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2 years ago

Air is warmed and moisten as it moves through both the mouth and nasal passages.

Physics

2 answers:

Serggg [28]2 years ago

8 0

Answer: The given statement is true.

Transport of oxygen from the air takes place through respiratory system. Oxygen from air ( that is taken from the atmosphere) enters via a system of tubes, into our lungs and then it is released into the bloodstream ( to serve in various organs) whereas opposite route is followed by CO2 ( carbon dioxide).

In the respiratory tract, the nose is lined by mucosal epithelium that possessed glands ( which secrete thick mucus). Due to this, air is moistened. Beneath the epithelium surface, there lies network of blood vessels that help in warming the nasal cavity and thus the air is warmed.

Alexxx [7]2 years ago

3 0

Well, in my own personal case, I can actually move the air in either
direction through both of those.

When I move the air inward from the outside, it's warmed and moistened.

When I move it outward from the inside, it's not.

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In the process of making a rainbow, light enters the water droplet at location 1 and exits at location 3. Which

elena-s [515]

Answer: D, Refraction, Reflection, Refraction.

Explanation: Refraction is a rainbow. See image for info.

6 0

2 years ago

Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),

salantis [7]

Answer:

The drag coefficient is $D_z = 1.30512$

Explanation:

From the question we are told that

The density of air is $\rho_a = 1.21 \ kg/m^3$

The diameter of bottom part is $d = 0.15 \ m$

The power trend-line equation is mathematically represented as

$F_{\alpha } = 0.9226 * v^{0.5737}$

let assume that the velocity is 20 m/s

Then

$F_{\alpha } = 0.9226 * 20^{0.5737}$

$F_{\alpha } = 5.1453 \ N$

The drag coefficient is mathematically represented as

$D_z = \frac{2 F_{\alpha } }{A \rho v^2 }$

Where

$F_{\alpha }$ is the drag force

$\rho$ is the density of the fluid

$v$ is the flow velocity

A is the area which mathematically evaluated as

$A = \pi r^2 = \pi \frac{d^2}{4}$

substituting values

$A = 3.142 * \frac{(0.15)^2}{4}$

$A = 0.0176 \ m^2$

Then

$D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }$

$D_z = 1.30512$

3 0

2 years ago

A 32-n force acts on a small mass in the positive x-direction. a 26-n force also acts on it in the negative x-direction. what is

motikmotik

Equilibrium force is the force that will keep the small mass in place, hence no movement must be made. So we know that 32 N of force is acted towards the positive direction so +32 N. Which is counteracted by 26 N force so:

32 N – 26 N = 6 N (positive)

Since positive 6 is left, therefore this must be acted by an equilibrant negative 6 N.

Answer:

6 0

2 years ago

A ball rolls for 8 seconds and travels 24 meters. How fast was it traveling?

belka [17]

Answer:

The speed of the ball was, v = 3 m/s

Explanation:

Given data,

The time period of the ball, t = 8 s

The distance the ball rolled, d = 24 m

The velocity of an object is defined as the object's displacement to the time taken. The formula for the velocity is,

v = d / t m/s

Substituting the given values in the above equation,

v = 24 / 8

= 3 m/s

Hence, the speed of the ball was, v = 3 m/s

8 0

3 years ago

A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.

LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

$\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}$

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

$W = \frac{1}{2}(50)(5.61^2) = 786.8025 J$

Now, we can define work:

$\large\boxed{W = Fdcos\theta}}$

Now, plug in the values:

$786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta$

Solve for theta:

$cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}$

4 0

2 years ago