Ask question

Ask question

All categories

3 years ago

8

The random thermal motion of molecules from a region of higher concentration to a region of lower concentration is called diffus

ion. The rate of diffusion across a barrier is defined by Fick’s law of diffusion:_________

1 answer:

Makovka662 [10]3 years ago

8 0

Answer:

$\frac{dq}{dt} =-DA\frac{dc}{dx}$

Explanation:

Fick's law of diffusion describes the times course of transfer of matter from a region of high concentration to a region of lower concentration separated by a thin barrier membrane.

Mathematically:

$\frac{dq}{dt} =-DA\frac{dc}{dx}$

where:

q = quantity of solute
t = time
A = membrane surface area
c = concentration
D = diffusion coefficient
dx = membrane thickness
$\frac{dc}{dx}$ = concentration gradient

You might be interested in

The only force acting on a 3.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a

bekas [8.4K]

Answer:

The work done on the canister by the 5.0 N force during this time is

54.06 Joules.

Explanation:

Let the initial kinetic energy of the canister be

KE₁ = $\frac{1}{2} mv_1^{2}$ = $\frac{1}{2} *3*3.6^{2}$ = 19.44 J in the x direction

Let the the final kinetic energy of the canister be

KE₂ = $\frac{1}{2} mv_2^{2}$ = $\frac{1}{2} *3*7.0^{2}$ = 73.5 J in the y direction

Therefore from the Newton's first law of motion, the effect of the force is the change of momentum and the difference in energy between the initial and the final

= 73.5 J - 19.44 J = 54.06 J

3 0

3 years ago

Read 2 more answers

SP 15The magnetic field in a region of space is measured to be:This field is known to be caused by a cluster of long-straight wi

tino4ka555 [31]

Answer:

i = 0.477 10⁴ B

the current flows in the counterclockwise

Explanation:

For this exercise let's use the Ampere law

∫ B . ds = μ₀ I

Where the path is closed

Let's start by locating the current vines that are parallel to the z-axis, so it must be exterminated along the x-axis and as the specific direction is not indicated, suppose it extends along the y-axis.

From BiotSavart's law, the field must be perpendicular to the direction of the current, so the magnetic field must go in the x direction.

We apply the law of Ampere the segment parallel to the x-axis is the one that contributes to the integral, since the other two have an angle of 90º with the magnetic field

Segment on the y axis

L₀ = (y2-y1)

L₀ = 3-0 = 3 cm

Segment on the point x = 2 cm

L₁ = 3-0

L₁ = 3cm

B L = μ₀ I

B 2L = μ₀ I

i = 2 L B /μ₀

i= 2 0.03 / 4π 10⁻⁷ B

i = 4.77 10⁴ B

The current is perpendicular to the magnetic field whereby the current flows in the counterclockwise

8 0

3 years ago

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

3 years ago

Solve for the BMI weight 58kg Height 1.61

vaieri [72.5K]

Answer:

Explanation:

BMI= weight/(height × height) ; weight in kilogram and height in metter

= 58kg / (1.61m × 1.61m )

= (58/ 2.5921) kg/ $m^{2}$

= 22.375 kg/ $m^{2}$

≈ 22.4 kg/ $m^{2}$

7 0

2 years ago

Assume that the blocks are accelerating, and the x components of their accelerations at a certain moment are a1x and a2x. find t

Serga [27]

we know that center of mass is given as

r = (m₁ $r_{1x}$ + m₂ $r_{2x}$ )/(m₁ + m₂)

taking derivative both side relative to "t"

dr/dt = (m₁ d $r_{1x}$ /dt + m₂ d $r_{2x}$ /dt)/(m₁ + m₂)

v = (m₁ $v_{1x}$ + m₂ $v_{2x}$ )/(m₁ + m₂)

taking derivative again relative to "t" both side

dv/dt = (m₁ d $v_{1x}$ /dt + m₂ d $v_{2x}$ /dt)/(m₁ + m₂)

a= (m₁ $a_{1x}$ + m₂ $a_{2x}$ )/(m₁ + m₂)

3 0

3 years ago

Read 2 more answers