By permanently locking in stakeholder requirements during a project's planning phase. -through highly detailed process documentation that is updated following every work cycle.
Answer:
=>> 167.3 kpa.
=>> 60° from horizontal face.
Explanation:
So, we are given the following data or parameters or information which is going to assist us in solving this kind of question;
=>> "A soil element is subjected to a minor principle stress of 50 kPa on a plane rotated 20 ° counterclockwise from vertical. "
=>>"If the deviator stress is 120 kPa and the shear strength parameters are a friction angle of 30° and a cohesion of 5 kPa."
The orientation of this plane with respect to the major principle stress plane = 50 tan^2 (45 + 30/2) + 10 tan ( 45 + 30/2).
magnitude of the stresses on the failure plane = 167.3 kpa.
The orientation of this plane with respect to the major principle stress plane => x = 60 cos 60° = 30kpa.
y = 60 sin 60° = 30√3 = sheer stress.
the orientation of this plane with respect to the major principle stress plane.
Theta = 45 + 15 = 60°.
Answer:
its bad and to drive drunk is even worse
Answer:
A
Explanation:
Option A was the only option that did not have a sentence fragment. All the other choices were fragments because they interrupted the flow of the sentence. A sentence fragment means that if you started reading where the period is, you would understand everything about the sentence and it would be clear. You cannot have an incomplete sentence that does not have background information. An example would be: And that is how you install it. This leaves information unanswered, such as what the subject of the sentence is, or in other words, what you were installing. Another indicator that this was a sentence fragment was that the sentence began with the word "And." Sentence fragments very commonly start with the word and, because that word shows that it is a continuation of a topic.
Answer:
\epsilon = 0.028*0.3 = 0.0084
Explanation:
\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2
where P_1 = P_2 = 0
V1 AND V2 =0
Z1 =0
h_P = \frac{w_p}{\rho Q}
=\frac{40}{9.8*10^3*0.2} = 20.4 m
20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10
we know thaTV =\frac{Q}{A}
V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec
20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10
f = 0.0560
Re =\frac{\rho v D}{\mu}
Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5
fro Re = 7.53*10^5 and f = 0.0560
\frac{\epsilon}{D] = 0.028
\epsilon = 0.028*0.3 = 0.0084
