Answer:
a) μ = [kg / m s]
b) D = m²/s
Explanation:
Viscosity is equivalent to a friction force that opposes the movement of fluids, it is defined by the relation
Shear stress = μ speed gradient
Fr/A = μ v₀ / h
Where Fr is the shear force, A the area, h the height and v₀ the velocity of the fluid layer
μ = [kg / m s]
The diffusion coefficient characterizes the ability of a material to move in a given solvent, depending on the size of the solute, the viscosity of the solvent, temperature
J = - D dΘ / dx
J is the broadcast glare
D = [m² / s]
Mars is the focus of much scientific study about possible human colonization. Its surface conditions and the presence of water on Mars make it arguably the most hospitable of the planets in the Solar System, other than Earth. Mars requires less energy per unit mass (delta-v) to reach from Earth than any planet except Venus.
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Many organizations support the colonization of Mars. They have also given different reasons and ways humans can live on Mars. One of the oldest organizations is the Mars Society. They promote a NASA program that supports human colonies on Mars. The Mars Society have set up Mars analog research stations in Canada and the United States. All other organizations include MarsDrive, who wants to help fund settlements on Mars, and Mars to Stay. Mars to Stay advocates settlements on Mars. In June 2012, Mars One released a statement that they believe could help start a colony on Mars by 2023.
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Answer:
2 1/2 m/s^2 or 2.5m/s^2
Explanation:
From the question
final velocity v =25m/s
Initial velocity u =10m/s
time = 6 seconds
acceleration a = ?
Using the equation for linear motion
v = u + at
25 = 10 + a x 6
25 = 10 + a6
Subtract 10 from both sides
25-10 = 10 + a6 -10
25 - 10 = 10 -10 + a6
15 = a6
Divide both sides by 6
15/6 = a6/6
5/2 = a
a = 2 1/2 m/s^2 or 2.5m/s^2
Given the values to proceed to solve the exercise, we resort to the solution of the exercise through differential equations.
The problem can be modeled through a linear equation, in the form:

With the initial conditions as,


Where Q(t) is the charge.
<em>The general solution of a linear equation is given as:</em>
<em>
</em>
Applying this definiton in our differential equation we have that

To find b and a we use the first equation and find the roots:


Then we have

To find the values of the Constant we apply the initial conditions, then

And for the derivate:



We have a system of 2x2:


Solving we have:


The we can replace at the equation and we have that the Charge at any moment is given by,

If we obtain the derivate we find also the Current, then
