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:
D
Explanation:
took test failed question D is the right answer
Answer:
Class of fit:
Interference (Medium Drive Force Fits constitute a special type of Interference Fits and these are the tightest fits where accuracy is important).
Here minimum shaft diameter will be greater than the maximum hole diameter.
Medium Drive Force Fits are FN 2 Fits.
As per standard ANSI B4.1 :
Desired Tolerance: FN 2
Tolerance TZone: H7S6
Max Shaft Diameter: 3.0029
Min Shaft Diameter: 3.0022
Max Hole Diameter:3.0012
Min Hole Diameter: 3.0000
Max Interference: 0.0029
Min Interference: 0.0010
Stress in the shaft and sleeve can be considered as the compressive stress which can be determined using load/interference area.
Design is acceptable If compressive stress induced due to selected dimensions and load is less than compressive strength of the material.
Explanation:
In Engineering, the thrust angle is checked by referencing: C. vehicle centerline.
<h3>What is a
thrust angle?</h3>
A thrust angle can be defined as an imaginary line which is drawn perpendicularly from the centerline of the rear axle of a vehicle, down the centerline.
This ultimately implies that, the thrust angle is a reference to the centerline (wheelbase) of a vehicle, and it confirms that the two wheels on both sides are properly angled within specification.
Read more on thrust angle here: brainly.com/question/13000914
#SPJ1
Answer:
k = 4.21 * 10⁻³(L/(mol.s))
Explanation:
We know that
k = Ae
------------------- euqation (1)
K= rate constant;
A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;
E = activation energy = 93.1kJ/mol;
R= ideal gas constant = 8.314 J/mol.K;
T= temperature = 332 K;
Put values in equation 1.
k = 4.36*10¹¹(M⁻¹s⁻¹)e![^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}](https://tex.z-dn.net/?f=%5E%7B%5B%28-93.1%2A10%5E3%29%28J%2Fmol%29%5D%2F%5B%288.314%29%28J%2Fmol.K%29%28332K%29%7D)
k = 4.2154 * 10⁻³(M⁻¹s⁻¹)
here M =mol/L
k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)
or
k = 4.21 * 10⁻³((L/mol)s⁻¹)
or
k = 4.21 * 10⁻³(L/(mol.s))
