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quester [9]
3 years ago
15

A 5000 kg African elephant has a resting metabolic rate of 2500 W. On a hot day, the elephant's environment is likely to be near

ly the same temperature as the animal itself, so cooling by radiation is not effective. The only plausible way to keep cool is by evaporation, and elephants spray water on their body to accomplish this. If this is the only possible means of cooling, how many kilograms of water per hour must be evaporated from an elephant's skin to keep it at a constant temperature?
Physics
1 answer:
alina1380 [7]3 years ago
3 0

Answer:

36 kg

Explanation:

To answer this question, a few assumptions have to be made:

  • That the temperature on the day is 35 °C
  • That all the heat from the elephant is goes to warming/evaporating the water on the surface of the elephant

Energy released per hour = 2500 J/s * 3600 s = 9 000 000 J

Q = mcΔT

9 000 000 J= m *4.186 J/g-K * (373K - 308K) + m*2260 J/g

m =  36 000 g = 36 kg

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bixtya [17]

Answer:

u_{s}=0.56

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

F_{s.max}=\frac{mv^2}{r} \\

Where the r is the radius of merry-go-round and v is the tangential speed

but

F_{s.max}=u_{s}F_{N}=u_{s}mg

So we have

u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}

Substitute the given values

So

u_{s}=\frac{4\pi^2 5.5m}{(9.8m/s^2)(6.3s)^2} \\u_{s}=0.56

6 0
3 years ago
Explain why sound is more clear at night than the day​
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5 0
2 years ago
How does changes in distance affect the gravitational pull between two objects? Describe and give one example.
maxonik [38]
The formula is

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As the distance decreases, the gravitational force will Increase.

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gizmo_the_mogwai [7]

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