An proton-antiproton pair is produced by a 2.20 × 10 3 mev photon. what is the kinetic energy of the antiproton if the kinetic energy of the proton is 242.85 mev? use the following joules-to-electron-volts conversion 1ev = 1.602 × 10-19 j. the rest mass of a proton is 1.67 × 10 − 27 kg.

Now that we have a feel for the state of the circuit in its steady state, let us obtain the expression for the current in the circuit as a function of time. note that we can use the loop rule (going around counterclockwise): e−vr−vl=0. note as well that vr=ir and vl=ldidt. using these equations, we can get, after some rearranging of the variables and making the subsitution x=er−i, dxx=−rldt. integrating both sides of this equation yields x=x0e−rt/l. use this last expression to obtain an expression for i(t). remember that x=er−i and that i0=i(0)=0. express your answer in terms of e, r, and l. you may or may not need all these variables. use the notation exp(x) for ex.