Respuesta :
The heaviest bowling ball that will float in a fluid of specific gravity is 77.1066N
r = Radius of ball = 0.134 m
ρ = Density of fluid = 1.1× [tex]10^{3}[/tex](Assumed)
g = Acceleration due to gravity = 9.81 m/s²
m = Mass of ball
V = Volume of ball = 4/3 π [tex]r^{3}[/tex]
The weight of the bowling ball will balance the buoyant force
W = Fb
mg = V ρg
m = Vρ
m = 4/3 π 0.134^3 × 1.1× [tex]10^{3}[/tex]
m = 7.156 kg
The mass of the bowling ball will be 7.156 kg
Weight will be 7.156 × 9.81 = 77.1066N
A measurement of a substance's density in relation to water's density is its specific gravity, which is more officially known as relative density. In general, specific gravity is calculated for solids and liquids in respect to water in its densest condition (at a temperature of 4 Celsius or 39.2 Fahrenheit), and for gases in proportion to air at room temperature. It is not stated with units because it is a ratio. SG or Sp. Gr. are abbreviations for specific gravity.
Temperature and air pressure affect how dense an object is. For precise measurements of specific gravity, it is necessary for scientific reasons to regulate and display the temperature and pressure.
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