The weight of a hollow sphere is 100N.If it floats in water just fully submerged,what is the exterrnal diameter of the shere ?
W = density * g * V
100 = 1000 * 9.81 * (4/3 * pi*r^3)
r = .13900
2r = .269
26.9cm convertion
Since it barely floats, the sphere volume, filled with water, would weigh 100 N (not including the hollow sphere itself).
pi*D^3/6 * 1000 kg/m^3*g 9.8 m/s^2 = 100 N
pi D^3/6 = 100/9800 = 0.010204
D^3 = 0.19488
D = 0.269 m = 26.9 cm
To find the external diameter of the sphere, we can use the principle of buoyancy.
When the hollow sphere is fully submerged in water and floating, its weight is equal to the buoyant force acting on it. The buoyant force is equal to the weight of the water displaced by the sphere.
Given that the weight of the sphere is 100N, we can assume that this is also the weight of the water displaced by the sphere.
Let's denote:
W = weight of the water displaced = 100N
ρ = density of water = 1000 kg/m^3 (assuming water density is approximately 1000 kg/m^3)
The volume of the water displaced can be calculated using the formula:
V = W / ρ
V = 100N / 1000 kg/m^3
V = 0.1 m^3
Now, let's find the volume of the hollow sphere:
V = (4/3) * π * (R^3 - r^3)
Where R is the external radius of the sphere and r is the internal radius of the sphere.
Since the sphere is hollow, we can assume that the internal radius is negligible, and therefore, r = 0.
V = (4/3) * π * R^3
0.1 m^3 = (4/3) * π * R^3
Now, we can solve for R:
R^3 = (0.1 m^3) * (3/4π)
R^3 ≈ 0.25 / π
R ≈ (0.25 / π)^(1/3)
Using a calculator, we can determine that:
R ≈ 0.369 m
Finally, we can find the external diameter (D) by multiplying the external radius by 2:
D = 2 * R
D ≈ 2 * 0.369 m
D ≈ 0.739 m
Therefore, the approximate external diameter of the hollow sphere is 0.739 meters.
To find the external diameter of the hollow sphere, we need to use Archimedes' principle, which states that the buoyant force experienced by an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
Given that the weight of the hollow sphere is 100N, and assuming the sphere is fully submerged in water, we know that the buoyant force acting on the sphere is also 100N (since it is in equilibrium).
To find the buoyant force, we can use the formula:
Buoyant force = Weight of the displaced fluid
The density of water is approximately 1000 kg/m³, and the volume of the displaced fluid can be expressed as the volume of the hollow sphere. The volume of a hollow sphere with external radius R and internal radius r can be calculated as:
Volume = (4/3) * π * (R^3 - r^3)
However, we need the external diameter of the hollow sphere, so we will express the radius in terms of the diameter:
R = D/2
Where D is the external diameter of the hollow sphere.
Substituting this into the volume formula:
Volume = (4/3) * π * [(D/2)^3 - r^3]
Now, we can set the buoyant force equal to the weight of the hollow sphere:
100N = (4/3) * π * [(D/2)^3 - r^3] * 1000 kg/m³ * 9.8 m/s²
Simplifying the equation, we can solve it for D:
D^3 - r^3 = (300N * 3 / (π * 1000 kg/m³ * 9.8 m/s²))
D^3 - r^3 ≈ 30.61
To find the external diameter D, we need to know the internal radius r. Without that information, it is not possible to provide a specific answer to the question.