A private plane pilot is what kind of individual transportation position? professional level mid-level entry-level EPA-certified
1 answer:
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Answer:
b)1.08 N
Explanation:
Given that
velocity of air V= 45 m/s
Diameter of pipe = 2 cm
Force exerted by fluid F
So force exerted in x-direction
F=0.763 N
So force exerted in y-direction
F=0.763 N
So the resultant force R
R=1.079
So the force required to hold the pipe is 1.08 N.
Answer:
The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero
Explanation:
The expression for the maximum shear stress is given:
Where
σx = stress in vertical plane = 20 ksi
σy = stress in horizontal plane = -30 ksi
τM = 32 ksi
Replacing:
Solving for τxy:
τxy = ±19.98 ksi
The principal stress is:
Where
σp1 = 20 ksi
σp2 = -30 ksi
(equation 1)
equation 2
Solving both equations:
σp1 = 27 ksi
σp2 = -37 ksi
The shear stress on the vertical plane is zero
Answer:
Explanation:
The mass inside the rigid tank before the high pressure stream enters is:
The final mass inside the rigid tank is:
The supplied air mass is: