Physics, 18.10.2019 18:00 igotdiss2178

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 19minutes.

determine the drag force on the runner during the race. suppose that the cross section area of the runner is 0.72 m2 and the density of air is 1.2 kg/m3.

what is this force as a fraction of the runner's weight?

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Answer from: bedsaul12345

Part A.1) The drag force on the runner is 8.33N .

Part A.2 ) The fraction of the drag force to the runner’s weight is 0.014 .

Explanation:

Drag force acts opposite to the direction of the of motion of an object or body. It tends to oppose the motion of the body. The drag force on the runner acts to oppose the motion of runner while he runs. This drag force depends on the speed of runner.

Solution:

Consider the drag coefficient for the runner is approximately 11 .

Write the expression for the speed of the runner.

v= d/t

Substitute 5 km for d, and 19 min. for t;

V= $\frac{5*1000}{19}$

V=4.39 m/s

Write the expression for the drag force on the runner.

$F_{d}$ = 0.5ρAv² $C_{d}$

subtitute A=0.72 m², ρ=1.2 kg/m³ , $C_{d}$ =1 and v= 4.39 m/s

$F_{d}$ = (0.5)*1.2*0.72* 4.39*1

$F_{d}$ = 8.33 N

Part A.1

The drag force on the runner is 8.33N .

(A.2)

Write the expression for the weight of the runner

w = mg

Here, w is the weight of the runner, g is acceleration due to the gravity, and mm is the mass of the runner.

m=60 kg ; 9.81m/s² for g

w= ( 60*9.81 )

w=588.6 N

The fraction of the drag force to the runner’s weight can be calculated as,

f= ( $F_{d}$ )/ (w)

Here, ff is the drag force as a fraction of runner’s weight.

f= (8.33) / (588.6)

f=0.014

Answer from: humblemalak

The drag force on the runner during the race is 8.31 N.

The fraction of the drag force on the runner's weight is 0.0141.

The given parameters;

mass of the runner, m = 60 kgdistance covered by the runner, s = 5 km = 5000 mtime of motion of the runner, t = 19 minutes = 1140 scross-sectional area of the runner, A = 0.72 m²density of air, ρ = 1.2 kg/m³

The speed of the runner is calculated as follows;

$v = \frac{s}{t} \\\\v = \frac{5000}{1140} \\\\v = 4.386 \ m/s$

The drag force on the runner is calculated as follows;

$F_D = \frac{1}{2} \rho v^2 C_DA$

where;

C_D is drag coefficient

Let the drag coefficient = 1

$F_D = 0.5 \times 1.2 \times 4.386^2 \times 1 \times 0.72\\\\F_D = 8.31 \ N$

The weight of the runner is calculated as follows;

W = mg

W = 60 x 9.8

W = 588 N

The fraction of the drag force on the runner's weight is calculated as follows;

$= \frac{F_D}{W} \\\\= \frac{8.31}{588} \\\\= 0.0141$

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5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000

Answer from: Quest

No because scientist are still changing there mines today so they are not correct to this day

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