i believe it's C but i'm not completely sure
Answer:
33 N
Explanation:
v = Velocity of fluid = 8+2 = 10 m/s
= 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

The drag force on the athlete is 33 N
<span>hydrocarbon (but im not 100% sure)</span>
Answer:
a)

b) 
Explanation:
The net force on the car must produce the centripetal acceleration necessary to make this circle, which is
. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

Which means:

The limit for falling off would be
, so the minimum speed would be:
