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Karolina [17]
3 years ago
10

When you drink cold water, your body must expend metabolic energy in order to maintain normal body temperature (37° C) by warmin

g up the water in your stomach. Could drinking ice water, then, substitute for exercise as a way to "burn calories?" Suppose you expend 390 kilocalories during a brisk hour-long walk. How many liters of ice water (0° C) would you have to drink in order to use up 390 kilocalories of metabolic energy? For comparison, the stomach can hold about 1 liter.
Physics
1 answer:
Papessa [141]3 years ago
5 0

Answer:

The number of liters of ice water is 11 L

Explanation:

Given data:

normal body temperature = 37°C

temperature of the ice water = 0°C

Cwater = specific heat of water = 4186 J/kg °C

Suppose the person drinks 1 L of cold water, then, the mass is 1 kg

The heat is:

Q_{water} =mC_{water} (T_{ice} -T_{body} )=1*4186*(0-37)=-154882J

The sign (-) indicates the energy lost by the metabolic process. If the Qwalk is 390 kilocalories, then the number of liters of ice water is equal to:

n=\frac{Q_{walk} }{Q_{water} } =\frac{390*4186}{154882} =10.54=11L

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gtnhenbr [62]

Answer:

"2Ω" is the net resistance in the circuit.

Explanation:

The given resistors are:

R1 = 3Ω

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The net resistance will be:

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On substituting the values, we get

⇒  \frac{1}{R_{net}} =\frac{1}{3} +\frac{1}{6}

On taking L.C.M, we get

⇒  \frac{1}{R_{net}} =\frac{2+1}{6}

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On applying cross-multiplication, we get

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2 years ago
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svetlana [45]
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F_{net} = m_w g
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m_w g = (m+m_w)a
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3 years ago
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tresset_1 [31]

Given:

Height of tank = 8 ft

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Total height = 8+13 =21 ft

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Now,

Work done required = \int_{0}^{8} 52\times 16\pi (21 - y)dy

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P = 100 J/s or 100 Watts

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