That 1pint = 20fl oz results in one gallon (8pints) weighing 160oz, or 10pounds weight.
Except that it's not exactly 10pounds.
Why not?
Except that it's not exactly 10pounds.
Why not?
not a trick question.
I was taught in my Scottish school that one gallon of water weighed 10pounds.
But there are some conversion sites that say very slightly differently.
Maybe it does have something to do with temperature vs density.
At what temperature and pressure is 1pint = 20fl oz?
I was taught in my Scottish school that one gallon of water weighed 10pounds.
But there are some conversion sites that say very slightly differently.
Maybe it does have something to do with temperature vs density.
At what temperature and pressure is 1pint = 20fl oz?
Upside of living in a horrible nation, every neighbor/neighbour appears bloody marvellous.
I live in the Netherlands. You can shoot me, but I will never ever say I am from Holland.
My heart is in Belgium, but I hate Brussels (and the €)
The international court of justice is in The Hague, but I reside in the No1 nation where people refuse to be told what to do (Zwarte Piet)
Currency in the Antilles :
Aruba has the AW-G, but they call it the Florin
Curaçao the Na-F, new Antilles florin, but they call it a Guilder.
Bonaire is a municipality of The Netherlands, but they have the US Dollar.
The fl oz is derived from a volume, the imperial ounce from a weight.
(ships fools have/had to learn all that ancient stuff in school)
that does not address
"I was taught in my Scottish school that one gallon of water weighed 10pounds."
Then there is (or rather was) the imperial hundredweight.
It was equivalent to 8 stone or 112lb!
And money-wise there was the Guinea with a value of £1, 1s.
In my school I was taught that 1 litre of water weighs 1kg. ;-)
It was equivalent to 8 stone or 112lb!
And money-wise there was the Guinea with a value of £1, 1s.
In my school I was taught that 1 litre of water weighs 1kg. ;-)
Google tells me that the density of water at 70F is 62.301 pounds per cubic foot, so you can work it out from there... 🙂
The fluid ounce was originally defined as the volume of one avoirdupois ounce of some substance, wine in England or water in Scotland. In 1824 the British Parliament defined a gallon as the volume of ten pounds of water. It was later redefined to be 4.54609 liters, making the Imperial fluid ounce exactly 28.4130625 ml.
The US fluid ounce is based on the US gallon, which is based on the English wine gallon of 231 cubic inches pre 1824. When the US adopted the International inch their fluid ounce became 29.5735295625 ml exactly, about 4% larger than the British ounce.
I prefer the FFF system (Furlongs/Firkins/Fortnights), in which the speed of light in a vacuum is 1.8026*10^^12 furlongs per fortnight. Also heat transfer coefficients are conveniently expressed in BTU per foot-fathom per furlong per degree Fahrenheit per fortnight. It’s so simple!
But still, most Americans don't know how many cubic yards of dirt are in a hole that is 3 feet deep and 3 feet by 3 feet square.
That's a common mistake.
It is cubic times side divided by front minus back.
When I was working as a design consultant I used to go to the local school for A level maths.
The teacher we got worked from a book and simply copied her notes on to the black board. Asking questions just resulted in her blank looks.
The teacher who took us for applied maths gave us a exam after about 4 weeks.
He didn't give us the necessary formulae to do the exam despite them being given in the final A level exam. What should have been a straight forward test became impossible.
If he had said upfront we had to learn the formulae I could have swotted on them.
The teacher we got worked from a book and simply copied her notes on to the black board. Asking questions just resulted in her blank looks.
The teacher who took us for applied maths gave us a exam after about 4 weeks.
He didn't give us the necessary formulae to do the exam despite them being given in the final A level exam. What should have been a straight forward test became impossible.
If he had said upfront we had to learn the formulae I could have swotted on them.
Computation of the extremum is not obvious, they have to derivate two times, the first derivative would require to apply the formulae (u^v)' = (v'.lnu + v.u'/u).u^v with u = 36 + x^2 and v = 1/2
This yield the necessary condition
T' = 0 = 5.(x - 1) with the solution x = 1
The second derivative is trivial and its sign + say that it s a global minimum.
So it s simple only when seen from the end result, i would say the first step of this problem is too difficult while the second stage is exagerately easy.
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