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In-s [12.5K]
3 years ago
10

Why is it that dislocations play an important role in controlling the mechanical properties of metallic materials, however, they

do not play a role in determining the mechanical properties of glasses?
Engineering
1 answer:
klio [65]3 years ago
5 0

Answer:

dislocations play an important role in controlling as

Explanation:

As dislocations plays an important role in the ductility, elasticity and plurality of materials

  • The elastic and elastic deflections play a large role in their properties as the metallic materials, because the dislocation of a glass material does not play a major role in their properties.

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Check the level of motor oil in your engine by ?
Usimov [2.4K]

Answer:

cpct gvxjjxjhdfjokjdzfjiyddzzsjhxf

6 0
2 years ago
An ideal vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated va
Ksju [112]

Explanation:

Note: Refer the diagram below

Obtaining data from property tables

State 1:

\left.\begin{array}{l}P_{1}=1.25 \text { bar } \\\text { Sat - vapour }\end{array}\right\} \begin{array}{l}h_{1}=234.45 \mathrm{kJ} / \mathrm{kg} \\S_{1}=0.9346 \mathrm{kJ} / \mathrm{kgk}\end{array}

State 2:

\left.\begin{array}{l}P_{2}=5 \text { bor } \\S_{2}=S_{1}\end{array}\right\} \quad h_{2}=262.78 \mathrm{kJ} / \mathrm{kg}

State 3:

\left.\begin{array}{l}P_{3}=5 \text { bar } \\\text { Sat }-4 q\end{array}\right\} h_{3}=71-33 \mathrm{kJ} / \mathrm{kg}

State 4:

Throttling process  h_{4}=h_{3}=71.33 \mathrm{kJ} / \mathrm{kg}

(a)

Magnitude of compressor power input

\dot{w}_{c}=\dot{m}\left(h_{2}-h_{1}\right)=\left(8 \cdot 5 \frac{\mathrm{kg}}{\min } \times \frac{1 \mathrm{min}}{\csc }\right)(262.78-234 \cdot 45)\frac{kj}{kg}

w_{c}=4 \cdot 013 \mathrm{kw}

(b)

Refrigerator capacity

Q_{i n}=\dot{m}\left(h_{1}-h_{4}\right)=\left(\frac{g \cdot s}{60} k_{0} / s\right) \times(234 \cdot 45-71 \cdot 33) \frac{k J}{k_{8}}

Q_{i n}=23 \cdot 108 \mathrm{kW}\\1 ton of retregiration =3.51 k \omega

\ Q_{in} =6 \cdot 583 \text { tons }

(c)

Cop:

\beta=\frac{\left(h_{1}-h_{4}\right)}{\left(h_{2}-h_{1}\right)}=\frac{Q_{i n}}{\omega_{c}}=\frac{23 \cdot 108}{4 \cdot 013}

\beta=5 \cdot 758

3 0
3 years ago
While playing a game of catch on the quadrangle, you throw a ball at an initial velocity of 17.6 m/s (approximately 39.4 mi/hr),
MAXImum [283]

Answer:

a) The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) The ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

Explanation:

a) The ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical motion at constant acceleration. First, we calculate the time taken by the ball to hit the ground:

y = y_{o} + (v_{o}\cdot \sin \theta) \cdot t+\frac{1}{2}\cdot g\cdot t^{2} (1)

Where:

y_{o}, y - Initial and final vertical position, measured in meters.

v_{o} - Initial speed, measured in meters per second.

\theta - Launch angle, measured in sexagesimal degrees.

g - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that y_{o} = 2\,m, y = 0\,m, v_{o} = 17.6\,\frac{m}{s}, \theta = 50^{\circ} and g = -9.807\,\frac{m}{s^{2}}, then the time taken by the ball is:

-4.904\cdot t^{2}+13.482\cdot t +2 = 0 (2)

This second order polynomial can be solved by Quadratic Formula:

t_{1} \approx 2.890\,s and t_{2} \approx -0.141\,s

Only the first root offers a solution that is physically reasonable. That is, t \approx 2.890\,s.

The vertical velocity of the ball is calculated by this expression:

v_{y} = v_{o}\cdot \sin \theta +g\cdot t (3)

Where:

v_{o,y}, v_{y} - Initial and final vertical velocity, measured in meters per second.

If we know that v_{o} = 17.6\,\frac{m}{s}, \theta = 50^{\circ}, g = -9.807\,\frac{m}{s^{2}} and t \approx 2.890\,s, then the final vertical velocity is:

v_{y} = -14.860\,\frac{m}{s}

The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) From a) we understand that ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball (x) is determined by the following expression:

x = (v_{o}\cdot \cos \theta)\cdot t (4)

If we know that v_{o} = 17.6\,\frac{m}{s}, \theta = 50^{\circ} and t \approx 2.890\,s, then the distance covered by the ball is:

x = 32.695\,m

The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before hitting the ground (v), measured in meters per second, is determined by the following Pythagorean identity:

v = \sqrt{(v_{o}\cdot \cos \theta )^{2}+v_{y}^{2}} (5)

If we know that v_{o} = 17.6\,\frac{m}{s}, \theta = 50^{\circ} and v_{y} = -14.860\,\frac{m}{s}, then the magnitude of the velocity of the ball is:

v \approx 18.676\,\frac{m}{s}.

The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is defined by the following trigonometric relationship:

\tan \theta = \frac{v_{y}}{v_{o}\cdot \cos \theta_{o}}

If we know that v_{o} = 17.6\,\frac{m}{s}, \theta_{o} = 50^{\circ} and v_{y} = -14.860\,\frac{m}{s}, the angle of the total velocity of the ball just before hitting the ground is:

\theta \approx -52.717^{\circ}

The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

3 0
3 years ago
Read 2 more answers
Transmission cleaners are used: A) Only in conjunction with fuel system cleanersB) Only in the colder monthsC) By themselvesD) I
valentina_108 [34]

Answer: D

Explanation: When you change your transmission fluid you change your filter. When you do that you clean where the filter was. Then you can put the new filter on and the new fluid.

5 0
3 years ago
Read 2 more answers
Consider a thin suspended hotplate that measures 0.25 m × 0.25 m. The isothermal plate has a mass of 3.75 kg, a specific heat of
Orlov [11]

Answer:

Heat losses by convection, Qconv = 90W

Heat losses by radiation, Qrad = 5.814W

Explanation:

Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:

1. Radiation

2. Conduction

3. Convection

Convection is defined as the transfer of heat through the actual movement of the molecules.

Qconv = hA(Temp.final - Temp.surr)

Where h = 6.4KW/m2K

A, area of a square = L2

= (0.25)2

= 0.0625m2

Temp.final = 250°C

Temp.surr = 25°C

Q = 64 * 0.0625 * (250 - 25)

= 90W

Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.

Qrad = E*S*(Temp.final4 - Temp.surr4)

Where E = emissivity of the surface

S = boltzmann constant

= 5.6703 x 10-8 W/m2K4

Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)

= 5.814 W

7 0
3 years ago
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