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Review Test Submission: Online Test 1 Question 1 PLANE 1 Acceleration is the unknown V = 48.5m/s; U = 0 m/s But v2 = u2 + 2.a.S 48.52 = 0 + 2 (a) (644) 48.52 = 1288a a= 1.826m2/s. Plane 2 has the same acceleration as plane 1 Therefore; a = 1.826m2/s. But v2 = u2 + 2.a.S 22.72= 0 + 2 (1.826) (S) 22.72= 3.652S Distance = 141.08 m Question 2 ( need more information to identify the answer) Question 3 Solution; Acceleration of the Free fall; ∑𝐹 = 𝑚𝑎 𝐹𝑔 = 𝑚𝑔 20𝑁 = 9.8𝑚/𝑠 2 𝑚 𝑚= 20 = 2.04𝑘𝑔 9.8 As such, acceleration is calculated as; 𝑎 = 60𝑁 − 8𝑁/(2.04𝑘𝑔) Acceleration = 25.50𝑚/𝑠 2 Question 4; (Done) Question 5 Kindly attach the diagram because that’s where the values are. Question 6 Tangential speed Vt = ω r. Vt = (2.8 rad/s) 2.0 m Tangential speed = 5.6m/s. Centripetal acceleration Mass = 36.0 kg. Radius = 2.0 m  = 2.8 rad/s ac= R2 = 2.0 m (2.8 rad/s) 2 = 15.68m/s2. Question 7 A. The magnitude of the displacement of an object is always less than or equal to the distance traveled by the object. E. An object cannot move in a circle with a constant speed unless there is a net force acting on it. I. If two external forces that are both equal in magnitude and opposite in direction act on the same object, the two force can never be an action-reaction force pair. J. A car driving down a long straight incline at constant speed has zero acceleration Question 8 Calculating the speed; 𝑣 = √(15)2 + 2(9.8) ...
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