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* g (acceleration due to gravity) - not the letter "g" ==Acceleration due to gravity==

...ce]]''. The pound-force is the force of 0.45359237 kilogram subject to a [[Acceleration due to gravity|standard gravity]] of 9.80665 m/s², or approximately 4.448 [[New

...e other factors here, and that explains why (near the earth's surface) the acceleration due to gravity (basically, the potential) is a constant 9.8 meters per second squared (or

The [[acceleration due to gravity]] of a body in the vicinity of the earth is approximately 9.8 m s<sup>&minu

*{{pl|Acceleration due to gravity}} (2189)

where ''g'' is the [[Gravitational acceleration|acceleration due to gravity]]. These two forces add to make a resultant at an angle ''φ'' from the ver

== Please look at the last edit of [[Acceleration due to gravity]] == Paul, would you please look at last edit (by Richard Moore) of the [[Acceleration due to gravity]] article. Is it correct? And again, can it be simplified? At the very leas

(here ''h''₀ is the initial height, and ''g'' is the [[acceleration due to gravity]]). Infinitely many corresponding values of ''t'' and ''h'' are embraced by

* [[Acceleration due to gravity]]

9.81 [[joule]] (see [[Acceleration due to gravity]]). Energy used to