Convert foot-pound to megaelectron-volt
Please provide values below to convert foot-pound [ft*lbf] to megaelectron-volt [MeV], or Convert megaelectron-volt to foot-pound.
How to Convert Foot-Pound to Megaelectron-Volt
1 ft*lbf = 8462350038241.79 MeV
Example: convert 15 ft*lbf to MeV:
15 ft*lbf = 15 × 8462350038241.79 MeV = 126935250573627 MeV
Foot-Pound to Megaelectron-Volt Conversion Table
| foot-pound | megaelectron-volt |
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Foot-Pound
A foot-pound is a unit of energy or work equal to the work done when a force of one pound-force is exerted over a distance of one foot.
History/Origin
The foot-pound has been used primarily in the United States and the United Kingdom as a customary unit of energy, originating from imperial and US customary measurement systems, before the widespread adoption of the SI system.
Current Use
Today, the foot-pound is still used in certain fields such as engineering, automotive, and aerospace industries in the US for expressing torque and energy, although the SI unit joule is more common internationally.
Megaelectron-Volt
A megaelectron-volt (MeV) is a unit of energy equal to one million electron-volts, commonly used in nuclear and particle physics to express the energy of particles.
History/Origin
The megaelectron-volt was introduced as a convenient unit for expressing high-energy particles in physics, particularly after the development of particle accelerators in the mid-20th century, replacing the electron-volt for larger energy scales.
Current Use
Today, the MeV is widely used in nuclear physics, particle physics, and astrophysics to quantify the energy of subatomic particles, nuclear reactions, and cosmic phenomena.