阿摩線上測驗
4. The potential energy of a 2.0-kg particle moving along the x axis is given by U(x)=5.0J/m²)x4 J.
When the particle is at x = 1.0 m it is traveling in the positive x direction with a speed of 5.0 m/s. It next stops momentarily to turn around at about x =
(A) 0
(B) 1.56 m
(C) -1.56 m
(D) 3.02 m
(E) -3.02 m.
When the particle is at x = 1.0 m it is traveling in the positive x direction with a speed of 5.0 m/s. It next stops momentarily to turn around at about x =
(A) 0
(B) 1.56 m
(C) -1.56 m
(D) 3.02 m
(E) -3.02 m.
答案：登入後查看
統計： A(0), B(1), C(0), D(0), E(0) #3332063
統計： A(0), B(1), C(0), D(0), E(0) #3332063
