Answer:
33 N
Explanation:
v = Velocity of fluid = 8+2 = 10 m/s
[tex]\rho[/tex] = Density of fluid = 1.2 kg/m³
C = Coefficient of drag = 1.1
A = Cross sectional area = 0.5 m²
Drag force is given by
[tex]F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2\times 1.1\times 0.5\times (8+2)^2\\\Rightarrow F=33\ N[/tex]
The drag force on the athlete is 33 N