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Solved: An object moving in a liquid experiences a linear
Chapter 6, Problem 76CP(choose chapter or problem)
An object moving in a liquid experiences a linear drag force: \(\vec{D}=(b v\), direction opposite the motion ), where \(b\) is a constant called the drag coefficient. For a sphere of radius \(R\), the drag constant can be computed as \(b=6 \pi \eta R\), where \(\eta\) is the viscosity of the liquid.
a. Use what you've learned in calculus to prove that
\(a_{x}=v_{x} \frac{d v_{x}}{d x}\)
b. Find an algebraic expression for \(v_{x}(x)\), the \(x\)-component of velocity as a function of distance traveled, for a spherical particle of radius \(R\) and mass \(m\) that is shot horizontally with initial speed \(v_{0}\) through a liquid of viscosity \(\eta\).
c. Water at \(20^{\circ} \mathrm{C}\) has viscosity \(\eta=1.0 \times 10^{-3} \mathrm{Ns} / \mathrm{m}^{2}\). Suppose a 1.0-cm-diameter,1.0 g marble is shot horizontally into a tank of \(20^{\circ} \mathrm{C}\) water at \(10 \mathrm{~cm} / \mathrm{s}\). How far will it travel before stopping?
Equation Transcription:
,direction opposite the motion)
Text Transcription:
^vec D=(bv,direction opposite the motion)
b
R
b=6 pi eta R
eta
a_x=v_x dv_x/dx
v_x(x)
x-components
R
m
v_0
eta
20^circ C
eta=1.0 x 10^-3 N s/m^2
20^circ C
Questions & Answers
QUESTION:
An object moving in a liquid experiences a linear drag force: \(\vec{D}=(b v\), direction opposite the motion ), where \(b\) is a constant called the drag coefficient. For a sphere of radius \(R\), the drag constant can be computed as \(b=6 \pi \eta R\), where \(\eta\) is the viscosity of the liquid.
a. Use what you've learned in calculus to prove that
\(a_{x}=v_{x} \frac{d v_{x}}{d x}\)
b. Find an algebraic expression for \(v_{x}(x)\), the \(x\)-component of velocity as a function of distance traveled, for a spherical particle of radius \(R\) and mass \(m\) that is shot horizontally with initial speed \(v_{0}\) through a liquid of viscosity \(\eta\).
c. Water at \(20^{\circ} \mathrm{C}\) has viscosity \(\eta=1.0 \times 10^{-3} \mathrm{Ns} / \mathrm{m}^{2}\). Suppose a 1.0-cm-diameter,1.0 g marble is shot horizontally into a tank of \(20^{\circ} \mathrm{C}\) water at \(10 \mathrm{~cm} / \mathrm{s}\). How far will it travel before stopping?
Equation Transcription:
,direction opposite the motion)
Text Transcription:
^vec D=(bv,direction opposite the motion)
b
R
b=6 pi eta R
eta
a_x=v_x dv_x/dx
v_x(x)
x-components
R
m
v_0
eta
20^circ C
eta=1.0 x 10^-3 N s/m^2
20^circ C
ANSWER:
Step 1 of 5
In this problem, we have to prove the given equation, then find an algebraic expression for \(v x(t)\) and also find the distance traveled before stopping.