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OLEGan [10]
3 years ago
7

Simple diffusion and facilitated diffusion are related in that both

Physics
1 answer:
kaheart [24]3 years ago
4 0

Answer:

allow the downward movement of the concentration gradient by passive transport

Explanation:

Passive transport is a process of substance transport, which is carried out spontaneously, without energy expenditure and in favor of the concentration gradient, that is, from a medium where the molecules are more concentrated towards a medium where their concentration is lower.

Three types of passive transport are distinguished: osmosis, simple diffusion and facilitated diffusion

<u>Simple diffusion</u>

It is the passage, through the plasma membrane, of small molecules without charge soluble in the lipid bilayer, such as some gases (oxygen and carbon dioxide). For a molecule to diffuse through the membrane it is necessary that there is a difference in concentration between the external and the internal environment.

<u>Diffusion facilitated </u>

There are molecules such as amino acids, glucose and small ions that, due to their chemical and size characteristics, cannot diffuse through the lipid bilayer and require transport proteins for diffusion.

The transport proteins are immersed in the plasma membrane and can be of two types: protein channels, formed by proteins that generate a channel in the membrane, and permeases, which are proteins that, when joined to the molecule to be transported, change their shape by carrying them into the cell.

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Consider a projectile launched with an initial velocity of v0 = 120 ft/s, inclined at an angle, θ with the horizontal. Let us as
Natali [406]

Answer:

How to find the maximum height of a projectile?

if α = 90°, then the formula simplifies to: hmax = h + V₀² / (2 * g) and the time of flight is the longest. ...

if α = 45°, then the equation may be written as: ...

if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion.

Explanation:

4 0
3 years ago
An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to
mafiozo [28]

Answer:

    wo = 18.75 rev / s

Explanation:

This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²

         w_f = w₀ - a t

         w₀ = w_f + a t

         w₀ = 109 + 4.7 1.87

         w₀ = 117.8 rad / s

let's reduce to revolutions / s

         w₀ = 117.8 rad / s (1 rev / 2pi rad)

         w₀ = 18.75 rev / s

8 0
3 years ago
Can anyone pls help me with this I’ll appreciate it. I’m trying to study for a test but I don’t know the answer to this.
forsale [732]

Answer:

okay so what you will do is where is says red giant you will write all about what it talks about red giants only, and the box plantary nebulas you will write about what is says about only planetary nebulas. x- hope this helps :)

Explanation:

7 0
3 years ago
Mr. hitch taught us about sedimentary, metamorphic, and igneous rocks. he described how they were formed, what they contain, and
Tems11 [23]
Mr. Hitch taught us about sedimentary, metamorphic, and igneous rocks. He described how they were formed, what they contain, and showed us samples of each. He is a good geologist. 

The missing word and answer is: geologist.
3 0
3 years ago
Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl
melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar

The boundary layer thickness is equal to:

\delta=\frac{4.91*1}{Re^{0.5} }  =\frac{4.91*1}{246507^{0.5} } =0.0098 ft

The shear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{1}  )(246507)^{0.5} =0.012

b) If the railing edge is 2 ft, the Reynold number is:

Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

The boundary layer is equal to:

\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

The sear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{2}  )(493015.6^{0.5} )=0.0082

c) The drag coefficient is equal to:

C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

The friction drag is equal to:

F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf

7 0
3 years ago
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