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exis [7]
3 years ago
7

Approximately 80% of the energy used by the body must be dissipated thermally. The mechanisms available to eliminate this energy

are radiation, evaporation of sweat, evaporation from the lungs, conduction, and convection. In this question, we will focus on the evaporation of sweat alone, although all of these mechanisms are needed to survive. The latent heat of vaporization of sweat at body temperature (37 °C) is 2.42 x 10^6 J/kg and the specific heat of a body is approximately 3500 J/(kg*°C).
(A) To cool the body of a jogger of mass 90 kg by 1.8°C , how much sweat has to evaporate?
O 130 g
O 230 g
O 23 g
O 13 g
Physics
1 answer:
Semenov [28]3 years ago
8 0

Answer:

the correct answer is c) 23 g

Explanation:

The heat lost by the runner has two parts: the heat absorbed by sweat in evaporation and the heat given off by the body

     Q_lost = - Q_absorbed

     

The latent heat is

      Q_absorbed = m L

The heat given by the body

      Q_lost = M c_{e} ΔT

       

where m is the mass of sweat and M is the mass of the body

       m L = M c_{e} ΔT

        m = M c_{e} ΔT / L

let's replace

        m = 90  3.500  1.8 / 2.42 10⁶

 

        m = 0.2343 kg

reduced to grams

        m = 0.2342 kg (1000g / 1kg)

        m = 23.42 g

 the correct answer is c) 23 g

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   H = mcpdT

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5 0
3 years ago
A 13.0 kg wheel, essentially a thin hoop with radius 1.80 m, is rotating at 469 rev/min. It must be brought to a stop in 16.0 s.
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Answer:

Explanation:

Given

mass of wheel m=13 kg

radius of wheel=1.8 m

N=469 rev/min

\omega =\frac{2\pi \times 469}{60}=49.11 rad/s

t=16 s

Angular deceleration in 16 s

\omega =\omega _0+\alpha \cdot t

\alpha =\frac{\omega }{t}=\frac{49.11}{16}=3.069 rad/s^2

Moment of Inertia I=mr^2=13\times 1.8^2=42.12 kg-m^2

Change in kinetic energy =Work done

Change in kinetic Energy=\frac{I\omega ^2}{2}-\frac{I\omega _0^2}{2}

\Delta KE=\frac{42.12\times 49.11^2}{2}=50,792.34 J

(a)Work done =50.79 kJ

(b)Average Power

P_{avg}=\frac{E}{t}=\frac{50.792}{16}=3.174 kW

7 0
3 years ago
A charge of 31.0 μC is to be split into two parts that are then separated by 24.0 mm, what is the maximum possible magnitude of
miskamm [114]

Answer:

1.72 x 10³ N.

Explanation:

When a charge is split equally and placed at a certain distance , maximum electrostatic force is possible.

So the charges will be each equal to

31/2 = 15.5 x 10⁻⁶ C

F = K Q q / r²

= \frac{9\times10^9\times(10.5)^2\times10^{-12}}{(24\times10^{-3})^2}

= 1.72 x 10³ N.

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Explanation:

5 0
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Read 2 more answers
A forklift lifts 5 boxes from the ground to a height of 2 meters (m). The boxes push down with a force of 1000 newtons (N). How
Nata [24]
Hello!

Answer:

2000 J

Explanation

Work equation is expressed as:

W=F.d.Cos \alpha

Where:

F: Applied force
d: traveled distance
α: Angle between the direction of the force and the direction of the movement. (in this case, both of the direction are the same, so the angle is 0°)

By substituting:

F=1000N.2m.Cos(0)=2000N.m=2000 J

Have a nice day!
8 0
3 years ago
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