Answer:
90 degrees
Explanation:
In the case when the sheer stress acts in the one plane of an element so it should be equal and opposite also the shear stress acted on a plan i.e. 90 degrees from the plane
Therefore as per the given situation it should be 90 degrees from the plane
hence, the same is to be considered and relevant too
Answer:
Mechanical average of a wheel = 3
Explanation:
Given:
Radius of wheel = 1.5 ft = 1.5 x 12 = 18 inches
Radius of axle = 6 inches
Find:
Mechanical average of a wheel
Computation:
Mechanical average of a wheel = Radius of wheel / Radius of axle
Mechanical average of a wheel = 18 / 6
Mechanical average of a wheel = 3
Answer:
3.03 INCHES
Explanation:
According to ASTM D198 ;
Modulus of rupture = ( M / I ) * y ----- ( 1 )
M ( bending moment ) = R * length of span / 2
= (120 * 10^3 ) * 48 / 2 = 288 * 10^4 Ib-in
I ( moment of inertia ) = bd^3 / 12
= ( 2 )*( d )^3 / 12 = 2d^3 / 12
b = 2 in , d = ?
length of span = 4 * 12 = 48 inches
R = P / 2 = 240 * 10^3 / 2 = 120 * 10^3 Ib
y ( centroid distance ) = d / 2 inches
back to equation ( 1 )
( M / I ) * y
940.3 ksi = ( 288 * 10^4 / 2d^3 / 12 ) * d / 2
= ( 288 * 10^4 * 12 ) / 2d^3 ) * d / 2
940300 = 34560000* d / 4d^3
4d^3 ( 940300 ) = 34560000 d ( divide both sides with d )
4d^2 = 34560000 / 940300
d^2 = 9.188 ∴ Value of d ≈ 3.03 in
Answer:
Option C (practice by..............technical) would be the appropriate choice.
Explanation:
- A prerogative seems to be the discipline or nature of the design process to ensure.
- All certificates must, therefore, throughout compliance with current requirements including professional competence, express one's right and privilege of providing offers perhaps throughout the areas of qualified professionals.
Some other alternatives that are given aren't connected to a particular scenario. But that's the best solution above.
The air flow necessary to remain at the lower explosive level is 4515. 04cfm
<h3>How to solve for the rate of air flow</h3>
First we have to find the rate of emission. This is solved as
2pints/1.5 x 1min
= 2/1.5x60
We have the following details
SG = 0.71
LEL = 1.9%
B = 10% = 0.1 a constant
The molecular weight is given as 74.12
Then we would have Q as
403*100*0.2222 / 74.12 * 0.71 * 0.1
= Q = 4515. 04
Hence we can conclude that the air flow necessary to remain at the lower explosive level is 4515. 04cfm
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