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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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What inflammatory molecule aids with muscle building and is released when muscles experience microscopic damage?
Reika [66]

Answer:

Cytokines

Explanation:

Cytokines are known as inflammatory molecules which are also proteinous and aid signaling of certain processes and conditions in the body.

They are also normally involved in aiding muscle building and are released when muscles experience microscopic damage which may lead to the muscles being sore.

3 0
3 years ago
Please help me! I don’t get this
goldfiish [28.3K]

The answer will be

(1) correct

(2) correct

(3) the force of the soccer ball on the net

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Hope this help

4 0
3 years ago
Read 2 more answers
f an arrow is shot upward on the moon with velocity of 35 m/s, its height (in meters) after t seconds is given by h(t)=35t−0.83t
inysia [295]

Answer:

The velocity of the arrow after 3 seconds is 30.02 m/s.

Explanation:

It is given that,

An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:

h(t)=35t-0.83t^2

We know that the rate of change of displacement is equal to the velocity of an object.

v(t)=\dfrac{dh(t)}{dt}\\\\v(t)=\dfrac{d(35t-0.83t^2)}{dt}\\\\v(t)=35-1.66t

Velocity of the arrow after 3 seconds will be :

v(t)=35-1.66t\\\\v(t)=35-1.66(3)\\\\v(t)=30.02\ m/s

So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.

7 0
3 years ago
You have a double slit experiment, with the distance between the two slits to be 0.025 cm. A screern is 120 cm behind the double
dimulka [17.4K]

Answer:

The wavelength of the light is 633 nm.

Explanation:

Given that,

Distance between the two slits d= 0.025 cm

Distance between the screen and slits D = 120 cm

Distance between the slits y= 1.52 cm

We need to calculate the angle

Using formula of double slit

\tan\theta=\dfrac{y}{D}

Where, y = Distance between the slits

D = Distance between the screen and slits

Put the value into the formula

\tan\theta=\dfrac{1.52}{120}

\theta=\tan^{-1}\dfrac{1.52}{120}

\theta=0.725

We need to calculate the wavelength

Using formula of wavelength

d\sin\theta=n\lambda

Put the value into the formula

0.025\times\sin0.725=5\times\lambda

\lambda=\dfrac{0.025\times10^{-2}\times\sin0.725}{5}

\lambda=6.326\times10^{-7}\ m

\lambda=633\ nm

Hence, The wavelength of the light is 633 nm.

4 0
3 years ago
a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit
ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)

We have

\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}

\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}

Then the total work is

W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right)  \approx \boxed{36.8\,\rm J}

5 0
2 years ago
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