Calculate the speed a spherical rain drop would achieve
falling from 5.00 km (a) in the absence of air drag (b) with air
drag. Take the size across of the drop to be 4 mm, the density
to be 1.00×103 kg/m3 , and the surface area to be πr2
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To calculate the speed of a spherical raindrop falling with and without air drag, we can use the principles of free fall and the forces acting upon the drop.
(a) Speed without air drag:
In the absence of air drag, the only force acting on the raindrop will be gravity. We can use the equations of motion to find the final speed.
1. Start by converting the given distance of 5.00 km to meters:
1 km = 1000 m
So, 5.00 km = 5.00 x 1000 m = 5000 m
2. The acceleration due to gravity is approximately 9.8 m/s². Since the drop is initially at rest, we can use the equation:
v² = u² + 2as
where v is the final speed, u is the initial speed (which is 0 m/s), a is the acceleration (9.8 m/s²), and s is the vertical distance (5000 m).
Plugging in the values:
v² = 0² + 2 x 9.8 x 5000
v² = 0 + 98,000
v = √98,000
v ≈ 313.23 m/s
Therefore, the speed of the raindrop falling from 5.00 km without air drag is approximately 313.23 m/s.
(b) Speed with air drag:
In the presence of air drag, we need to consider an additional force that opposes the motion of the raindrop. The drag force depends on the velocity of the object, the size and shape of the object, and the density of the fluid (air in this case).
The drag force acting on a spherical object can be calculated using the equation:
F_drag = (1/2) * ρ * A * C * v²
Where:
- F_drag is the drag force
- ρ is the density of the fluid (air) = 1.00 × 10³ kg/m³
- A is the cross-sectional area of the object = π * r²
- C is the drag coefficient (depends on the shape of the object)
- v is the velocity of the object (in this case, the speed of the raindrop)
The drag coefficient for a smooth sphere is approximately 0.47.
Now, let's calculate the speed of the raindrop with air drag:
1. Convert the size across of the drop to meters:
4 mm = 4 x 10⁻³ m
2. Calculate the cross-sectional area:
A = π * (r²)
A = π * (2² x 10⁻³ m) (radius is half of the size)
A = 4π x 10⁻⁶ m²
3. Calculate the drag force using the given values and equations mentioned above.
F_drag = (1/2) * ρ * A * C * v²
F_drag = (1/2) * (1.00 x 10³ kg/m³) * (4π x 10⁻⁶ m²) * (0.47) * v²
4. Set the drag force equal to the weight of the raindrop to find the terminal velocity where the drag force equals the gravitational force acting on the drop:
F_drag = weight of the drop
(1/2) * (1.00 x 10³ kg/m³) * (4π x 10⁻⁶ m²) * (0.47) * v² = m * g
(1/2) * (1.00 x 10³ kg/m³) * (4π x 10⁻⁶ m²) * (0.47) * v² = (4/3) * π * (2 x 10⁻³ m)³ * (1.00 x 10³ kg/m³) * g
5. Simplify the equation and solve for v²:
(1/2) * (1.00 x 10³ kg/m³) * (4π)(4 x 10⁻⁶)(0.47) * v² = (4/3) * π * (2 x 10⁻³)³ * (1.00 x 10³ kg/m³) * 9.8 m/s²
v² = [(4/3) * π * (2 x 10⁻³)³ * (1.00 x 10³ kg/m³) * 9.8 m/s²] / [(1/2) * (1.00 x 10³ kg/m³) * (4π)(4 x 10⁻⁶)(0.47)]
6. Calculate v:
v = √( [(4/3) * π * (2 x 10⁻³)³ * (1.00 x 10³ kg/m³) * 9.8 m/s²] / [(1/2) * (1.00 x 10³ kg/m³) * (4π)(4 x 10⁻⁶)(0.47)] )
The final value of v will be the speed of the raindrop falling from 5.00 km with air drag.