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3 years ago

What must the condition of the alveoli be in order for gas exchange to occur

Physics

1 answer:

aliya0001 [1]3 years ago

8 0

Answer:

There are 5 main conditions of the alveoli for proper gaseous exchange to in the lungs.

Explanation

Alveoli is found in the lungs of mammals, birds and reptiles. It is the part of the lungs where gaseous exchange really occurs.

Alveoli has a shape of that of balloon but indeed smaller than the real balloon. It allows the passage of oxygen and carbon dioxide

The five main conditions includes:

The Alveoli must have large surface to volume ratio which increase the gases that could be exchanged
The wall must be thin. This can shorten diffusion distance.
Alveoli must be very moist so that oxygen and carbon dioxide can pass through the solution
Alveoli must be well supplied with blood
Alveoli must be permeable

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Unpolarized light passes through two polarizers whose transmission axes are at an angle # with respect to each other. What shoul

Yuki888 [10]

Answer:

$63.4^{\circ}$

Explanation:

When unpolarized light passes through the first polarizer, the intensity of the light is reduced by a factor 1/2, so

$I_1 = \frac{1}{2}I_0$ (1)

where I_0 is the intensity of the initial unpolarized light, while I_1 is the intensity of the polarized light coming out from the first filter. Light that comes out from the first polarizer is also polarized, in the same direction as the axis of the first polarizer.

When the (now polarized) light hits the second polarizer, whose axis of polarization is rotated by an angle $\theta$ with respect to the first one, the intensity of the light coming out is

$I_2 = I_1 cos^2 \theta$ (2)

If we combine (1) and (2) together,

$I_2 = \frac{1}{2}I_0 cos^2 \theta$ (3)

We want the final intensity to be 1/10 the initial intensity, so

$I_2 = \frac{1}{10}I_0$

So we can rewrite (3) as

$\frac{1}{10}I_0 = \frac{1}{2}I_0 cos^2 \theta$

From which we find

$cos^2 \theta = \frac{1}{5}$

$cos \theta = \frac{1}{\sqrt{5}}$

$\theta=cos^{-1}(\frac{1}{\sqrt{5}})=63.4^{\circ}$

6 0

3 years ago

Mike is a runner. Mike runs 200m in 25s; what is his average velocity during the run? assume that units are in m/s. (asap, will

maw [93]

Velocity is displacement/time
(Displacement is the overall change in distance)

So you’ll want to divide 200 by 25, which should give you:

8 m/s

5 0

3 years ago

You are designing a delivery ramp for crates containing exercise equipment. The 1890 N crates will move at 1.8 m/s at the top of

Mashcka [7]

Answer:

K = 588.3 N/m

Explanation:

From a forces diagram, and knowing that for the maximum value of K, the crate will try to rebound back up (Friction force will point downward):

Fe - Ff - W*sin(22) = 0 Replacing Fe = K*X and then solving for X:

$X = \frac{Ff + W*sin(22)}{K}=\frac{1223}{K}$

By conservation of energy:

$\frac{K*X^{2}}{2}-mg*d*sin(22)-\frac{m*V^{2}}{2}=-Ff*d$

Replacing our previous value for X and solving the equation for K, we get maximum value to prevent the crate from rebound:

K = 588.3 N/m

6 0

3 years ago

A vibrating tuning fork is held over a water column with one end closed and the other open. As the water level is allowed to fal

DiKsa [7]

Answer:

630 Hz.

Explanation:

As we are considering the one end open pipe. So for the sound wave there will be a pressure node at the open end of the tube as at that place the molecules can not move back and forth. However on the closed end there will be a flow node as the water molecules their are moving back and forth. So it will produces the resonance at the positions 1/4, 3/4.......

we can find the wavelength by multiplying the levels distance by 2.

λ = 2 × 0.27 m = 0.54

f = Vs/λ

= 340/0.54

= 630 Hz

6 0

3 years ago

A 61.5 KG student sits at a desk 1.2 5M away from a 70.0 KG student. What is the magnitude of the gravitational force between th

muminat

mass of two students are

$m_1 = 61.5 kg$