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The mass of the solid with a volume of 1.68 ft and density of 9.2g/ml is 14.85 kilograms.

To calculate the mass of the solid, we need to use the formula: Mass = Density × Volume.

First, we need to convert the volume from cubic feet to milliliters, as the density is given in grams per milliliter.

1 cubic foot is equal to 28,316.8466 milliliters (ml). So, the volume of the solid is 1.68 ft^3 × 28,316.8466 ml/ft^3 = 47,594.768 ml.

Now, we can calculate the mass by multiplying the density (9.2 g/ml) by the volume (47,594.768 ml).

Mass = 9.2 g/ml × 47,594.768 ml = 437,186.38 grams.

Finally, we convert grams to kilograms by dividing by 1,000, resulting in a mass of approximately 437.19 kilograms or rounded to 14.85 kilograms.

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We can deduce here that our galaxy, the Milky Way is about 100 million times larger than the solar system.

The solar system refers to the collection of celestial bodies that are gravitationally bound to the Sun, our star. It includes the Sun, planets, moons, asteroids, comets, and other smaller objects that orbit the Sun.

The solar system formed about 4.6 billion years ago from a rotating cloud of gas and dust called the solar nebula. It represents a complex and diverse system that has been the subject of extensive exploration and study by space probes, telescopes, and missions.

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In 10 billion years, the peak of the spectrum emitted from the cosmic microwave background radiation (CMB) will shift to longer wavelengths.

What is cosmic microwave background radiation (CMB)?

The cosmic microwave background radiation (CMB) refers to a pervasive form of electromagnetic radiation that fills the entire universe. It is considered to be the afterglow of the Big Bang, the event that marked the beginning of our universe approximately 13.8 billion years ago.

This phenomenon is known as cosmological redshift. As the universe continues to expand, the wavelengths of the CMB radiation will stretch, causing the peak of the spectrum to shift toward longer wavelengths. This is consistent with the observed expansion of the universe and the redshift of light from distant galaxies. Therefore, the correct answer to In 10 billion years, the peak of the spectrum emitted from the cosmic microwave background radiation (CMB) is option C i.e. shift to longer wavelengths.

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If the resistors are in series, it's 75 volts.

If they're in parallel, it's 150 volts.

You never told us series or parallel.

Answer:

1) four lines

2) three lines

3) fives lines

4) six lines

Answer:

4

Explanation:

Static friction keeps a sled weighing 200 N in place on a 15.

(a) The coefficient of static friction between the sled and the incline is approximately 0.27.

(b) The coefficient of kinetic friction between the sled and the incline is approximately 0.443.

To solve this problem, we'll use the following formulas:

For static friction:

[tex]\[F_\text{static friction} = \mu_s \cdot N\][/tex] = μ_s * N

For kinetic friction:

[tex]\[F_\text{kinetic friction} = \mu_k \cdot N\][/tex]

Where:

[tex]\[F_\text{static friction}[/tex] is the force of static friction,

[tex]\[F_{\text{kinetic friction}}[/tex] is the force of kinetic friction,

[tex]\[\mu_s\][/tex] is the coefficient of static friction,

[tex]\[\mu_k\][/tex] is the coefficient of kinetic friction, and

N is the normal force.

(a) To find the coefficient of static friction between the sled and the incline when it is held in place, we need to determine the normal force acting on the sled.

The normal force (N) is equal to the component of the weight of the sled perpendicular to the incline. In this case, the incline is at an angle of 15 degrees, so the normal force can be calculated as:

N = mg * cos(theta)

where m is the mass of the sled and g is the acceleration due to gravity (approximately 9.8 m/s²).

Given that the weight of the sled is 200 N, we can find its mass (m) using the formula:

weight = mass * gravity

200 N = m * 9.8 m/s²

Solving for m:

[tex]m = \frac{200 N}{9.8 m/s^2} \approx 20.41 kg[/tex]

Now, we can calculate the normal force:

N = 20.41 kg * 9.8 m/s² * cos(15 degrees)

N ≈ 195.43 N

Next, we can use the formula for static friction to find the coefficient of static friction ([tex]\ensuremath{\mu s}[/tex]):

[tex]F_\text{static friction} = \mu_s \cdot N[/tex]

The force of static friction is equal to the component of the weight of the sled parallel to the incline, which is given by:

[tex]F_\text{parallel} = mg \cdot \sin(\theta)[/tex]

[tex]F_parallel = 20.41 kg * 9.8 m/s² * sin(15 degrees)[/tex]

[tex]F_parallel[/tex] ≈ 52.87 N

Since the sled is held in place, the force of static friction is equal to the force parallel to the incline:

[tex]F_static_friction[/tex] = 52.87 N

Plugging this into the formula:

52.87 N = [tex]\ensuremath{\mu s}[/tex] * 195.43 N

Solving for [tex]\ensuremath{\mu s}[/tex]:

[tex]\begin{equation}\mu_s = \frac{52.87\text{ N}}{195.43\text{ N}} \approx 0.27\end{equation}[/tex]

Therefore, the coefficient of static friction between the sled and the incline is approximately 0.27.

(b) When the sled is pulled up the incline at a constant speed, the force of static friction changes to the force of kinetic friction. The force of kinetic friction is given by:

[tex]\begin{equation}F_\text{kinetic friction} = \mu_k N\end{equation}[/tex]

In this case, the force pulling the sled up the incline is 100 N, and the angle between the rope and the incline is 30 degrees. We can calculate the force parallel to the incline:

[tex]F_parallel = 100 N * cos(30 degrees) = 86.60 N[/tex]

To find the coefficient of kinetic friction ([tex]$\mu_k$[/tex]), we need to determine the normal force (N) acting on the sled.

The normal force can be calculated as before:

[tex]$N = mg \cos(\theta)$[/tex]

[tex]$N = 20.41\ \text{kg} \times 9.8\ \text{m/s}^2 \times \cos(15^\circ)$[/tex]

N ≈ 195.43 N

Now, we can plug in the values into the formula for kinetic friction:

86.60 N = [tex]$\mu_k$[/tex] * 195.43 N

[tex]\[\mu_k = \frac{86.60 \text{ N}}{195.43 \text{ N}} \approx 0.443\][/tex]

Therefore, the coefficient of kinetic friction between the sled and the incline is approximately 0.443.

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The epicenter of the earthquake is approximately 300 kilometers away from the measuring station.

How to solve for the distance

To determine the distance to the epicenter of the earthquake, we can use the formula:

Distance = Velocity × Time

First, let's calculate the time it took for the P-waves to reach the measuring station:

Time (P-wave) = Distance / Velocity (P-wave) = ? / 15 km/s

Next, we'll calculate the time it took for the S-waves to reach the measuring station:

Time (S-wave) = Distance / Velocity (S-wave) = ? / 10 km/s

Given that there is a time delay of 10 seconds between the arrival of the P-waves and S-waves, we can set up the following equation:

Time (S-wave) - Time (P-wave) = 10 seconds

Now, let's substitute the formulas for time and solve for distance:

(Distance / 10 km/s) - (Distance / 15 km/s) = 10 seconds

To simplify the equation, we can find the common denominator, which is 30 km/s:

[(3 * Distance) - (2 * Distance)] / (30 km/s) = 10 seconds

Distance / (30 km/s) = 10 seconds

Multiplying both sides of the equation by 30 km/s:

Distance = 10 seconds * 30 km/s

Distance = 300 kilometers

Therefore, the epicenter of the earthquake is approximately 300 kilometers away from the measuring station.

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Therefore, the electric field (magnitude only) at a distance of 8.0 cm from the wire is approximately 3.24 x 10^4 N/C.

The electric field of a long, thin wire can be determined by Coulomb's law. Coulomb's Law states that the electric force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

For a long, thin wire, the electric field is given by;

E = λ/2πε₀r

Where;

λ = linear charge density = 91 µC/m,

ε₀ = permittivity of free space = 8.85 x 10^-12 C^2/Nm^2

r = distance from the wire = 8.0 cm = 0.08 m.

Substitute the given values into the formula to find the electric field;

E = (91 x 10^-6)/(2 x π x 8.85 x 10^-12 x 0.08)

E≈ 32433.8 N/C

E≈ 3.24 x 10^4 N/C.

Electric field refers to the force per unit charge that one object exerts on another object due to the electric charge present in the objects. It is a vector quantity and is measured in newtons per coulomb (N/C).

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Answer:

Absorbed

Explanation:

Hope this helped!!!

A 3.00-kilogram mass is thrown vertically upward with an initial speed of 9.80 meters per second. [Neglect friction.]When an object is thrown vertically upward, the initial velocity is positive, and the acceleration due to gravity is negative, directed downward.

We can use the following formula to calculate the maximum height, also known as the maximum displacement, reached by the object:

[tex]v_f^2 = v_i^2 + 2ad[/tex]

where v_f is the final velocity, [tex]v_i[/tex] is the initial velocity, a is the acceleration, and d is the displacement. At the maximum height, the final velocity is zero, so we can simplify the equation to:

[tex]d = (v_f^2 - v_i^2) / (2a)[/tex]

Substituting the given values:

[tex]d = (0 - 9.80^2) / (2 x -9.81)d = 4.90 m[/tex]

Therefore, the maximum height reached by the object is 4.90 m.

Hence, the correct option is (2) 4.90 m.Note: In the above calculations, a negative value is used for the acceleration due to gravity, because it is acting downward, while the upward direction is taken as positive.

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Answer:

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

Explanation:

El mango experimenta un movimiento de caída libre, es decir, un movimiento uniformemente acelerado debido a la gravedad terrestre, despreciando los efectos de la viscosidad del aire y la rotación planetaria. Entonces, la velocidad final del mango, es decir, la velocidad con la que llega al suelo, se puede determinar mediante la siguiente fórmula cinemática:

[tex]v = v_{o}+g\cdot t[/tex] (1)

Donde:

[tex]v_{o}[/tex] - Velocidad inicial, en metros por segundo.

[tex]v[/tex] - Velocidad final, en metros por segundo.

[tex]g[/tex] - Aceleración gravitacional, en metros por segundo al cuadrado.

[tex]t[/tex] - Tiempo, en segundos.

Si sabemos que [tex]v_{o} = -75\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex] y [tex]t = 26\,s[/tex], entonces la velocidad final del mango es:

[tex]v = v_{o}+g\cdot t[/tex]

[tex]v = -75\,\frac{m}{s}+\left(-9.807\,\frac{m}{s} \right)\cdot (26\,s)[/tex]

[tex]v = -329.982\,\frac{m}{s}[/tex]

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

Answer:

because it get the energy from the heat and the cold mixed together

Explanation:

The force of friction is parallel to the ground surface and opposes the motion of the object. The force of friction acting on the object is determined by the equation:

f=f(k)FN

where FN is the normal force, f(k) is the coefficient of kinetic friction, and f is the force of friction acting on the object.

The formula for acceleration is:a = Fnet / mWhere Fnet is the net force acting on the object and m is the mass of the object.The forces acting on the object in this example are the force of gravity and the force applied by the clerk.

[tex]F_gravity = mg = (35.0 kg) (9.81 m/s^2) = 343.5 N[/tex]

The force applied by the clerk can be resolved into horizontal and vertical components:

[tex]F_applied_horiz = F_applied * cos(25.0) = (185.0 N) cos(25.0) = 166.8 NF_applied_vert = F_applied * sin(25.0) = (185.0 N) sin(25.0) = 78.9 N[/tex].

The normal force is equal and opposite to the force of gravity acting on the object:

[tex]FN = F_gravity = 343.5 N[/tex]

The force of friction acting on the object is:

[tex]f = f(k) * FN = (0.450) (343.5 N) = 154.6 N[/tex]

The net force acting on the object is:

[tex]Fnet = F_applied_horiz - f = 166.8 N - 154.6 N = 12.2 N[/tex]

The acceleration of the object is:

[tex]a = Fnet / m = 12.2 N / 35.0 kg = 0.349 m/s^2[/tex]

Therefore, the acceleration of the box is 0.349 m/s².

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Answer:

1.heat is a form of 1.temperature is a form

energy that gives of energy that is used to

sensation of measure hotness or

warmth. or coldness of body.

2.its si unit is 2.its si unit is kelvin.

joule.

1. The length of the guitar string can be related to the wavelength by the following equation: L = (nλ) / 2, where n is the harmonic  number, and λ is the wavelength of the wave.

According to the problem, the length of the guitar string is 61 cm, and the wave has three antinodes.

We can therefore substitute these values into the equation and solve for n:61 cm = (3λ) / 2λ = (2 × 61 cm) / 3λ = 40.7 cm (rounded to one significant figure)

Therefore, the wavelength is 40.7 cm (rounded to two significant figures).

2. The wavelength: We can now use the above value of λ and the formula

v = fλ to calculate the frequency of the wave.

However, the velocity of a wave in a string is given by the formula

v = √(T/μ), where T is the tension in the string and μ is its linear mass density (mass per unit length).  

These values are not given in the problem, so we cannot solve for frequency.

Instead, we can use another equation that relates the wavelength to the length of a string:λ = 2L / n,

where L is the length of the string and n is the harmonic number. Substituting the given values: L = 61 cm, n = 3λ = 40.7 cm (from part a).

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Answer:

6

Explanation:

120/30=6

The uncertainty of energy of the gamma ray is 3.34 x 10⁻²⁵MeV.

Energy of the gamma ray emitted by the nucleus, E = 0.511 MeV

Lifetime of the nucleus, Δt = 1 ns = 10⁻⁹s

a) The expression for the uncertainty of energy of the gamma ray is given by,

ΔE = h/(2Δt)

ΔE = 6.67 x 10⁻³⁴/(2 x 10⁻⁹)

ΔE = 3.34 x 10⁻²⁵MeV

b) Detection of gamma rays is carried out photon by photon. By studying the impact they have on materials, gamma rays can be found.

Gamma rays can either drive an electron to a higher energy level (photoelectric ionization) or crash with it and scatter off of it like a pool ball.

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1. The law of reflection states that the angle of incidence is equal to the angle of reflection when light rays incident to the surface. This law applies to both smooth and rough surfaces. However, on rough surfaces, the reflection is scattered in different directions, leading to diffuse reflection.

2. Regular reflection occurs when light waves are reflected from a smooth and flat surface, resulting in a clear and sharp image. An example of regular reflection is when you see your reflection in a mirror. Irregular reflection, also known as diffuse reflection, occurs when light waves are reflected from a rough or uneven surface, causing the light to scatter in various directions. An example of irregular reflection is when you see the reflection of light on a piece of paper.

3. Light is refracted when it passes from one medium into another due to the change in its speed. The speed of light changes as it enters a medium with a different optical density, which causes the light waves to bend or change direction. This bending of light is understood as refraction.

4. An image created by reflection can be either real or virtual. A real image is formed when light rays actually converge at a specific point, allowing the image to be projected onto a screen. Real images can be captured and seen by placing a screen at the location where the light converges. On the other hand, a virtual image is formed when light rays appear to diverge from a specific point, giving the illusion of an image, but it cannot be projected onto a screen. Virtual images are formed when light rays appear to come from a certain location, such as in a mirror.

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Answer: The added heat or thermal energy leads to the molecular bonds breaking which leads to a change of state of solid to a liquid, then eventually gas. Solids melt when they absorb enough thermal energy.

The 170 g mass will need to be located 23.53 cm from the fulcrum to keep the system in balance.

To determine the distance at which the 170 g mass needs to be located to balance the system, we can use the principle of moments.

The principle of moments states that the sum of the clockwise moments about a point is equal to the sum of the counterclockwise moments about the same point.

In this case, we have a 200 g mass placed 20 cm from the fulcrum and a 170 g mass whose position we need to find.

Let's call the distance of the 170 g mass from the fulcrum x cm.

The moment of the 200 g mass is given by the product of its mass (0.2 kg) and its distance from the fulcrum (20 cm):

Moment1 = 0.2 kg × 20 cm

Moment1 = 4 kg·cm.

The moment of the 170 g mass will be equal and opposite to the moment of the 200 g mass to keep the system in balance:

Moment2 = -4 kg·cm.

We can express the moment of the 170 g mass in terms of its mass and its distance from the fulcrum:

Moment2 = (0.17 kg) × (x cm).

Setting the moments equal to each other, we have:

-4 kg·cm = (0.17 kg) × (x cm).

Solving for x, we find:

x cm = -4 kg·cm / (0.17 kg)

x cm ≈ -23.53 cm.

Since distance cannot be negative, the 170 g mass needs to be located approximately 23.53 cm from the fulcrum to keep the system in balance.

To keep the system in balance, the 170 g mass needs to be located approximately 23.53 cm from the fulcrum.

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The magnitude of the net gravitational force exerted on the spacecraft by the Earth and Moon is approximately 4.60 x 10^12 N, and the direction of the net gravitational force is towards the center of the equilateral triangle, forming an angle of 60 degrees with the line connecting the Earth and the spacecraft.

A. The magnitude of the net gravitational force exerted on the spacecraft by the Earth and Moon can be calculated using the formula for gravitational force:

Gravitational force (F) = G * ((m1 * m2) / r^2)

Where G is the gravitational constant (6.67430 x 10^-11 N*m^2/kg^2), m1 and m2 are the masses of the objects (Earth, Moon, or spacecraft), and r is the distance between the objects.

Given:

Mass of Earth (mE) = 5.972 × 10^24 kg

Mass of Moon (mM) = 7.348 × 10^22 kg

Mass of spacecraft (mS) = 1470 kg

Length of the sides of the equilateral triangle (s) = km = 1,000 m

To find the magnitude of the net gravitational force on the spacecraft, we need to consider the gravitational forces between the spacecraft and both the Earth and the Moon. Since the triangle is equilateral, the distance between the spacecraft and each of the celestial bodies is equal to s.

F_Earth = G * ((mE * mS) / s^2)

F_Moon = G * ((mM * mS) / s^2)

Net gravitational force (F_net) = F_Earth + F_Moon

B. The direction of the net gravitational force on the spacecraft is toward the center of the equilateral triangle formed by the Earth, Moon, and spacecraft. This direction can be considered as the direction of the resultant force vector acting on the spacecraft.

C. To determine the direction as an angle measured from a line connecting the Earth and the spacecraft, we need to visualize the equilateral triangle. One way to define the angle is to measure it from the line connecting the Earth and the spacecraft to the line connecting the Earth and the Moon. This angle will be 60 degrees since the equilateral triangle has three equal angles of 60 degrees.

D. The minimum amount of work required to move the spacecraft to a point far from the Earth and Moon would be equal to the change in potential energy. As the spacecraft moves far away, the potential energy decreases. The work done is given by the formula:

Work (W) = ΔPE = PE_final - PE_initial

Since the potential energy depends on the distance from the Earth and Moon, moving the spacecraft to a point far away where the gravitational influence is negligible would result in a significant decrease in potential energy. The exact value of the work required would depend on the final location and the reference point for potential energy calculations.

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Answer:

This is true.

The situation with the greatest magnitude of net force along the incline is when the cart is going uphill (Option 2).

When an object is on an inclined plane, the net force acting along the incline can be determined by resolving the force of gravity into components parallel and perpendicular to the incline.

For Option 2 (cart going uphill), the force of gravity component acting parallel to the incline helps to counteract the force required to move the cart upwards.

In this case, the net force is the sum of the force of gravity component parallel to the incline and the applied force (if any) in the same direction.

The magnitude of the net force (|F|) in this case can be calculated using the formula:

|F| = |mgsinθ + F_applied|

where m is the mass of the cart, g is the acceleration due to gravity, θ is the angle of the incline, and F_applied is any additional applied force.

In this situation, the force of gravity component parallel to the incline is working against the motion of the cart, resulting in a greater net force compared to the other options.

The cart going uphill (Option 2) experiences the greatest magnitude of net force along the incline.

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Answer: Radioactive decay is the breakdown of a material into stable isotopes which are used for determining the age of the ancient material.

Explanation:

The radioactive decay is a natural process in which an ancient or old material whether in the form of rock, object or fossil break down into elements. Carbon 14 is an unstable isotope which decays to produce stable elements, the dating procedure uses these stable elements and the rate of decay of the isotopes to determine the age of absolute ancient of the objects but exact age cannot be determined just an approximation can be accepted.

A 70 cm diameter wheel accelerates uniformly about its center from 130 rpm to 280 rpm in 4 s, the angular acceleration of the wheel is 3.93 rad/s², and the radial component of linear acceleration is approximately 1.375 m/s², and the tangential component is approximately 165.86 m/s².

(a) The angular acceleration of the wheel can be determined using the formula:
Angular acceleration (α) = (Final angular velocity - Initial angular velocity) / Time
Given:

Initial angular velocity (ω₁) = 130 rpm

Final angular velocity (ω₂) = 280 rpm

Time (t) = 4 s
First, we need to convert the angular velocities from rpm to radians per second (rad/s):
ω₁ = 130 rpm * (2π rad/1 min) * (1 min/60 s) = 13.61 rad/s

ω₂ = 280 rpm * (2π rad/1 min) * (1 min/60 s) = 29.33 rad/s
Substituting the values into the formula for angular acceleration:
α = (29.33 rad/s - 13.61 rad/s) / 4 s = 3.93 rad/s²
Therefore, the angular acceleration of the wheel is 3.93 rad/s².

(b) To determine the radial and tangential components of the linear acceleration of a point on the edge of the wheel after 2 s, we can use the following formulas:
Radial acceleration (ar) = r * α

Tangential acceleration (at) = r * ω²
Given:

Radius of the wheel (r) = 70 cm / 2 = 35 cm = 0.35 m

Angular acceleration (α) = 3.93 rad/s²

Angular velocity (ω) at t = 2 s can be found using the formula:

ω = ω₁ + α * t
Substituting the values:
ω = 13.61 rad/s + 3.93 rad/s² * 2 s = 21.47 rad/s
Now we can calculate the radial and tangential components of linear acceleration:

ar = r * α = 0.35 m * 3.93 rad/s² ≈ 1.375 m/s²

at = r * ω² = 0.35 m * (21.47 rad/s)² ≈ 165.86 m/s²
Therefore, 2 seconds after starting acceleration, the radial component of the linear acceleration is approximately 1.375 m/s², and the tangential component is approximately 165.86 m/s².

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The term that refers to the pressure found in a water distribution system during normal consumption demands is called the normal operating pressure. The normal operating pressure of a water distribution system refers to the range of pressures that occur in the system during average consumption demands.

It is measured in pounds per square inch (psi).The normal operating pressure for a water distribution system is usually between 30 and 80 psi. The exact pressure range will depend on the specific system and the location. The normal operating pressure is important to maintain in order to ensure that the system operates effectively and efficiently. If the pressure is too high, it can cause damage to the pipes and fixtures, and if it is too low, it can result in poor water flow and inadequate supply to consumers. Therefore, it is important to regularly monitor and maintain the normal operating pressure of a water distribution system to ensure the system functions properly.

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Answer: 2

Explanation:

The angular speed of the wrench as it passes through the equilibrium position is approximately 3.17 radians per second.

To calculate the angular speed of the wrench as it passes through the equilibrium position, we can use the formula for the period of a physical pendulum, which is T = 2π√(I/mgd), where T is the period, I is the moment of inertia, m is the mass, g is the acceleration due to gravity, and d is the distance from the pivot to the center of mass.

Given:

Mass of the wrench (m): 1.80 kg

Period of small-angle oscillations (T): 0.940 s

Displacement from equilibrium (θ): 0.400 rad

First, we need to find the moment of inertia (I) of the wrench. The correct answer provided is 0.099 kg·m^2.

Now, we can use the formula T = 2π√(I/mgd) to solve for the angular speed (ω).

Rearranging the formula:

T = 2π√(I/mgd)

√(I/mgd) = T / (2π)

I/mgd = (T / (2π))^2

ω = √(gd/I)

Substituting the given values:

g = 9.8 m/s^2 (acceleration due to gravity)

d = 0.250 m (distance from pivot to center of mass)

I = 0.099 kg·m^2 (moment of inertia)

ω = √(9.8 * 0.250 / 0.099) ≈ 3.17 rad/s

Therefore, the angular speed of the wrench as it passes through the equilibrium position is approximately 3.17 radians per second.

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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.