Running on a treadmill is slightly easier than running outside because there is no drag force to work against. Suppose a 60 kg runner completes a 5.0 km race in 23 minutes.(a) Determine the drag force on the runner during the race. Suppose that the cross section area of the runner is 0.72 m^2 and the density of air is 1.2 kg/m^3.(b) What is this force as a fraction of the runner's weight?

a) Drag force = 5.67 N

b) Fraction of (drag force/weight) = 0.00964 = 0.964% (just less than 1%)

Step-by-step explanation:

Drag force is given by the expression

Drag force = CρAv²/2

C = Drag coefficient of air for an average human moving through air = 1.0

ρ = density of air = 1.2 kg/m³

A = Cross sectional Area of runner facing the air = 0.72 m²

v = velocity of the runner = (5km/23min) = (5000/23×60) = 3.623 m/s

Drag force = 1(1.2)(0.72)(3.623²)/2

Drag force = 5.67 N

b) Runner's weight = mg = 60×9.8 = 588 N

Fraction of (drag force/weight) = (5.67/588) = 0.00964 = 0.964% (just less than 1%)

idk

step-by-step explanation:

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)