A student studied the motion of a sphere falling through a liquid to determine the viscosity of the liquidand suggested that the viscosity is dependent on the square of the velocity. The viscosity of the fluid, nf,can be calculatedusing2grg(Ps – Pr)V2Given thatr is the radius of the sphere = (1.2±0.1) mmg is the acceleration due to gravity = 9.81 m. s-2Ps is the density of the sphere = 8615 kg.mis the density of the fluid = 1015 kg.m-3-3PfV; is the terminal velocity of the sphere through the fluid = (15+2) mm. s-1use this information to calculate the best estimate of the viscosity of the fluidand the uncertainty in the best estimate, assuming errors in the radius andterminal velocity are uncorrelated. Assume the uncertainties in the otherquantities are negligible.NB: USE THE CALCULUS APPROACH TO ESTIMATE THE UNCERTAINTY

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A student studied the motion of a sphere falling through a liquid to determine the viscosity of the liquid and suggested that the viscosity is dependent on the square of the velocity. The viscosity of the fluid, nf, can be calculated using Given that r is the radius of the sphere=(1.2±0.1) mm g is the acceleration due to gravity = 9.81 m. s-2 Ps is the density of the sphere = 8615 kg.m-3 Ps is the density of the fluid = 1015 kg.m-3 Ve is the terminal velocity of the sphere through the fluid = (15±2) mm. s-1 use this information to calculate the best estimate of the viscosity of the fluid and the uncertainty in the best estimate, assuming errors in the radius and terminal velocity are uncorrelated. Assume the uncertainties in the other quantities are negligible.

A student studied the motion of a sphere falling through a liquid to determine the viscosity of the liquid and suggested that the viscosity is dependent on the square of the velocity. The viscosity of the fluid, nf, can be calculated using Given that r is the radius of the sphere=(1.2±0.1) mm g is the acceleration due to gravity = 9.81 m. s-2 Ps is the density of the sphere = 8615 kg.m-3 Ps is the density of the fluid = 1015 kg.m-3 Ve is the terminal velocity of the sphere through the fluid = (15±2) mm. s-1 use this information to calculate the best estimate of the viscosity of the fluid and the uncertainty in the best estimate, assuming errors in the radius and terminal velocity are uncorrelated. Assume the uncertainties in the other quantities are negligible.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Properties of Fluids

The term fluid represents a type of substance that can easily change its shape under any external pressure/force and can flow. Fluids can acquire any shape in which it is stored and can resist normal stresses but cannot resist shear stress in rest condition. Whenever external shear stress is applied to fluids, it starts to flow. The fluids may be liquid or gas. Example- water, air, etc.

Question

A student studied the motion of a sphere falling through a liquid to determine the viscosity of the liquid
and suggested that the viscosity is dependent on the square of the velocity. The viscosity of the fluid, nf,
can be calculated
using
2
grg(Ps – Pr)
V
2
Given that
r is the radius of the sphere = (1.2±0.1) mm
g is the acceleration due to gravity = 9.81 m. s-2
Ps is the density of the sphere = 8615 kg.m
is the density of the fluid = 1015 kg.m
-3
-3
Pf
V; is the terminal velocity of the sphere through the fluid = (15+2) mm. s-1
use this information to calculate the best estimate of the viscosity of the fluid
and the uncertainty in the best estimate, assuming errors in the radius and
terminal velocity are uncorrelated. Assume the uncertainties in the other
quantities are negligible.
NB: USE THE CALCULUS APPROACH TO ESTIMATE THE UNCERTAINTY

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