To solve this problem, we can use the following expression for terminal velocity:
v = √[(2 * m * g) / (ρ * A * C)]
where v is the terminal velocity, m is the mass of the object, g is the acceleration due to gravity (approximately 9.81 m/s²), ρ is the density of the fluid (air) in kg/m³, A is the cross-sectional area of the object, and C is the drag coefficient.
First, we need to find the mass of the wooden sphere. The volume of a sphere is given by the formula:
V = (4 / 3) * π * r³
where r is the radius of the sphere.
For a sphere with a radius of 9.00 cm, the volume is:
V = (4 / 3) * π * (0.09 m)³ ≈ 3.05 x 10⁻³ m³
Next, we need to convert the density of wood from g/cm³ to kg/m³:
density_wood = 0.770 g/cm³ * (1 kg / 1000 g) * (100 cm / 1 m)³ = 770 kg/m³
Now we can find the mass of the sphere:
mass = density_wood * volume ≈ 770 kg/m³ * 3.05 x 10⁻³ m³ ≈ 2.35 kg
The cross-sectional area of a sphere is given by the formula:
A = π * r²
For a sphere with a radius of 9.00 cm, the area is:
A = π * (0.09 m)² ≈ 0.0254 m²
Now we can plug the values into the terminal velocity equation:
v = √[(2 * 2.35 kg * 9.81 m/s²) / (1.20 kg/m³ * 0.0254 m² * 0.500)] ≈ 47.9 m/s
The terminal speed of the wooden sphere falling through air is approximately 47.9 m/s.
(b) To find the height from which an object would reach this speed in the absence of air resistance, we can use the following equation:
v² = 2 * g * h
where v is the terminal velocity, g is the acceleration due to gravity, and h is the height.
Rearrange the equation to solve for h:
h = v² / (2 * g) ≈ (47.9 m/s)² / (2 * 9.81 m/s²) ≈ 116.2 m
In the absence of air resistance, a freely falling object would reach the terminal speed of 47.9 m/s from a height of approximately 116.2 m.
Estimate the terminal speed of a wooden sphere (density 0.770 g/cm3) falling through air, if its radius is 9.00 cm and its drag coefficient is 0.500. (The density of air is 1.20 kg/m3.)
(b) From what height would a freely falling object reach this speed in the absence of air resistance?
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