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shepuryov [24]
2 years ago
11

When you think of the word "respiration," you might think about the process of breathing, which is actually called ventilation.

(The respiratory system consists of the windpipe, lungs, etc.) How is breathing related to cellular respiration?
Physics
1 answer:
Alenkinab [10]2 years ago
6 0

Answer:

Breathing and cellular respiration are complementary processes that enable the body to produce energy by taken in oxygen which is required for the chemicals contained in food to be broken down there by producing, energy, water and carbon dioxide. The breathing and cellular respiration process also enables the removal of the produced carbon dioxide finally through nose and/or mouth

Explanation:

In cellular respiration, glucose molecules in the presence of oxygen gas are broken down into carbon dioxide and water aerobically in living cells, to release energy and produce ATP as follows;

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

During breathing, oxygen is inhaled into the lungs from the atmosphere  and carbon dioxide is exhaled from the longs into the atmosphere, such that the carbon dioxide produced during cellular respiration is transported out of the body through the veins respiratory system, from where is passes out through the nose, while oxygen used in cellular respiration comes from breathing in oxygen into the respiratory system

The oxygen is then transported to the cells through by blood in the blood vessels of the circulatory system to the cells, where the cells use the oxygen for cellular respiration to release energy.

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Answer:

\dot{m_{2}}=0.865 kg/s

Explanation:

\dot{m_1}= 0.5kg/s

from steam tables , at 250 kPa, and at

T₁ = 80⁰C ⇒ h₁ = 335.02 kJ/kg

T₂ = 20⁰C⇒ h₂ = 83.915 kJ/kg

T₃ = 42⁰C ⇒ h₃ = 175.90 kJ/kg

we know

\dot{m_{in}}=\dot{m_{out}}

\dot{m_{1}}+\dot{m_{2}}=\dot{m_{3}}

according to energy balance equation

\dot{m_{in}}h_{in}=\dot{m_{out}}h_{out}

\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}

\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=(\dot{m_{1}}+\dot{m_{2}})h_{3}\\(0.5\times 335.02)+(\dot{m_{2}}\times 83.915)=(0.5+\dot{m_{2}})175.90\\\dot{m_{2}}=0.865 kg/s

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You run<br> completely around a 400m track in<br> 80s. What was your average velocity?
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A speaker is designed for wide dispersion for a high frequency sound. What should the diameter of the circular opening be for a
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To solve this problem we will apply the concepts related to wavelength, as well as Rayleigh's Criterion or Optical resolution, the optical limit due to diffraction can be calculated empirically from the following relationship,

sin\theta = 1.22\frac{\lambda}{d}

Here,

\lambda = Wavelength

d= Diameter of aperture

\theta = Angular resolution or diffraction angle

Our values are given as,

\theta = 11\°

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The speed of the sound is v = 343 m/s

The wavelength of the sound is

\lambda = \frac{v}{f}

Here,

v = Velocity of the wave

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Replacing,

\lambda = \frac{(343 m/s)}{(9100 Hz)}

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