Bowling balls are roughly the same size, but come in a variety of weights. Given its official radius of roughly 0.109 m, calculate the heaviest bowling ball that will float in water. Constants & GeometryFluid Staticsg = 9.81 m/s²1000 kg/m917 kg/m³1025 kg/mp = M/VP = F/APbottom = Ptop + pghFB = WDL = (Pfluid)(Vsubmerged)IF, = W = (pobject)(Vtotal)gGeometryAcircle3PwaterTr23PiceVsphere = Tr33PseawaterPatm101, 325 Pa

Constants & Geometry Fluid Statics g = 9.81 m/s² 1000 kg/m 917 kg/m³ 1025 kg/m p = M/V P = F/A Pbottom = Ptop + pgh FB = WDL = (Pfluid)(Vsubmerged)I F, = W = (pobject)(Vtotal)g Geometry Acircle 3 Pwater Tr2 3 Pice Vsphere = Tr3 3 Pseawater Patm 101, 325 Pa

Constants & Geometry Fluid Statics g = 9.81 m/s² 1000 kg/m 917 kg/m³ 1025 kg/m p = M/V P = F/A Pbottom = Ptop + pgh FB = WDL = (Pfluid)(Vsubmerged)I F, = W = (pobject)(Vtotal)g Geometry Acircle 3 Pwater Tr2 3 Pice Vsphere = Tr3 3 Pseawater Patm 101, 325 Pa

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter1: A Physics Toolkit

Section1.2: Mathematics And Physics

Problem 13SSC

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Question

Bowling balls are roughly the same size, but come in a variety of weights. Given its official radius of roughly 0.109 m, calculate the heaviest bowling ball that will float in water.

Constants & Geometry
Fluid Statics
g = 9.81 m/s²
1000 kg/m
917 kg/m³
1025 kg/m
p = M/V
P = F/A
Pbottom = Ptop + pgh
FB = WDL = (Pfluid)(Vsubmerged)I
F, = W = (pobject)(Vtotal)g
Geometry
Acircle
3
Pwater
Tr2
3
Pice
Vsphere = Tr3
3
Pseawater
Patm
101, 325 Pa

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