water at 20°C has viscosity n=1.0*10^-3 Ns/m^2. sand grains have density 2400kg/m^3. suppose a 1.1 mm diameter sand grain is dropped into a 51m deep lake whose water is a constant 20°C. If the sand grain reaches terminal speed almost instantly (a quite good approximation), how long will it take the sand grain to settle to the bottom of the lake?
Use Stokes' drag equation for the terminal velocity.
Stoke's equation says, for a sphere in a viscous fluid,
F=6(pi)RnVc,
where F is the force, R is the radius of the sphere, n is the viscosity, and V is the velocity through the fluid.
For the force F, use the weight minus the buoyancy force.
Divide the lake depth by that velocity.
Make that last term V, not Vc
a) Vterm = (mg)/(6pi N R)
b)V term = ((4/3)pi R^3)pg))/(6pi N R)
To determine the time it takes for the sand grain to settle to the bottom of the lake, we can use the concept of terminal velocity. Terminal velocity is the maximum speed at which an object falls due to the balance between the gravitational force pulling it downward and the drag force resisting its motion through a fluid.
To begin, let's calculate the terminal velocity of the sand grain using the following formula:
v_terminal = (2 * m * g) / (ρ * C * A)
Where:
v_terminal is the terminal velocity,
m is the mass of the sand grain,
g is the acceleration due to gravity,
ρ is the density of the water,
C is the drag coefficient, and
A is the cross-sectional area of the sand grain.
Given Information:
- Diameter of the sand grain (d) = 1.1 mm
- Density of the sand grain (ρ_s) = 2400 kg/m^3
- Density of water (ρ) = 1000 kg/m^3 (approximate value at 20°C)
- Gravitational acceleration (g) = 9.8 m/s^2
- Water viscosity (n) = 1.0 * 10^-3 Ns/m^2
First, we need to calculate the mass and cross-sectional area of the sand grain:
1. Calculating the mass of the sand grain:
The mass (m) can be calculated using the formula:
m = ρ_s * V
Where V is the volume of the sand grain. The volume of a sphere can be calculated using the formula:
V = (4/3) * π * (d/2)^3
Substituting the given values:
d = 1.1 mm
ρ_s = 2400 kg/m^3
Converting the diameter to meters:
d = 1.1 mm = 1.1 * 10^-3 m
Substituting the values into the volume formula:
V = (4/3) * π * (1.1 * 10^-3 / 2)^3
≈ 1.267 × 10^-9 m^3
Substituting the volume into the mass formula:
m = ρ_s * V
= 2400 kg/m^3 * 1.267 × 10^-9 m^3
≈ 3.04 × 10^-6 kg
2. Calculating the cross-sectional area of the sand grain:
The cross-sectional area (A) of the sand grain can be calculated using the formula:
A = π * (d/2)^2
Substituting the given value of the diameter:
d = 1.1 * 10^-3 m
Substituting the value into the area formula:
A = π * (1.1 * 10^-3 m/2)^2
≈ 9.487 × 10^-7 m^2
Now we have all the values required to calculate the terminal velocity.
Substituting the given values into the terminal velocity formula:
v_terminal = (2 * m * g) / (ρ * C * A)
Substituting the remaining known values:
g = 9.8 m/s^2
ρ = 1000 kg/m^3
C = ? (drag coefficient)
At this point, we are missing the drag coefficient to calculate the terminal velocity accurately. To find the drag coefficient, we would need additional information about the shape and characteristics of the sand grain.