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26 days ago

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A world class runner can run long distances at a pace of 15 km/hour. That runner expends 800 kilocalories of energy per hour. a)

Find the power (in Watts [W]) the runner is exerting while running. b) Find the total energy (in Joules [J]) exerted by the runner in a 15 km run. c) How many Milky Way (Original Single 52.2g) chocolate bars does the runner need to buy to supply the amount of energy to complete a half-marathon (13.1 miles)? (Cite your source for the number of calories in a Milky Way bar)

1 answer:

pantera1 [230]26 days ago

7 0

Response: a) 1.05 kW b) 3.78 MJ c) 5.3 bars

Clarification:

A)

Conversion results in 900 kcal equating to 900000 x 4.2 J/cal {4.2 J/cal is the conversion factor}

= 3780 kJ

Knowing 1 hour = 3600 seconds, we find:

Thus, Power in watts = 3780/3600 = 1.05 kW = 1050 W

B)

If running at a speed of 15 km/hour, completing a 15 km distance takes 1 hour.

1 hour equals 3600 seconds during which the runner expends 1050 joules each second.

Therefore, energy consumed in 1 hour = 3600 x 1050 J/s

= 3780000 J or 3.78 MJ

C)

1 mile is 1.61 km so 13.1 miles converts to 13.1 x 1.61 = 21.1 km

As 15 km requires 3.78 MJ, 21.1 km necessitates 3.78 x 21.1/15 = 5.32 MJ = 5320 kJ

Ultimately,

1 Milky Way equals 240000 calories; hence, 4.2 multiplied by 240000 J results in 1008000 J or 1008 kJ

This implies the runner must acquire 5320/1008 = 5.3 bars.

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Explanation:

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mass flow rate, $\dot{m} = 60.6 kg/s$

flow density, $\rho = 1000 kg/m^3$

efficiency of the pump, $\eta = 0.74$

output gauge pressure, $p_o = 344.75 kPa$

input gauge pressure, $p_i = 68.95 kPa$

cross-sectional area of output pipe, $A_o = 0.069 m^2$

cross-sectional area of input pipe, $A_i = 0.093 m^2$

height of discharge, $z_o = 1.22 m - 0.61 m = 0.61 m$ (evaluated at pump’s maximum height of 1.22 m)

input height, $z_i = 0 m$

hydraulic power of the pump, $P =? kW$

Initially, the volumetric flow (Q) needs to be determined

$Q = \frac{\dot{m}}{\rho}$

$Q = \frac{60.6 kg/s}{1000 kg/m^3}$

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Next, compute the velocity (v) for both input and output

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$v_o = \frac{0.0606 m^3/s}{0.069 m^2}$

$v_o = 0.88 m/s$

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$v_i = 0.65 m/s$

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Answer:

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