Physics

First, find the initial velocity (v) of the ball as it leaves the machine, using the kinematic equation for maximum height:
v² = u² + 2as ⇒ 0 = v² - 2gh_max
v² = 2gh_max = 2 * 10 * 20 = 400
v = 20 m/s.
Now, apply the work-energy theorem to the ball while it is being pushed by the machine. The work done by the constant force F equals the change in kinetic energy of the ball.
Work done W = F * d
Change in K.E. = (1/2)mv² - 0
F * 0.2 = (1/2) * 0.15 * (20)²
F * 0.2 = (1/2) * 0.15 * 400 = 0.15 * 200 = 30
F = 30 / 0.2 = 150 N.

The height of capillary rise is given by the formula:
h = (2T cosθ) / (rρg)
Given: h = 15 cm = 0.15 m, r = 0.015 cm = 1.5 * 10? m, ρ = 900 kg/m ³, g = 10 m/s², and θ ≈ 0° (so cosθ ≈ 1).
We need to find the surface tension, T.
T = (h r ρ g) / (2 cosθ)
T = (0.15 * 1.5 * 10? * 900 * 10) / 2
T = 0.10125 N/m
The answer is required in milliNewton m? ¹, so we multiply by 1000.
T = 101.25 mN/m.

By the principle of conservation of angular momentum, the total angular momentum of the system (platform + person) remains constant.
L_initial = L_final
I_initial * ω_initial = I_final * ω_final
Initial state: Person is on the rim.
I_initial = I_platform + I_person_rim = (1/2)M_platform R² + M_person R²
Final state: Person is at the center.
I_final = I_platform + I_person_center = (1/2)M_platform R² + 0
[ (1/2) (200)R² + (80)R²] * 5 rpm = [ (1/2) (200)R²] * ω_final
(100R² + 80R²) * 5 = (100R²) * ω_final
180R² * 5 = 100R² * ω_final
ω_final = (180 * 5) / 100 = 9 rpm.