Final answer:
The weight of the heaviest bowling ball is approximately 8.723 kg. To calculate the weight of the heaviest bowling ball that will float in a fluid of density 1.100 x 10³ kg/m³, we need to consider the buoyant force. By using the formula for buoyant force and equating it to the weight of the ball, we can solve for the weight.
Step-by-step explanation:
To calculate the weight of the heaviest bowling ball that will float in a fluid of density 1.100 x 10³ kg/m³, we need to consider the buoyant force acting on the ball.
The buoyant force can be calculated using the formula F_b = V * ρ * g, where V is the volume of the ball, ρ is the density of the fluid, and g is the acceleration due to gravity.
Since the ball is floating, the buoyant force must be equal to the weight of the ball.
Therefore, we can equate the formulas for weight and buoyant force: mg = V * ρ * g.
Solving for m gives us m = V * ρ, where V is the volume of the ball and ρ is the density of the fluid.
Using the given information that the official radius of a bowling ball is roughly 0.110 m, we can calculate the volume of the ball using the formula V = (4/3) * π * r³.
Plugging in the value for the radius, we get V = (4/3) * π * 0.110³ = 0.00793 m³.
Now we can calculate the weight of the heaviest bowling ball using the formula m = V * ρ, where ρ is the density of the fluid.
Plugging in the values, we get m = 0.00793 m³ * 1.100 x 10³ kg/m³ = 8.723 kg.