A cylinder with a radius of 1.8 cm is floating in water as shown below. The mass of the cylinder is 1.80 kg. Calculate the depth of the bottom end of the cylinder(Density of water = 1000 kg/m3; one atmosphere = 1.013 X 105 Pa., 1meter = 100 cm) Intensity level:mT = 2n,kB=10logI.Frequency:f =TIntensity level difference:Angular frequency:kw = 2n fAB =10logI,mPV = nRT, where R = 8.31 J/mole-KMaximum Velocity, Acceleration:A2n fAo = A(2nf}Umax = A@ =a max =PV = constant,PIV1 = P2V2A = (27 f)а maxIntensity:Simple pendulum:I= Pav4ar?T= 2n=Doppler Shift:Moving observer: f' = f(v+vo)/vSpeed sound temperature:Moving source: f' = f(v)/ ( v-vs)v= f2Where vo = velocity of observer, vs is thespeed of the source, and v is thespeed of soundTU = (331m/s),273KIntensity of a wave:powerBeat frequency fB\f2- filI=%3DareaAbol Area of circle: A = Tr?Rate of flowQ = Avt = F(sin0)rT = Fl, wherel= lever armBernoulli's equation:1P++ pgy, = P, +1pv,´ + pgy,Equilibrium:Te = T - 273.15°CΣτ=09T =-T. +325Non-equilibrium:5Tc =-(T, - 32)Στ= ΙαТетреrature:Moment of inertial for a point massrevolving about a center at a distance r:ΔL L,ΔΤAA = y A,AT Y = 2aAV = BVAT B = 3aI= Mr?Kinetic Energy:Specific Heat:Translational: ½ mv²Rotational: ½ Im?(J/kg. 'C)mATQ = mc(T, –T,)Angular speed o =v/rQcold=-QhotTensile strain:EQ = 0Latent Heat:FALL.Y -Young's modulusphase change: Q=±mLEnergy Transfer:ΔΤP =AtDensity:ArMp =V(kg/m²)Thermal conductivity:Pressure:(T, – T.)P = kAL.FP=A(Ра)Hooke's Law:F =-kxP = P, + pghAcceleration in simple harmonicmotion:Archimedes' principle:ka =B = P fuidV famfluid8Elastic potential energy:РЕ, %3D-kx²2Equation of flow continuity:Av, = A,v,Period:

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Description: A cylinder with a radius of 1.8 cm is floating in water as shown below. The mass of the cylinder is 1.80 kg. Calculate the depth of the bottom end of the cylinder(Density of water = 1000 kg/m3; one atmosphere = 1.013 X 105 Pa., 1meter = 100 cm) Inte

Given: The mass of the cylinder is 1.8 kg. The radius of the cylinder is 1.8 cm. The density of the…

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A cylinder with a radius of 1.8 cm is floating in water as shown below. The mass of the cylinder is 1.80 kg. Calculate the depth of the bottom end of the cylinder (Density of water = 1000 kg/m3; one atmosphere = 1.013 X 105 Pa., 1meter = 100 cm)

A cylinder with a radius of 1.8 cm is floating in water as shown below. The mass of the cylinder is 1.80 kg. Calculate the depth of the bottom end of the cylinder (Density of water = 1000 kg/m3; one atmosphere = 1.013 X 105 Pa., 1meter = 100 cm)

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter14: Fluid Mechanics

Section: Chapter Questions

Problem 94P: Using the equation of the previous problem, find the viscosity of motor oil in which a steel ball of...

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