Complete the sentence to identify a useful advance in the culinary arts.

Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle gra

Vlada [557]

Answer:

Explanation:

Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries because there is not as much crystallographic misalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction.

Low angle grain boundaries (quasi-coherent) are formed by the dislocation network positioned along the geometric plane with small tilt angle differences between successive peers that is tilt boundary made up edge dislocations therefore it may only divert the slip direction of the incoming gliding dislocation with very little frictional stresses. And on the other hand, a high angle grain boundary region because of their disordered almost liquid like structure which acts as a strong barrier against dislocation slip motion and causes actually formation of dislocations file-up against it by arresting their motion unless that the stress concentration at the leading dislocation becomes high enough to go though the barrier.

5 0

3 years ago

Different Gauss quadrature formulae predict different values for the same integral a. True b. False

scoray [572]

Answer: False

Explanation:

The given statement are false as, the gauss quadrature predicted same value only when their is a minute error at one point and two point rule. It basically given the more precise answer and the answer are only changed by the decimal place. The gauss quadrature method are basically used for calculating the certain integral.

5 0

3 years ago

Find the thickness of the material that will allow heat transfer of 6706.8 *10^6 kcal during the 5 months through the rectangle

Vinvika [58]

Answer:

The thickness of the material is 6.23 cm

Explanation:

Given;

quantity of heat, Q = 6706.8 *10⁶ kcal

duration of the heat transfer, t = 5 months

thermal conductivity of copper, k = 385 W/mk

outside temperature of the heater, T₁ = 30° C

inside temperature of the heater, T₂ = 50° C

dimension of the rectangular heater = 450 cm by 384 cm

1 kcal = 1.163000 Watt-hour

6706.8 *10⁶ kcal = 7800008400 watt-hour

I month = 730 hours

5 months = 3650 hours

Rate of heat transfer, P = $\frac{7800008400 \ Watt-Hour}{3650 \ Hours} = 2136988.6 \ W$

Rate of heat transfer, $P = \frac{K*A *\delta T}{L}$

where;

P is the rate of heat transfer (W)

k si the thermal conductivity (W/mk)

ΔT is change in temperature (K)

A is area of the heater (m²)

L is thickness of the heater (m)

$P = \frac{KA(T_2-T_1)}{L} \\\\L = \frac{KA(T_2-T_1)}{P}\\\\L = \frac{385(4.5*3.84)(50-30)}{2136988.6}\\\\L = 0.0623 \ m$

L = 6.23 cm

Therefore, the thickness of the material is 6.23 cm

8 0

3 years ago

Find the differential and evaluate for the given x and dx: y=sin2xx,x=π,dx=0.25

Sedaia [141]

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

<h3>How to determine the differential of a one-variable function</h3>

Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:

dy = y'(x) · dx (1)

If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:

$y' = -\frac{1}{x^{2}}\cdot \sin 2x + \frac{2}{x}\cdot \cos 2x$

$y' = \frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}}$

$dy = \left(\frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}} \right)\cdot dx$

$dy = \left(\frac{2\pi \cdot \cos 2\pi -\sin 2\pi}{\pi^{2}} \right)\cdot (0.25)$

$dy = \frac{1}{2\pi}$

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

To learn more on differentials: brainly.com/question/24062595

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4 0

2 years ago

A mass of 5 kg of saturated liquid-vapor mixture of water is contained in a piston-cylinder device at 125 kPa. Initially, 2 kg o

arlik [135]

A because it is the best one

5 0

3 years ago