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Mazyrski [523]
3 years ago
7

What is the major drawback to use whiskers as a dispersed agents in composites? a)- High price b)- Large length to diameter rati

os c)- Difficult to disperse d)- a & c
Engineering
1 answer:
artcher [175]3 years ago
7 0

Answer: C) a&c

Explanation: Composite materials are the materials which are made up of the two or more material of different properties.They are usually having properties like high hardness,low in density,no dissolving into each other etc. Whiskers as a dispersing agent in composite materials are usually not preferred because they are expensive as well as they are not easy to disperse in the composite material.Thus option (c) is the correct answer.

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A wire is 0.92 m long and 1.2 mm2 in cross-sectional area. It carries a current of 5.0 A when a 2.2 V potential difference is ap
Flura [38]

The solution is in the attachment

3 0
3 years ago
A pipe is insulated such that the outer radius of the insulation is less than the critical radius. Now the insulation is taken o
irina [24]

Answer:

the heat transfer from the pipe will decrease when the insulation is taken off for r₂< r_{cr}

where;

r₂ = outer radius

r_{cr} = critical radius

Explanation:

Note that the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h .

r_{cr} =\frac{k}{h}

The rate of heat transfer from the cylinder increases with the addition of insulation for outer radius less than  critical radius (r₂< r_{cr}) 0,  and reaches a maximum when r₂ = r_{cr}, and starts to decrease for r₂< r_{cr}. Thus, insulating the pipe may actually increase the rate of heat transfer from the pipe instead of decreasing it when r₂< r_{cr} .

7 0
3 years ago
A thick steel slab (rho= 7800 kg/m3 , cp= 480 J/kg K, k= 50 W/m K) is initially at 300 °C and is cooled by water jets impinging
dimaraw [331]

Answer:

t = 2244.3 sec

Explanation:

calculate the thermal diffusivity

\alpha = \frac{k}{\rho c}

           = \frac{50}{7800\times 480} = 1.34 \times 10^{-5} m^2/s

                   

Temperature at 28 mm distance after t time  = =  50 degree C

we know that

\frac[ T_{28} - T_s}{T_i -T_s} = erf(\frac{x}{2\sqrt{at}})

\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]

0.909 = erf{\frac{3.8245}{\sqrt{t}}}

from gaussian error function table , similarity variable w calculated as

erf w = 0.909

it is lie between erf w = 0.9008  and erf w = 0.11246 so by interpolation we have

w = 0.08073

erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]

0.08073 = \frac{3.8245}{\sqrt{t}}

solving fot t we get

t = 2244.3 sec

3 0
2 years ago
Does the music contribute to the performance of its production?
gizmo_the_mogwai [7]

Answer:

Of course music plays crucial role

6 0
3 years ago
A 5-in.-diameter pipe is supported every 9 ft by a small frame consisting of two members asshown. Knowing that the combined weig
jarptica [38.1K]

Answer:

AC: at D , M_max = 12.25 lb-ft

BC: at E , M_max = 8.75 lb-ft

Explanation:

Given:

- The diameter of the pipe d = 5-in

- The pipe is supported every L = 9 ft of pipe in length

- The weight if the pipe + contents W = 10 lb/ft

Find:

determine the magnitude and location of the maximum bending moment in members AC and BC.

Solution:

- The figure (missing) is given in the attachment.

- We will first determine the external forces acting on each member:

             Section: 9-ft section of pipe.

                     Sum of forces perpendicular to member AC = 0

                     F_d - 0.8*W*L = 0

                     F_d = 0.8*10*9 = 72 lb

                     Sum of forces perpendicular to member BC = 0

                     F_e - 0.6*W*L = 0

                     F_e = 0.6*10*9 = 54 lb

              F_d = 72 lb ,  F_e = 54 lb

- Then we will determine the support reactions for each member AC point A and BC point B.

              Section: Entire Frame.

                    Sum of moments about point B = 0

                    -A_y*(18.75/12) + F_d*(d /2*12) + F_e*((11.25-2.5)/12) = 0

                    -A_y*(1.5625) + 15 + 39.375 = 0

                    A_y = 34.8 lb  

                   Sum of forces in vertical direction = 0

                     A_y + B_y - 0.8*F_d - 0.6*F_e = 0

                     B_y = 0.8*(72) + 0.6*(54) - 34.8

                     B_y = 55.2 lb  

                   Sum of forces in horizontal direction = 0

                     A_x + B_x - 0.6*F_d + 0.8*F_e = 0

                     A_x + B_x = 0

               Section: Member AC

                    Sum of moments about point C = 0

                     F_d*(2.5/12) - A_y*(12/12) - A_x*(9/12) = 0

                     72*2.5 - 34.8*12 - 9*A_x = 0

                     A_x = -237.6 / 9 = - 26.4 lb

                     B_x = - A_x = 26.4 lb

                     A_x = -26.4 lb  ,  B_x = 26.4 lb

- Now we will calculate bending moment for each member at different sections.

               Member AC:

                    From point A till just before point D

                     -0.6*A_x*x - A_y*0.8*x + M = 0

                     15.84*x - 27.84*x + M = 0

                      M = 12*x   ..... max value at D, x = 12.25 in

                      M_max = 12*12.25/12 = 12.25 lb-ft

               Member BC:

                    From point B till just before point E

                     -0.8*B_x*x + B_y*0.6*x + M = 0

                     -21.12*x + 33.12*x + M = 0

                      M = -12*x   ..... max value at E, x = 11.25 - 2.5 = 8.75 in

                      M_max = -12*8.75/12 = -8.75 lb-ft

- The maximum bending moments and their locations are:

                      AC: at D , M_max = 12.25 lb-ft

                      BC: at E , M_max = 8.75 lb-ft

5 0
2 years ago
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