Answer:
Outdoors
Explanation:
Construction workers perform outdoors.
—By the maximum shear stress is then calculated by: where b = 2 (ro − ri) is the effective width of the cross section, Ic = π (ro4 − ri4) / 4 is the centroidal moment of inertia, and A = π (ro2 − ri2) is the area of the cross section.
Answer:
The overview of the given scenario is explained in explanation segment below.
Explanation:
- The inception of cavitation, that further sets the restriction for high-pressure and high-free operation, has always been the matter of substantial experimental study over the last few generations.
- Cavitation inception would be expected to vary on the segment where the local "PL" pressure mostly on segment keeps falling to that are below the "Pv" vapor pressure of the fluid and therefore could be anticipated from either the apportionment of the pressure.
⇒ A cavitation number is denoted by "σ" .
Answer:
the correct distance is 202 ft
Explanation:
The computation of the correct distance is shown below:
But before that correction to be applied should be determined
= (101 ft - 100 ft) ÷ (100 ft) × 200 ft
= 2 ft
Now the correct distance is
= 200 ft + 2 ft
= 202 ft
Hence, the correct distance is 202 ft
The same would be relevant and considered too
Answer:
33.4
Explanation:
Step 1:
\sumMo=0 (moment about the origin)
Fb(15)-Fc(15)=0
Fb=Fc
Step 2:
\sumFx=0
-Fb-Fccos\theta+Ncsin\theta=0
Fc=0.3Nc=Fb
-0.3Nc-0.3Nccos\theta+Ncsin\theta=0
(-0.3-cos\theta+sin\theta)Nc=0----(1)
\sumFy=0
Nccos\theta+Fcsin\theta-Nb=0
Nccos\theta+0.3Ncsin\theta-Nc=0
Nc[cos\theta+0.3sin\theta-1]=0--------(2)
Solving eq (1) and eq (2)
\theta=33.4
Step 3:
As the roller is a two force member
2(90-\phi)+\theta=180
\phi=\theta/2
\phi=Tan(\muN/N)-1
\phi=16.7
\theta=2x16.7=33.4
