1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Alex Ar [27]
3 years ago
13

(An easy problem which will be graded). Later in the quarter we will spend some time solving the diffusion equation Op(r, t) = D

V4p(r, t), at where D is called the diffusion constant. For now, let us consider a solution which represents a steady state solution (i.e apat = 0) and which has spherical symmetry. In that case the above reduces to 1d (20p) = 0 p2 dr dr / a) Obtain the solution of this second-order differential equation. It will have two arbitrary constants. b) Application: diffusion from a large distance to a spherical absorber. Consider p(r) to represent the density of some particles which are diffusing through some media. Think of them as food for a cell. Their density at large distance (infinity) is po. There is a spherical absorber of radius R centered at the origin, which means p(R) = 0. (This absorber represents the cell.) Show that the particular solution that satisfies these two condition is p(r) = po(1 – R/r). An interesting result: The current density (the number of particles passing through unit area per unit time) is given by j(r) = -Ddp/dr = -DpoR/r2. (The minus sign means that the flow is inward.) The magnitude of the current at a distance r is I = 47r2|j(r)= 47 DPOR. Note that this result says that the rate at which particles are absorbed is proportional to the radius of the sphere, not to the area of the sphere as would be the case if we were dealing with light being absorbed. Of course light does not travel by diffusion!

Physics
1 answer:
alexandr402 [8]3 years ago
7 0

Answer:

Explanation:

Answer is in the attachment below:

You might be interested in
An object is located 10.0 cm from a convex mirror. The magnitude of the
Sunny_sXe [5.5K]

Answer:

B. 6.00 cm

Explanation:

6 0
2 years ago
At which position is the LOWEST potential energy?
Ad libitum [116K]
The answer is position 3, because it is at its lowest point.

Potential Energy is “stored energy.” It is energy that is ready to be converted or released as another type of energy. We most often think of potential energy as gravitational potential energy. When objects are higher up, they are ready to fall back down. When you stretch an object and it has a tendency to return to its original shape, it is said to have elastic potential energy. Chemical potential energy is the stored energy in a substance’s chemical structure that can be released in a chemical reaction or as heat.

7 0
2 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
A natural resource that is in limited supply is known as a _________________.
goblinko [34]

Answer:

Limited Resources

6 0
3 years ago
Read 2 more answers
What is one way to describe elements
Talja [164]

Answer:

They are the smallest pancise of any matere

4 0
3 years ago
Other questions:
  • If a person and wheelchair have a combined weight of 185 lb, how much ideal effort
    5·1 answer
  • un motor electric efectueaza un lucru mecanic de 864 j in 0.5 min.tensiunea la bornele sale este de 12v si este parcurs de un cu
    7·1 answer
  • 3. Name and describe 3 things that can occur when light hits and object. Provide an example of each.
    11·1 answer
  • What is the difference between the velocity and the speed of an object?
    6·2 answers
  • a car is rolling backward when it hits the gas. after 8.25 s it is moving forward at 8.62m/s, and is 12.9m to the right of its s
    15·2 answers
  • What is the speed of a truck that travels 60 km in 20 minutes
    12·1 answer
  • 1. The unit of power is called a derived unit, why?
    9·1 answer
  • Add these two velocity vectors to find the magnitude of their resultant vector.
    6·1 answer
  • In general, a _______ intake manifold produces high torque at _______ engine rpm because of the increased amount of time between
    15·1 answer
  • Each of the gears a and b has a mass of 675 g and has a radius of gyration of 40 mm, while gear c has a mass of 3. 6 kg and a ra
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!