Question

The drag force on a sphere of radius 2 cm that falls through a fluid with speed 10 cm/s is 8 N. What is the drag force on a sphere of radius 4 cm that falls through the same fluid with a speed of 5 cm/s?

Answer 1

0.10 N

Step-by-step explanation:

The drag force can be expressed using the following formula:

F = 1/2 * rho * A * Cd * v^2

Let's calculate the cross-sectional area of the sphere:

A1 = pi * r1^2

A1 = pi * (0.02 m)^2

A1 = 0.00126 m^2

Converting the speed to meters per second:

v1 = 0.1 m/s

Now we can use the formula to solve for the density times the drag coefficient:

rho * Cd = 2 * F / (A1 * v1^2)

rho * Cd = 2 * 8 N / (0.00126 m^2 * (0.1 m/s)^2)

rho * Cd ≈ 1.592 kg/m

For the second sphere with a radius of 4 cm and a speed of 5 cm/s, we can calculate the cross-sectional area and the speed in the same way:

A2 = pi * r2^2

A2 = pi * (0.04 m)^2

A2 = 0.00502 m^2

v2 = 0.05 m/s

Using the same value of the density times the drag coefficient:

F2 = 1/2 * rho * Cd * A2 * v2^2

F2 = 1/2 * (1.592 kg/m) * 0.00502 m^2 * (0.05 m/s)^2

F2 ≈ 0.10 N