If a proton with mass 1.67 × 10^-27kg moves in an accelerator such that its total energy is three times its rest energy, what is its speed? (c = 3.00 × 108m/s)
Added by Raul W.
Step 1
So, the rest energy of the proton is E = (1.67 × 10^-27kg) * (3.00 × 10^8m/s)^2 = 1.50 × 10^-10 Joules. Since the total energy is three times the rest energy, the total energy of the proton is 3 * 1.50 × 10^-10 Joules = 4.50 × 10^-10 Joules. The total energy Show more…
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A proton (rest mass $1.67 \times 10^{ \times 27} \mathrm{kg} )$ has total energy that is 4.00 times its rest energy. What are (a) the kinetic energy of the proton; (b) the magnitude of the momentum of the proton; (c) the speed of the proton?
\bullet A proton (rest mass $1.67 \times 10^{-27} \mathrm{kg}$ ) has total energy that is 4.00 times its rest energy. What are (a) the kinetic energy of the proton; (b) the magnitude of the momentum of the proton; (c) the speed of the proton?
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