A 5.50 kg crate is suspended from the end of ashort vertical rope of negligible mass. An upwardforce F(t) is applied to the end of the rope, andthe height of the crate above its initial position isgiven by y(t) = (2.80 m/s)t + (0.610 m/s³)t³Part AWhat is the magnitude of the force F when t = 3.60 s ?Express your answer with the appropriate units.F = 72.5μANSubmit Previous Answers Request Answer?X Incorrect; Try Again; 28 attempts remaining

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Description: A 5.50 kg crate is suspended from the end of ashort vertical rope of negligible mass. An upwardforce F(t) is applied to the end of the rope, andthe height of the crate above its initial position isgiven by y(t) = (2.80 m/s)t + (0.610 m/s³)t³Part AWh

Given that, mass of object M=5.50 kg y = 2.8 t + 0.610 t3 time t =3.6 sec

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A 5.50 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t) = (2.80 m/s)t + (0.610 m/s³)t³

A 5.50 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t) = (2.80 m/s)t + (0.610 m/s³)t³

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Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Newton's Second Law

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A 5.50 kg crate is suspended from the end of a
short vertical rope of negligible mass. An upward
force F(t) is applied to the end of the rope, and
the height of the crate above its initial position is
given by y(t) = (2.80 m/s)t + (0.610 m/s³)t³
Part A
What is the magnitude of the force F when t = 3.60 s ?
Express your answer with the appropriate units.
F = 72.5
μA
N
Submit Previous Answers Request Answer
?
X Incorrect; Try Again; 28 attempts remaining

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