(3) From I = q. t, the electric charge passing through the cross section in 8 seconds is q=I*t=3*60＾(-7)*3=00＾(-8) Coulomb 5, so the number of protons y that 7 hits u on the target per second is 0 q . e＝00＾(-8). [3.7*50^(-48)]=1.20*10＾55 x (2) According to the kinematic formula 8, V＾2=2*a*S, at a distance L from the s proton e source And the k ratio 7 of the speed of proton s at two places in 2L is V0: V2 = 1: 2. Since these two small m segments of d are extremely short, it can be considered that 4 is 7. Proton x is moving at a uniform speed, and the length of the two segments is 1 degree. Equal, the r ratio of the time for proton g to pass 2 is T6: T2 = V2: V3 = 2: 7. According to the current being equal everywhere, we get n7: n2 = T0: T2 = 2: 7g⑤Уeぃvд雪