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
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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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![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
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Transcribed Image Text: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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