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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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A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on th
ira [324]

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a)  Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10⁶ N.m²/C

8 0
3 years ago
The force that probably propels the movement of the earth s gigantic rock plates is
il63 [147K]

The correct answer is:

c. convection.

The heating of magma and the continuous cycle of evolution of the magma creating a convection current is the reason for the evolution of Earths tectonic plates.

Explanation:

Tectonic plates are ready to move because the Earth's lithosphere has higher strength than the underlying asthenosphere. Lateral density changes in the mantle appear in convection. Plate movement is believed to be driven by a succession of the motion of the seafloor apart from the extended ridge (due to variations in topography and density of the crust.

4 0
3 years ago
The table below shows some common animals and their hearing range in Hertz (Hz). Animal Hearing Range in Rate of Vibrations per
kap26 [50]

Answer:

D

Explanation:

If you list and compare, the elephants hearing range is the closest to a humans.

Humans: 20-20,000

Elephant: 16-12,000

While the others are either too much more or less than a humans.

Bat: 2,000-110,000

Dolphin: 90-105,000

Chicken: 125-2,000

5 0
3 years ago
Which best describes a difference between laser light and regular light?
Mashcka [7]

Answer:

B

Explanation:

6 0
3 years ago
The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from
grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

\epsilon=-N\dfrac{d\phi}{dt}

Where

\dfrac{d\phi}{dt} is the rate of change of magnetic flux,

And \phi=BA

\epsilon=-NA\dfrac{dB}{dt}

\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}

\epsilon=28.32\ V

Using Ohm's law, \epsilon=I\times R

Induced current, I=\dfrac{\epsilon}{R}

I=\dfrac{28.32}{1.5}

I = 18.88 A

So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

5 0
3 years ago
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