How many foot pounds in 1 megaelectron volt? The answer is 1.1817053329112E-13. We assume you are converting between foot-pound force and megaelectronvolt. You can view more details on each measurement unit: foot pounds or megaelectron volt The SI derived unit for energy is the joule. 1 joule is equal to 0.73756214927727 foot pounds, or 6241506479963.2 megaelectron volt. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between foot pounds and megaelectronvolts. Type in your own numbers in the form to convert the units!
1 foot pounds to megaelectron volt = 8462346510160.9 megaelectron volt
2 foot pounds to megaelectron volt = 16924693020322 megaelectron volt
3 foot pounds to megaelectron volt = 25387039530483 megaelectron volt
4 foot pounds to megaelectron volt = 33849386040644 megaelectron volt
5 foot pounds to megaelectron volt = 42311732550804 megaelectron volt
6 foot pounds to megaelectron volt = 50774079060965 megaelectron volt
7 foot pounds to megaelectron volt = 59236425571126 megaelectron volt
8 foot pounds to megaelectron volt = 67698772081287 megaelectron volt
9 foot pounds to megaelectron volt = 76161118591448 megaelectron volt
10 foot pounds to megaelectron volt = 84623465101609 megaelectron volt
You can do the reverse unit conversion from megaelectron volt to foot pounds, or enter any two units below:
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 "mega" represents a factor of 106, or in exponential notation, 1E6.
So 1 megaelectronvolt = 106 .
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