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Potential energy of suspended mass expressed in kWh

CircutScoper

Mar 29, 2022
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Interesting. We've just tossed our kg, so I thought that the new standard for kg had been accepted.


Well, maybe to do with trade between nations. But with that aim in mind, there is significant science in the definitions to make as many standards as possible based solely on fundamental constants that can be measured or defined; getting away from that rather arbitrary block of Pt and such.

The science doesn't care a whit whether g = 9.8m/s^2 or 32.2ft/s^2, the foot can be defined in terms of the wavelength of krypton emitted light just as easily as the meter can, absolute zero is 0.0 degrees in both Kelvin and Rankin, and so on.

G = 3.19x10^13 cubits/fortnight^2 would work just as well, too.
 
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davenn

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I think your opinion makes sense but the physics' department has a limitless vault of details and bs with which to dazzle and baffle you. Please read post #49.


sadly a lot of the info in that post is incorrect
ohhhh and by the way, it's NOT my opinion, it's physics fact
 

roughshawd

Jul 13, 2020
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I think the lesson here is superfluouse joules! That being said, only a gravity in the force can keep the ship stranded here, and when they release that piece of aluminum sheet, be far from the direct path of its travel....please.
 

HANKMARS

Jul 28, 2019
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Geee! Meanwhile, 8.17kWh is just enough to raise 1000 tonnes (i.e., 1000 x 1000kg = 1Gg) 3m.[/QUOT
8.17 kWh is needed to raise a single tonne 3 meters, according to my calculations. Your calcs regarding raising a Gg with 8.17 kWh is off by a factor of 1000.
 

HANKMARS

Jul 28, 2019
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That is nonsense. 500 g are 500 g. In SI units. Or 0.5 kg.

Again nonsense and not related to either SI or imperial units.
1 M is the prefix for Mega which means 10^6.
So 1 Mg = 1000 kg = 1 000 000 g

Just because 1 kg is the base unit doesn't mean that the prefixes like k, M, m etc. lose or change their meaning. 1 kg = 1000 g because k ist the prefix for kilo meaning 1000.
Absolutely correct. So, when using SI units, the user obeys the SI convention and enters 1000 kg as 1000, and derives their result, never looking back, and looses 3 decimal places. Now if the original formula reads U = mgh and explicitly and in no uncertain terms states that m is to be expressed in kg, not to meet calculation convention, but rather to meet physical criteria, then I am at error, and yes Virginia, a Mg mass can be raised 3 m ( under the influence of Earth's gravity ) with the energy stored in 3 AA cells. So my question is currently, " In the explanation of the formula U = mgh, when the explainer states that mass is to be in kilograms, is that conversion made to satisfy a method of math or is it to adjust things to qualify the value into the real physical world?" Personally, I think that m was meant to be in grams, and the instructor, say working in the SI system, mentions that 1,000,000 grams is to be expressed in kg, ( 1000 kg ) which is further botched up by the fact that in the SI system assumes that mass is expressed in kg, therefore entering the value as 1000 g, and instructor leaves his explanation at that point. Leaving the student twisting in the wind. I agree that convention needs to be set in order that values can be used worldwide, but I contend that most people do not have a grasp upon the true affect of formula manipulation. So I need to know the exact intent of the formula, U = mgh. Further, if the quip, "expressed in kg," is true in the formula E = mc², I will need to mine a thousand times more uranium for the reactor in my basement !
 

HANKMARS

Jul 28, 2019
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I think the lesson here is superfluouse joules! That being said, only a gravity in the force can keep the ship stranded here, and when they release that piece of aluminum sheet, be far from the direct path of its travel....please.
Let's use the corresponding formula and/or just let the sheet fly and take physical readings.
 

Harald Kapp

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enters 1000 kg as 1000
Regardless of the system (SI, imperial, ???) you use, you should always include the units in your calculations (of course in the equations on paper or on the computer, the calculator will usually not be able to handle units). Keep the units throughout the calculations. This will alllow a final sanity check whether the resulting unit makes sense. If the resulting unit doesn't make sense, you can be sure to have made an error.

Example:
P = V × I (power equals voltage times current, an easy one)
Assume V = 3 V, I = 100 mA
When you simply calculate P = 3 × 100 = 300 you may easily come to the conclusion that power is 300 W - which is obviously wrong and can easily be spotted in this simple case but may be hard to spot in other cases.
When you use P = 3 V × 100 mA you get 300 V × mA = 300 mW which is correct.

Remember the issues with NASA's collaboration with ESA? NASA engineers using imperial units and ESA engineers using SI units? This is no problem at all as long as you keep th eunits within in your calculation because then you will easily see that you need to convert between units at some point or other because of a mismatch of units when they meet within the calculation. If, however, you simply use the numbers without prior correction of units, ylou're in deep mud.
 

HANKMARS

Jul 28, 2019
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Regardless of the system (SI, imperial, ???) you use, you should always include the units in your calculations (of course in the equations on paper or on the computer, the calculator will usually not be able to handle units). Keep the units throughout the calculations. This will alllow a final sanity check whether the resulting unit makes sense. If the resulting unit doesn't make sense, you can be sure to have made an error.

Example:
P = V × I (power equals voltage times current, an easy one)
Assume V = 3 V, I = 100 mA
When you simply calculate P = 3 × 100 = 300 you may easily come to the conclusion that power is 300 W - which is obviously wrong and can easily be spotted in this simple case but may be hard to spot in other cases.
When you use P = 3 V × 100 mA you get 300 V × mA = 300 mW which is correct.

Remember the issues with NASA's collaboration with ESA? NASA engineers using imperial units and ESA engineers using SI units? This is no problem at all as long as you keep th eunits within in your calculation because then you will easily see that you need to convert between units at some point or other because of a mismatch of units when they meet within the calculation. If, however, you simply use the numbers without prior correction of units, ylou're in deep mud.
Indeed. I try to maintain a pair of swamp buggies for just such muddy situations. I have a friend who believes it could be helpful if the term billion was stated as 1000 million. He believes the non-math folks may have a better comprehension of the actual value. A trillion termed as "a million million." Personally, when dollar amounts creep into a mathematical range that is typically used in respect to the cosmos, I can't help but to think things are skewed.
 

CircutScoper

Mar 29, 2022
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Indeed. I try to maintain a pair of swamp buggies for just such muddy situations. I have a friend who believes it could be helpful if the term billion was stated as 1000 million. He believes the non-math folks may have a better comprehension of the actual value. A trillion termed as "a million million." Personally, when dollar amounts creep into a mathematical range that is typically used in respect to the cosmos, I can't help but to think things are skewed.

I agree. But you misspelled "screwed."
 
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