Answer:
hello some parts of your question is missing attached below is the missing part
answer :
A) Determine changes in the 50-mm dimensions
The changes are : 0.006mm compression in y-direction
0.002 mm expansion in x and z directions
B) the stress required are evenly distributed
Explanation:
<em>Given data</em> :
50-mm cube of graphite fiber reinforced polymer matrix
subjected to 125-KN force in direction 2,
direction 2 is perpendicular to fiber direction ( direction 1 ) and cube is constrained against expansion in direction 3
A) Determine changes in the 50-mm dimensions
The changes are : 0.006mm compression in y-direction
0.002 mm expansion in x and z directions
B) the stress required are evenly distributed
attached below is the detailed solution
Answer:
<em>181 °C</em>
<em></em>
Explanation:
Initial pressure = 100 kPa
Initial temperature = 30 °C = 30 + 273 K = 303 K
Final pressure = 1200 kPa
Final temperature = ?
n = 1.2
For a polytropic process, we use the relationship
(/
) = (
/
)^γ
where γ = (n-1)/n
γ = (1.2-1)/1.2 = 0.1667
substituting into the equation, we have
(/303) = (1200/100)^0.1667
/303 = 12^0.1667
/303 = 1.513
= 300 x 1.513 = 453.9 K
==> 453.9 - 273 = 180.9 ≅ <em>181 °C</em>
Answer:
0.00650 Ib s /ft^2
Explanation:
diameter ( D ) = 0.71 inches = 0.0591 ft
velocity = 0.90 ft/s ( V )
fluid specific gravity = 0.96 (62.4 ) ( x )
change in pressure ( P ) = 0 because pressure was constant
viscosity = (change in p - X sin∅ ) / 32 V
= ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2 / 32 * 0.90
= - 59.904 sin (-90) * 0.0035 / 28.8
= 0.1874 / 28.8
viscosity = 0.00650 Ib s /ft^2