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pishuonlain [190]
3 years ago
6

How much sweat (in ml) would you have to evaporate per hour to remove the same amount of heat a 90.0 w light bulb produces? (1w=

1j/s.)?
Physics
1 answer:
Sveta_85 [38]3 years ago
6 0
1 W = 1 J/s
Therefore;
Total energy = 90*1*60*60 J = 324000 J = 324 kJ

Such an evaporation must take place at a temperature near the body temperature. Assuming a body temperature of 37°C at which heat of vaporization is approximately 2413.1 kJ/kg (again assuming sweet behaves like water).

Then,
mC = 324 kJ, where m = mass of sweet and C = heat of vaporization
Therefore,
m = 324/C = 324/2413.1 = 0.13427 kg

Density = m/v => volume = m/Density = 0.13427/1000 = 1.3427*10^-4 m^3
1 m^3 = 1000 liters
Then,
1.3427*10^-4 m^3 = 0.13427 liters of sweet = 134.27 ml
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Two men, Joel and Jerry, each pushes an object that are identical on a horizontal frictionless floor starting from rest. Joel an
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Answer:

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Explanation:

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W = F\cdot \Delta s

Where:

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F - Force exerted on the object, measured in newtons.

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During the first 10 minutes, the net work exerted on the object is zero. That is:

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W_{net} = (F-F)\cdot \Delta s

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W_{net} = 0\,J

In exchange, the net work in the next 5 minutes is the work done by Joel on the object:

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acceleration=\frac{\Delta\ velocity}{\Delta\ time}\\\\
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