Three forces acting on an object are given by F = (-1.61 +6.201) N, F2 =...

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Three forces acting on an object are given by F = (-1.61 +6.201) N, F2 = (5.2011.551) N, and F3 = (-471) N. The object experiences an acceleration of magnitude 3.60 m/s. (a) What is the direction of the acceleration? -6.14 X Remember that tan(+180) = tan(e), so you must check which quadrant the vector lies in before accepting your calculator answer for the inverse tangent. (counterclockwise from the +x-axis) (b) What is the mass of the object? kg (c) If the object is initially at rest, what is its speed after 20.0 s? m/s (d) What are the velocity components of the object after 20.0 s? (Let the velocity be denoted by v.) V = + i) m/s Need Help? Read It Watch It Submit Answer Three forces acting on an object are given by F = (-1.61 +6.201) N, F2 = (5.2011.551) N, and F3 = (-471) N. The object experiences an acceleration of magnitude 3.60 m/s. (a) What is the direction of the acceleration? -6.14 X Remember that tan(+180) = tan(e), so you must check which quadrant the vector lies in before accepting your calculator answer for the inverse tangent. (counterclockwise from the +x-axis) (b) What is the mass of the object? kg (c) If the object is initially at rest, what is its speed after 20.0 s? m/s (d) What are the velocity components of the object after 20.0 s? (Let the velocity be denoted by v.) V = + i) m/s Need Help? Read It Watch It Submit Answer