A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 4.50 s, it is at
point (5.80 m, 5.60 m) with velocity (3.50 m/s) j and acceleration in the positive x direction. At time t2 = 11.7 s, it has velocity (–3.50 m/s) i and acceleration in the positive y direction. What are the (a) x and (b) y coordinates of the center of the circular path? Assume at both times that the particle is on the same orbit.Expert Answer
1 answer:
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Answer:
3.52176 x 10^-10 N
Explanation:
Fg = 3.52176 x 10^-10 Newton
