Question

A sphere of radius 50 mm and mass 1 kg falling vertically through air of density 1.2 kg m-' attains a steady velocity of 11.0 m s-1. If the above equation is applied to its fall what is the value of k in this case?​

A) 0.0024 kg m^-1 s^-1
B) 0.012 kg m^-1 s^-1
C) 0.024 kg m^-1 s^-1
D) 0.12 kg m^-1 s^-1

Answer 1

The value of k in this case is 0.024 kg m^-1 s^-1.

Step-by-step explanation:

To find the value of k in this case, we can use the equation provided. The equation for the fall of an object through air with air resistance is given by:

Fair = -kv

Where Fair is the air resistance force, v is the velocity of the object, and k is a constant related to the properties of the object and the air.

In this case, the object is a sphere with a radius of 50 mm, a mass of 1 kg, and a steady velocity of 11.0 m/s. Plugging in these values, we have:

-k ⋅ 11.0 = 1.2 ⋅ (4/3) ⋅ π ⋅ (0.05)^3 ⋅ (11.0)

By solving for k, we find that k = 0.024 kg m-1 s-1. Therefore, the correct answer is C) 0.024 kg m-1 s-1.