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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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Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a
suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N

Part (b)

the magnitude of each charge, if they have equal magnitude

F = \frac{KQ^2}{r^2}

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q =  \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C

4 0
3 years ago
What are some practical applications for determining the motion of an object?
Mama L [17]
- rocket science
- automotive research
- space research

4 0
3 years ago
A hoop (I = MR2) of mass 3 kg and radius 1.1 m is rolling at a center-of-mass speed of 11 m/s. An external force does 842 J of w
Scilla [17]

Answer:

v_f = 20 m/s

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2

now for pure rolling condition we will have

v = R\omega

also we have

I = mR^2

now we will have

KE = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2}

KE = mv^2

now by work energy theorem we can say

W = KE_f - KE_i

842 J = mv_f^2 - mv_i^2

842 = 3(v_f^2) - 3\times 11^2

now solve for final speed

v_f = 20 m/s

3 0
3 years ago
The counter is 1.5 m high, and the bowl has a mass of 0.5 kg. How much gravitational energy is stored in the bowl-earth system?
alina1380 [7]

Answer:

7.35 J

Im assuming, upon answering the question, that the gravity in this scenario is 9.8? As 9.8 is the gravitational force upon the earth.

5 0
2 years ago
Please help me find the equations guys
Alexeev081 [22]

The line at the bottom of the picture ... probably the first line on a list of choices  .. is the correct equation.

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