A ball has a diameter of 3.77 cm and average density of 0.0839 g/cm3. What force is required to hold it completely submerged under water? magnitude N

The force required to hold it completely submerged under water is 0.252 N

As a result of the low density (ρ1 = 0.0839 g/cm3 = 83.9 kg/m3)of the ball compared to that of water (ρ2 =1000 kg/m3), the buoyant force that is acting on the ball is greater than its weight.

Therefore, the minimum force required to hold the ball submerged under water can be calculated using the relation

Radius = 3.77/2 cm = 0.0377/2 m = 0.01885 m

Volume of sphere = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ = 2.805 e-5 m³

Mass of sphere = 4/3 π r³ ρ1 = 4/3 * 3.142 * 0.01885³ * 83.9 = 0.0023 kg

Weight of sphere = 4/3 π r³ ρ1 g = 4/3 * 3.142 * 0.01885³ * 83.9 * 9.8 = 0.023 N

Volume of water displaced = 4/3 π r³ = 2.805 e-5

Buoyant force = weight of water displaced = 4/3 π r³ ρ2 g = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ * 1000 * 9.8 = 0.275 N

F = 0.275 - 0.023 = 0.252 N

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The force required to hold it completely submerged under water is 0.25 N

The density of the ball () = 0.0839 g/cm³ = 83.9 kg/m³

The density of water = 1000 kg/m³

Diameter = 3.77 cm = 0.0377 m

radius of ball = 0.0377/2 = 0.01885 m

The volume (V) =

Let us assume the acceleration due to gravity (g) = 9.8 m/s², Hence:

The force is required to hold it completely submerged under water (F) is:

F = 0.25 N

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explanation: