How many foot pound in 1 gigaelectron volt?
The answer is 1.1817053329112E-10.
We assume you are converting between foot-pound force and gigaelectronvolt.
You can view more details on each measurement unit:
foot pound or gigaelectron volt
The SI derived unit for energy is the joule.
1 joule is equal to 0.73756214927727 foot pound, or 6241506479.9632 gigaelectron volt.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between foot pounds and gigaelectronvolts.
Type in your own numbers in the form to convert the units!
1 foot pound to gigaelectron volt = 8462346510.1609 gigaelectron volt
2 foot pound to gigaelectron volt = 16924693020.322 gigaelectron volt
3 foot pound to gigaelectron volt = 25387039530.483 gigaelectron volt
4 foot pound to gigaelectron volt = 33849386040.644 gigaelectron volt
5 foot pound to gigaelectron volt = 42311732550.804 gigaelectron volt
6 foot pound to gigaelectron volt = 50774079060.965 gigaelectron volt
7 foot pound to gigaelectron volt = 59236425571.126 gigaelectron volt
8 foot pound to gigaelectron volt = 67698772081.287 gigaelectron volt
9 foot pound to gigaelectron volt = 76161118591.448 gigaelectron volt
10 foot pound to gigaelectron volt = 84623465101.609 gigaelectron volt
You can do the reverse unit conversion from gigaelectron volt to foot pound, or enter any two units below:
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foot pound to gigajoule
foot pound to joule
foot pound to erg
foot pound to exawatt-hour
foot pound to megalerg
In physics, a foot-pound force (symbol ft·lbf or ft·lbf), usually abbreviated to foot-pound, is an Imperial and U.S. customary unit of mechanical work, or energy, although in scientific fields one commonly uses the equivalent metric unit of the joule (J). There are approximately 1.356 J/ft·lbf, or exactly 1.355 817 948 331 400 4 J/ft·lbf.
The SI prefix "giga" represents a factor of 109, or in exponential notation, 1E9.
So 1 gigaelectronvolt = 109 .
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