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VLD [36.1K]
3 years ago
8

Assuming a typical efficiency for energy use by the body, how many slices of pizza must you eat to walk for 2.5 h at a speed of

5.0 km/h? Assume that your mass is 68 kg, metabolic power of 68 kg individual walking at 5 km/h is 380 W, and the energy content of slice of pizza is 1260 kJ.
Physics
1 answer:
larisa86 [58]3 years ago
6 0

Answer:

2.7 Pizzas.

Explanation:

The power required to walk through 5km in 1 hour is 380W.

A watt is basically Jules per second, then we need to standardized this measurement to second.

5km/hr is equal to,

\frac{5km}{hr}*\frac{1hr}{3600s}*\frac{1000m}{1km}=1.389m/s

Walking by 2.5 hours is equal to a distance of,

d=v*t=1.389*(2.5*3600) = 12500m

The total energy required then would be,

E = \frac{380J}{1.389m/s}(12500)=3.4199*10^6J

Then we know that one pizza slice gives 1260*10^3J of energy, the total pizza needed are,

\eta = \frac{3.4199*10^6}{1260*10^3} = 2.7142

<em>Then you need to buy 3 pizza.</em>

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Answer:

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C_w = specific heat capacity of water = 4200 J/kg.°C

C_a = specific heat capacity of aluminum = 890 J/kg.°C

\Delta T_c = change in temperature of copper = 283°C - T

\Delta T_w = change in temperature of water = T - 14.6°C

\Delta T_a = change in temperature of aluminum = T - 14.6°C

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(227\ g)(385\ J/kg.^oC)(283^oC-T)=(844\ g)(4200\ J/kg.^oC)(T-14.6^oC)+(155\ g)(890\ J/kg.^oC)(T-14.6^oC)\\\\24732785\ J - (87395\ J/^oC) T = (3544800\ J/^oC) T - 51754080\ J+ (137950\ J/^oC) T-2014070\ J\\\\24732785\ J +51754080\ J+2014070\ J = (3544800\ J/^oC) T+(137950\ J/^oC+(87395\ J/^oC) T\\\\78560935\ J = (3770145\ J/^oC) T\\\\T = \frac{78560935\ J}{3770145\ J/^oC}

<u>T = 20.84°C</u>

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(a) 4.06 cm

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