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larisa [96]
3 years ago
8

An average sleeping person metabolizes at a rate of about 80 W by digesting food or burning fat. Typically, 20% of this energy g

oes into bodily functions, such as cell repair, pumping blood, and other uses of mechanical energy, while the rest goes to heat. Most people get rid of all this excess heat by transferring it (by conduction and the flow of blood) to the surface of the body, where it is radiated away. The normal internal temperature of the body (where the metabolism takes place) is 37∘C37 ∘ C, and the skin is typically 7C∘7C ∘ cooler. By how much does the person’s entropy change per second due to this heat transfer?
Physics
1 answer:
Leto [7]3 years ago
8 0

Answer:

<em> -4.7 x 10^-3 J/K-s</em>

Explanation:

The Power generated by metabolizing food = 80 W

The watt W is equivalent to the Joules per sec J/s

therefor power = 80 J/s

20% of this energy is not used for heating, amount available for heating is

==> H = 80% of 80 = 0.8 x 80 = 64 J/s

The inner body temperature = 37 °C = 273 + 37 = 310 K

The entropy of this inner body ΔS = ΔH/T

ΔS = 64/310 = 0.2065 J/K-s

The skin temperature is cooler than the inner body by 7 °C

Temperature of the skin =  37 - 7 = 30 °C = 273 + 30 = 303 K

The entropy of the skin = ΔS = ΔH/T

ΔS = 64/303 = 0.2112 J/K-s

change in entropy of the person's body = (entropy of hot region: inner body) - (entropy of cooler region: skin)

==> 0.2065 - 0.2112 =<em> -4.7 x 10^-3 J/K-s</em>

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KengaRu [80]

Answer:

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E_i=E_f\\\frac{mv^2}{2}= E_{Friction}\\E_{Friction}=F_{Friction}*d\\F_{Friction}= \mu_kmg\\\frac{mv^2}{2}= \mu_kmgd\\ d=\frac{v^2}{2\mu_kg}

a)

Looking at the formula you can see that the mass doesn't affect the distance travelled, as lng as the initial velocity is constant (Which indicates that the force must be higher to push the block to the same speed) therefore the distance is the same.

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6 0
3 years ago
Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in un
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          K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}

  \gamma = 1,     K_{IC} = 24.1

  [/tex]\sigma_{y}[/tex] = 570

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Hence, we will calculate the critical crack length as follows.

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            Largest size = 2a

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A bungee cord is 30m long and when stretched a distance x itexerts a restoring force of magnitude kx. Your father-in-law (mass95
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Answer:

The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.

Explanation:

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the stretched length required is give as  y=l'-l=41-30=11m

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x=\dfrac{F}{k}\\x=\dfrac{380}{630.92}\\x=0.602 m

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