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Question
The potential energy of a 0.20-kg particle moving along the xaxis is given by
U(x) =(8.0 J/m2)x2 + (2.0 J/m4)x4.
When the particle is at x= 1.0 m it is traveling in the positive xdirection with a speed of 5.0 m/s. It next stops momentarily to turn around at x=
A) 0 m
B) -1.1 m
C) 1.1 m
D) -2.3 m
E) 2.3 m
Answer
This answer is hidden. It contains 1 characters.
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