Part CThe final velocity_of the train just after it travels for t =20 seconds will be V =m/sO 60120O 180>O 1200SubmitRequest AnswerPart DThe final kinetic energy of the train just after it travelled for t = 20 seconds will be KE =JoulesO 144,000O 14,400O 1,728,00017,280,000SubmitRequest AnswerO O O O A train car of mass%3DM%3D 2400 Kg is initially at rest, Pushed by a net force exerted by magnets, the train covers a distance D= 1200 m in a time T= 20 seconds.Part AThe acceleration of the train is a =m/s2O 12O 1/6O 60SubmitRequest AnswerPart BThe net force exerted on the train is Fnet =NewtonO 14,400O 2,400O 24,000O 1440SubmitRequest Answer

Answered: Image /qna-images/answer/78b0c524-2d9f-4a4c-ba64-dbcf43d6a6cd.jpg

A train car of mass%3DM%3D 2400 Kg is initially at rest, Pushed by a net force exerted by magnets, the train covers a distance D= 1200 m in a time T= 20 seconds. Part A The acceleration of the train is a = m/s2 O 12 O 1/6 O 60 Submit Request Answer Part B The net force exerted on the train is Fnet = Newton O 14,400 O 2,400 O 24,000 O 1440 Submit Request Answer O O O

A train car of mass%3DM%3D 2400 Kg is initially at rest, Pushed by a net force exerted by magnets, the train covers a distance D= 1200 m in a time T= 20 seconds. Part A The acceleration of the train is a = m/s2 O 12 O 1/6 O 60 Submit Request Answer Part B The net force exerted on the train is Fnet = Newton O 14,400 O 2,400 O 24,000 O 1440 Submit Request Answer O O O

College Physics

10th Edition

ISBN:9781285737027

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter4: The Laws Of Motion

Section4.5: Applications Of Newton's Laws

Problem 4.6QQ: For the woman being pulled forward on the toboggan in Figure 4.33, is the magnitude of the normal...

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