A toy submarine of mass m = 0.580 kg moves around a submerged circular track of radius R = 2.02 m. The submarine's engine provides a constant propulsion force of F = 5.13 N. When the sub is in motion, it is subject to a R. viscous drag force exerted by the water. This force is proportional to the sub's speed; the proportionality factor is C = 1.26 kg/s. Assuming it starts from rest at t = 0 s, the speed v(t) of the submarine at a later time t is given by F v (t) = (1 – e Culm) 000 where e is the base of the natural logarithm. 1600 How much time has passed when the submarine's speed reaches 64% of its terminal value? t = 0.4703 What is the magnitude of the submarine's acceleration a at this time? a = 3.184 m/s2 Incorrect

Principles of Physics: A Calculus-Based Text
5th Edition
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Author:Raymond A. Serway, John W. Jewett
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Chapter5: More Applications Of Newton’s Laws
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O Attempt 2
A toy submarine of mass m = 0.580 kg moves around a
R.
submerged circular track of radius R = 2.02 m. The
%3D
submarine's engine provides a constant propulsion force of
F = 5.13 N. When the sub is in motion, it is subject to a
viscous drag force exerted by the water. This force is
proportional to the sub's speed; the proportionality factor is
C = 1.26 kg/s. Assuming it starts from rest at t = 0 s, the
speed v(t) of the submarine at a later time t is given by
%3D
F
v(1) =-(1 – e-Cilm)
%3D
where e is the base of the natural logarithm.
How much time has passed when the submarine's speed
t 3D
0.4703
reaches 64% of its terminal value?
What is the magnitude of the submarine's acceleration a at
m/s?
3.184
a =
this time?
Incorrect
Transcribed Image Text:22 of 25 O Attempt 2 A toy submarine of mass m = 0.580 kg moves around a R. submerged circular track of radius R = 2.02 m. The %3D submarine's engine provides a constant propulsion force of F = 5.13 N. When the sub is in motion, it is subject to a viscous drag force exerted by the water. This force is proportional to the sub's speed; the proportionality factor is C = 1.26 kg/s. Assuming it starts from rest at t = 0 s, the speed v(t) of the submarine at a later time t is given by %3D F v(1) =-(1 – e-Cilm) %3D where e is the base of the natural logarithm. How much time has passed when the submarine's speed t 3D 0.4703 reaches 64% of its terminal value? What is the magnitude of the submarine's acceleration a at m/s? 3.184 a = this time? Incorrect
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