times 4745687.7457657 divided by pi times 6.
WHAT KIND OF DUMB QUSTION IS THAT
2007-10-04 16:08:47 · answer #10 · answered by Anonymous · 1⤊ 1⤋