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The acceleration of a particle moving only on a horizontal plane is given by a= 3ti +4tj, where a is in meters per secondsquared and t is in seconds. At t = 0s, the position vector r= (20.0 m)i + (40.0 m)j locates the particle,
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The acceleration of a particle moving only on a horizontal plane is given by a= 3ti +4tj, where a is in meters per secondsquared and t is in seconds. At t = 0s, the position vector r= (20.0 m)i + (40.0 m)j locates the particle,
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A particle leaves the origin with an initial velocity = (2.35 m/s) and moves with constant acceleration = (1.90 m/s2) + (2.60 m/s2)>. (a) How far does the particle move in the x direction before turning around? m (b) What is the
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A 1.49kg particle initially at rest and at the origin of an xy coordinate system is subjected to a timedependent force of F(t) = (4.00ti − 8.00j) N with t in seconds. (a) At what time t will the particle's speed be 14.0 m/s?
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A 1.49kg particle initially at rest and at the origin of an xy coordinate system is subjected to a timedependent force of F(t) = (4.00ti − 8.00j) N with t in seconds. (a) At what time t will the particle's speed be 14.0 m/s?
asked by joy on February 8, 2018 
Physics
A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 6.0 m, y = 3.0 m, and has velocity v = 4.0 m/s + 1.0 m/s . The acceleration is given by the vector a = 4.0 m/s2 + 0 m/s2 . (a) Find
asked by Anonymous on July 7, 2015 
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can somebody shows me how to get to the answer key a)=2.51s, b)=19.9m, and c)=11.8i16.1j A 1.57kg particle initially at rest and at the origin of an xy coordinate system is subjected to a timedependent force of F(t) =
asked by joy on February 8, 2018