A2 kg stone moves with a speed of 1 m/s. a second 2 kg stone is moving twice as fast. compare their kinetic energies.
a. the second stone has 0.5 times as much kinetic energy.
b. the second stone has 2 times as much kinetic energy.
c. the second stone has 0.25 times as much kinetic energy.
d. the second stone has 4 times as much kinetic energy.

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.