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Tresset [83]
2 years ago
11

In very cold weather, a significant mechanism for heat loss by the human body is energy expended in warming the air taken into t

he lungs with each breath. A) On a cold winter day when the temperature is -19.0 C, what is the amount of heat needed to warm to internal body temperature (37 C ) the 0.470 J (Kg *K) of air exchanged with each breath? Assume that the specific heat capacity of air is 1020 and that 1.0 L of air has a mass of 1.3 g . B) How much heat is lost per hour if the respiration rate is 21.0 breaths per minute?
Physics
1 answer:
Pie2 years ago
7 0

Answer:

A) Q_a=74256\ J

B) Q=93562560\ J

Explanation:

Given:

  • temperature of air, T_a=-19+273=254\ K
  • temperature of lungs, T_l=37+273=310\ K
  • specific Heat exchanged from the lungs , c_l=0.47\ J.kg^{-1}.K^{-1}
  • specific heat of air, c_a=1020\ J.kg^{-1}.K^{-1}
  • mass of 1 L air, m'=1.3\ kg
  • breath rate, b=21\ breath.min^{-1}

A)

Now,

amount of heat needed to warm the air of lungs to the body temperature:

Q_a=m'.c_a.\Delta T

Q_a=1.3\times1020\times (310-254)

Q_a=74256\ J

B)

Amount of heat lost per hour:

<u>No. of breaths per hour:</u>

B=b.60

B=21\times 60

B=1260

<u>Now the total loss of energy in 1 hr.:</u>

Q=Q_a.B

Q=74256\times 1260

Q=93562560\ J

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<u>Explanation</u>:

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                     v^{2} = (1.10 \times 10^42 \times 2) / (3.26 \times 10^31)

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eimsori [14]

The force of friction is 19.1 N

Explanation:

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m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

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Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

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F cos \theta is the horizontal component of the applied force, with

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mamaluj [8]

Answer:

The  charge on the dust particle is  q_d  = 6.94 *10^{-13} \  C

Explanation:

From the question we are told that

    The length is  l = 2.0 \ m

    The width is  w = 4.0 \ m

   The charge is  q =  -10\mu C= -10*10^{-6} \ C

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Generally the electric field on the carpet is mathematically represented as

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substituting values

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Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        F__{E}} =  F__{G}}

=>     q_d *  E  =  m * g

=>      q_d  =  \frac{m * g}{E}

=>      q_d  =  \frac{5.0 *10^{-9} * 9.8}{70621.5}

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Explanation:

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where:

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R is the radius of the planet

Here we want to find the acceleration of gravity on the surface of Moon B, which has the following data:

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4 0
2 years ago
Recall that the essential matrix E describing the stereo geometry of two calibrated cameras is a function only of the relative r
Mashcka [7]

Answer:

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Explanation:

The explanation is shown on the first uploaded image

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