Consider an infinite lattice with coordination number z in which every site is occupied by a molecule. (As a reminder, the coordination number is the number of neighbors each site has). Each molecule can be in one of two states: an 'up' state, and a 'down' state. A system interacts only with adjancet particles that share its state (i.e. ups interact with ups and downs interact with downs). Each interaction of this kind contributes an interaction energy of -epsilon to the system's energy. Using a mean-field approach, compute the following. (a) The fraction of molecules in the up state as a function of reduced temperature kT/(epsilon*z) (b) The mean energy per site as a function of reduced temperature. Here are some hints. In doing this, you will want to use the perturbation (mean field) approach retaining only F0 and F1. To obtain F0, just compute the entropy of the lattice in the absence of interactions. To obtain F1, compute the energy that is then introduced to the system by the attractions, using the configurational probabilities from the reference (noninteracting state). For this second part, you will have to use the idea I introduced in class that in a non-interacting lattice the site probability is just given by a mole fraction. When you determine the fraction of molecules in the up state, the easiest way will be to use the criterion that F is at a minimum at equilibrium, i.e. dF/dX = 0. In this case, you will want to minimize with respect to the fraction of up or down sites. This equation will require some numerical solution near the end. Note also that below some temperature, there should be two solutions to this minimization, in terms of two possible equilibrium values of the mole fraction of up sites (or down sites). This leads to the final part of this problem: (c) Determine the reduced temperature below which two different fractions of 'up' molecules are possible at the same temperature.

Answer:

ice water is pure depending on the purity of the water that was frozen

Answer:

See explanation

Explanation:

Since no figure was given, I'll explain how to do this problem theoretically. The formula for shear flow is [tex]q=\frac{VQ}{I}[/tex] where V is the shear force, Q is the moment of area (more on this later), and I is the moment of inertia.

The first step to solve this problem is to find the resultant internal forces of the beam. This can be done in several ways, but the easiest is to solve the beam statically and draw a shear diagram to determine the maximum shear force V.

The second step to solving this problem is to determine the location of the neutral axis of the cross section if it is not given. The formula for the neutral axis is  [tex]NA = \frac{\sum y*A}{\sum A}[/tex]. The y in this equation represents the middle of the small shapes that the web is divided into. An I-beam can be thought of as 3 rectangles, while a T-beam can be thought of as 2. The A in this formula represents the area of each of the rectangles (an I-beam will have 3 of these and a T-beam will have 2).

The third step for this problem is to find the moment of inertia. There are several formulas for moment of inertia depending on the shape of the cross section. I-beam's and T-beams both can be thought of as multiple rectangles, so they have the same base formula of [tex]I=\frac{1}{12}bh^3[/tex] where b is the base of the rectangles and h is the height. For I-beams, the easiest way to calculate moment of inertia is to think of the entire cross section as a big rectangle that had two smaller rectangles cut out of it. The formula for this moment of inertia becomes [tex]I=\frac{1}{12} b_{big}h^{3} _{big}-\frac{1}{6}b_{small}h^{3}_{small}[/tex]. Note that this form of moment of inertia already takes into account subtracting 2 small rectangles. For T-beams, this approach will not work, so the parallel axis theorem must be used. The moment of inertia for the T-beam becomes [tex]I=\frac{1}{12}b_{1} h^{3}_{1} +b_{1}h_{1}dy_{1}^{2} +\frac{1}{12}b_{2} h^{3}_{2} +b_{2}h_{2}dy_{2}^{2}[/tex] where the terms with the subscript 1 represent the first rectangle and the terms with the subscript 2 represent the second rectangle. The dy terms represent the distance from the center of that specific rectangle to the neutral axis.

The fourth step for this problem is to find Q. The formula to find Q is [tex]Q=\sum y'A'[/tex] where y' represents the distance from the neutral axis to the center of the "wanted" point and A' is the area of the rectangle that has the wanted point at its center. (This would be the area above or below the thickness (t) if you were solving for maximum shear [tex]\tau=\frac{VQ}{It}[/tex]).

The last step for this problem is to substitute the found values into the formula for shear flow [tex]q=\frac{VQ}{I}[/tex]. V came from step 1, Q came from step 4, and I came from step 3.

Answer:

89375 N

Explanation:

Rearrange the formula for normal stress for F:

[tex]\sigma=\frac{F}{A}[/tex]

[tex]F=\sigma*A[/tex]

Convert given values to base units:

275 MPa = [tex]275*10^{6}[/tex] Pa

325 [tex]mm^{2}[/tex] = 0.000325 [tex]m^{2}[/tex]

Substituting in given values:

F = [tex](275*10^{6})*(0.000325)=89375[/tex] N

Answer: Plants and animals can be agents of mechanical weathering. The seed of a tree may sprout in soil that has collected in a cracked rock. As the roots grow, they widen the cracks, eventually breaking the rock into pieces. Over time, trees can break apart even large rocks.

I searched it, hope it helps! Have a great day!

Answer:

There are three common methods of charging a battery; constant voltage, constant current and a combination of constant voltage/constant current with or without a smart charging circuit.

Constant voltage allows the full current of the charger to flow into the battery until the power supply reaches its pre-set voltage.  The current will then taper down to a minimum value once that voltage level is reached.  The battery can be left connected to the charger until ready for use and will remain at that “float voltage”, trickle charging to compensate for normal battery self-discharge.

Constant current is a simple form of charging batteries, with the current level set at approximately 10% of the maximum battery rating.  Charge times are relatively long with the disadvantage that the battery may overheat if it is over-charged, leading to premature battery replacement.  This method is suitable for Ni-MH type of batteries.  The battery must be disconnected, or a timer function used once charged.

Constant voltage / constant current (CVCC) is a combination of the above two methods.  The charger limits the amount of current to a pre-set level until the battery reaches a pre-set voltage level.  The current then reduces as the battery becomes fully charged.  The lead acid battery uses the constant current constant voltage (CC/CV) charge method. A regulated current raises the terminal voltage until the upper charge voltage limit is reached, at which point the current drops due to saturation.

Answer:

Textile Technologists work with a variety of materials including man-made and natural textiles, leather, fur, plastic and metals. They may be responsible for developing fabrics for furnishings, clothing, household items, medical supplies, or textiles for use within the automotive industry.

The misfire monitor continuously monitors either crankshaft speed or possibly combustion chamber feedback that indicates misfiring cylinders.

The misfire monitor continuously monitors either crankshaft speed or possibly combustion chamber.

The misfire detection monitor serves as a term that describes the running of normal engine operating and driving conditions.

It is used by the PCM to determine the functionality of an engine and give feedback that indicates misfiring cylinders.

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The three methods used to find the real roots of the function are,

graphically, the quadratic formula, and by iteration.

The correct vales are;

(a) Graphically, the roots obtained are; x ≈ -1.629, and 5.629

(b) Using the quadratic formula, the real roots of the given function are; x ≈ -1.62589, x ≈ 5.62859

(c) Using three iterations, we have; the bracket is [tex]x_l[/tex] = 5.625, and [tex]x_u[/tex] = 6.25

Reasons:

The given function is presented as follows;

f(x) = -0.6·x² + 2.4·x + 5.5

(a) The graph of the function is plotted on MS Excel, with increments in the

x-values of 0.01, to obtain the approximation of the x-intercepts which are

the real roots as follows;

[tex]\begin{array}{|c|cc|}x&&f(x)\\-1.63&&-0.00614\\-1.62&&0.03736\\5.62&&0.03736 \\5.63&&-0.00614\end{array}\right][/tex]

Checking for the approximation of x-value of the intercept, we have;

[tex]x = -1.63 + \dfrac{0 - (-0.00614)}{0.0376 - (-0.00614)} \times (-1.62-(-1.63)) \approx -1.629[/tex]

Therefore, based on the similarity of the values at the intercepts, the x-

values (real roots of the function) at the x-intercepts (y = 0) are;

x ≈ -1.629, and 5.629

(b) The real roots of the quadratic equation are found using the quadratic

formula as follows;

The quadratic formula for finding the roots of the quadratic equation

presented in the form f(x) = a·x² + b·x + c, is given as follows;

[tex]x = \mathbf{ \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}}[/tex]

Comparison to the given function, f(x) = -0.6·x² + 2.4·x + 5.5, gives;

a = -0.6, b = 2.4, and c = 5.5

Therefore, we get;

[tex]x = \dfrac{-2.4\pm \sqrt{2.4^{2}-4\times (-0.6)\times 5.5}}{2\times (-0.6)} = \dfrac{-2.4\pm\sqrt{18.96} }{-1.2} = \dfrac{12 \pm\sqrt{474} }{6}[/tex]

Which gives

The real roots are; x ≈ -1.62859, and x ≈ 5.62859

(c) The initial guesses are;

[tex]x_l[/tex] = 5, and [tex]x_u[/tex] = 10

The first iteration is therefore;

[tex]x_r = \dfrac{5 + 10}{2} = 7.5[/tex]

[tex]Estimated \ error , \ \epsilon _a = \left|\dfrac{10- 5}{10 + 5} \right | \times 100\% = 33.33\%[/tex]

[tex]True \ error, \ \epsilon _t = \left|\dfrac{5.62859 - 7.5}{5.62859} \right | \times 100\% = 33.25\%[/tex]

f(5) × f(7.5) = 2.5 × (-10.25) = -25.625

The bracket is therefore; [tex]x_l[/tex] = 5, and [tex]x_u[/tex] = 7.5

Second iteration:

[tex]x_r = \dfrac{5 + 7.5}{2} = 6.25[/tex]

[tex]Estimated \ error , \ \epsilon _a = \left|\dfrac{7.5- 5}{7.5+ 5} \right | \times 100\% = 20\%[/tex]

[tex]True \ error, \ \epsilon _t = \mathbf{\left|\dfrac{5.62859 - 6.25}{5.62859} \right | \times 100\%} \approx 11.04\%[/tex]

f(5) × f(6.25) = 2.5 × (-2.9375) = -7.34375

The bracket is therefore; [tex]x_l[/tex] = 5, and [tex]x_u[/tex] = 6.25

Third iteration

[tex]x_r = \dfrac{5 + 6.25}{2} = 5.625[/tex]

[tex]Estimated \ error , \ \epsilon _a = \left|\dfrac{5.625- 5}{5.625+ 5} \right | \times 100\% = 5.88\%[/tex]

[tex]True \ error, \ \epsilon _t = \mathbf{\left|\dfrac{5.62859 - 5.625}{5.62859} \right | \times 100\%} \approx 6.378 \times 10^{-2}\%[/tex]

f(5) × f(5.625) = 2.5 × (0.015625) = 0.015625

Therefore, the bracket is [tex]x_l[/tex] = 5.625, and [tex]x_u[/tex] = 6.25

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Answer:

F = ma

=3kg × 6/ms2

=18 N

Explanation:

Since, force is the product of mass and acceleration.

The force applied to an object is

Explanation:

mass=3 kg,

Acceleration=6/ ms2

force=?

Given

force= mass x Acceleration

= 3kg x 6

=18kg N ans

Answer:

sorry di ko alam

Explanation:

Answer:

their multiple moving parts.

Explanation:

Answer:

Metamorphic manufacturing

Explanation:

I don't know what form of an answer you wanted but I hope this helps :)

Answer:

Un motor de paso es un motor eléctrico CC sin escobillas que divide una rotación completa en varios pasos iguales. Rota una distancia incremental específica por cada paso. El número de pasos que se ejecutan controla el grado de rotación del eje del motor.

Los motores de paso tienen cierta capacidad inherente para controlar la posición, ya que tienen pasos de salida integrados. Pueden controlar con gran precisión cuán lejos y cuán rápido rotará el motor de paso. El número de pasos que ejecuta el motor es igual al número de comandos de pulsos del controlador. Un motor de paso rotará una distancia y a una velocidad proporcional al número de la frecuencia de sus comandos de pulso.

Explanation:

Answer:

if you're talking about the car b-post, the answer is "posts"

Explanation:

looked it up