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Vanyuwa [196]
3 years ago
9

Running on a treadmill is slightly easier than running outside because there is no drag force to work against. Suppose a 60 kg r

unner 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?
Mathematics
1 answer:
Nat2105 [25]3 years ago
5 0

Answer:

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%)

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Exchange rate means the rate at which one item exchanges hands with another, for instance, in this case, 6 jagged rocks are equivalent to 2 smooth pebbles.

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divide both sides by 2

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We can rewrite as below:

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30 jagged rocks=10 smooth pebbles

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Read more about exchange rate in the link below:

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dimaraw [331]

<u>Answer:</u>

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<u>Step-by-step explanation:</u>

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