Question

A sphere completely submerged in water is tethered to the bottom with Part A a string. The tension in the string is one-fifth the weight of the sphere. What is the density of the sphere? Express your answer as an integer and include the appropriate units.

Answer 1

To find the density of the sphere, use Archimedes' Principle and the equation Density = 15 * Tension / (4 * pi * g * r^3).

Step-by-step explanation:

To find the density of the sphere, we first need to understand Archimedes' Principle, which states that the buoyant force acting on a submerged object is equal to the weight of the fluid displaced by the object. In this case, the buoyant force is equal to the weight of the sphere because it is completely submerged. The tension in the string is one-fifth the weight of the sphere, so we can set up the equation: Tension = 1/5 * Weight of sphere.

Since Weight = mg, where m is the mass of the sphere and g is the acceleration due to gravity, we can rewrite the equation as: Tension = 1/5 * mg. Solving for m, we find that m = 5 * Tension / g. The density of the sphere can be calculated using the formula: Density = mass / volume.

Since the sphere is completely submerged in water, the volume of water displaced by the sphere is equal to the volume of the sphere, so we can rewrite the equation as: Density = mass / (4/3 * pi * r^3). Substituting the value of m from the previous equation, we get: Density = (5 * Tension / g) / (4/3 * pi * r^3). Simplifying further, we find that Density = 15 * Tension / (4 * pi * g * r^3).

For example, let's say the tension in the string is 10 N and the radius of the sphere is 2 m. Substituting these values into the equation, we find that Density = 15 * 10 / (4 * pi * 9.8 * 2^3) = 0.119 kg/m^3. Therefore, the density of the sphere is 0.119 kg/m^3 (rounded to three decimal places).