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shepuryov [24]
3 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]3 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.

You might be interested in
B. A 20kg wagon is pulled up a 30° incline at an acceleration of 2.5ms?. The force pulling the wagon is parallel to the incline
yawa3891 [41]

The value of the coefficient of kinetic friction between the wagon and inclined surface is 0.78.

<h3>Coefficient of the kinetic friction</h3>

The value of coefficient of kinetic friction is calculated as follows;

F - Ff = ma

F - μmgcosθ = ma

where;

  • F is applied force
  • μ is coefficient of kinetic friction
  • m is mass of the wagon
  • a is acceleration of the wagon

182 - μ(20 x 9.8 x cos30) = 20(2.5)

182 - 169.74μ = 50

182 - 50 = 169.74μ

132 = 169.74μ

μ = 132/169.74

μ = 0.78

Thus, the value of the coefficient of kinetic friction between the wagon and inclined surface is 0.78.

Learn more about coefficient of friction here: brainly.com/question/20241845

7 0
3 years ago
Can anyone solve this?
love history [14]

Answer:

F = 39.2 N

Explanation:

Since, the object is in uniform motion. Therefore, the frictional force on object will be:

Frictional Force = μk N = μk mg

where,

μk = coefficient of kinetic friction = 0.2

m = mass of crate = 10 kg

g = 9.8 m/s²

Therefore,

Frictional Force = (0.2)(10 kg)(9.8 m/s²)

Frictional Force = 19.6 N

The horizontal component of force must be equal to this frictional force to continue the uniform motion:

F Sin 30° = 19.6 N

F = 19.6 N/Sin 30°

<u>F = 39.2 N</u>

3 0
3 years ago
An amateur golfer swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for o
Rudiy27

Answer:

The average force ≅ 519.44 N.

Explanation:

Impulse = change in momentum of a body

i.e Ft = m(v - u)

where F is the force, t is the time, m is the mass of the body, v is the final velocity and u is the initial velocity.

m = 55.0 g (0.055 Kg), t = 0.00360 s, v = 34.0 m/s, since the ball was initially at rest; u = 0 m/s

So that,

F x 0.00360 = 0.055(34 - 0)

F x 0.00360 = 0.055 x 34

                    = 1.87

F = \frac{1.87}{0.0036}

 = 519.4444

The average force exerted on the ball by the club is approximately 519.44 N.

4 0
3 years ago
A force of 385 N is applied in pushing a stalled automobile at a constant speed for a distance of 150 m. How much work (in J) wa
Helga [31]

Answer:57,750J

Explanation:

3 0
2 years ago
PLEASE HELP!
lutik1710 [3]

Answer:

Thats her fault.........................b

Explanation:

6 0
3 years ago
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