Solution :
i. Slip plane (1 1 0)
Slip direction -- [1 1 1]
Applied stress direction = ( 1 0 0 ]
τ = 50 MPa ( Here slip direction must be perpendicular to slip plane)
τ = σ cos Φ cos λ




τ = σ cos Φ cos λ
∴ 
σ = 122.47 MPa
ii. Slip plane --- (1 1 0)
Slip direction -- [1 1 1]


τ = σ cos Φ cos λ
∴ 
σ = 122.47 MPa
iii. Slip plane --- (1 0 1)
Slip direction --- [1 1 1]


τ = σ cos Φ cos λ
∴ 
σ = 122.47 MPa
iv. Slip plane -- (1 0 1)
Slip direction ---- [1 1 1]


τ = σ cos Φ cos λ
∴ 
σ = 122.47 MPa
∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0
Answer:
hL = 0.9627 m
Explanation:
Given
Q = 0.040 m³/s (constant value)
D₁ = 15 cm = 0.15 m ⇒ R₁ = D₁/2 = 0.15 m/2 = 0.075 m
D₂ = 8 cm = 0.08 m ⇒ R₂ = D₂/2 = 0.08 m/2 = 0.04 m
P₁ = 480 kPa = 480*10³Pa
P₂ = 440 kPa = 440*10³Pa
α = 1.05
ρ = 1000 Kg/m³
g = 9.81 m/s²
h₁ = h₂
hL = ? (the irreversible head loss in the reducer)
Using the formula Q = v*A ⇒ v = Q/A
we can find the velocities v₁ and v₂ as follows
v₁ = Q/A₁ = Q/(π*R₁²) = (0.040 m³/s)/(π*(0.075 m)²) = 2.2635 m/s
v₂ = Q/A₂ = Q/(π*R₂²) = (0.040 m³/s)/(π*(0.04 m)²) = 7.9577 m/s
Then we apply the Bernoulli law (for an incompressible flow)
(P₂/(ρ*g)) + (α*v₂²/(2*g)) + h₂ = (P₁/(ρ*g)) + (α*v₁²/(2*g)) + h₁ - hL
Since h₁ = h₂ we obtain
(P₂/(ρ*g)) + (α*v₂²/(2*g)) = (P₁/(ρ*g)) + (α*v₁²/(2*g)) - hL
⇒ hL = ((P₁-P₂)/(ρ*g)) + (α/(2*g))*(v₁²-v₂²)
⇒ hL = ((480*10³Pa-440*10³Pa)/(1000 Kg/m³*9.81 m/s²)) + (1.05/(2*9.81 m/s²))*((2.2635 m/s)²-(7.9577 m/s)²)
⇒ hL = 0.9627 m
Answer:
Explanation:
Given that :
the diameter of the reservoir D = 18 mm
the diameter of the manometer d = 6 mm
For an equilibrium condition ; the pressure on both sides are said to be equal
∴


According to conservation of volume:



Replacing x into (1) ; we have;



Thus; the liquid deflection is : 
when the applied pressure is equivalent to 25 mm of water (gage); the liquid deflection is:




L = 27.21 mm
Answer:
foot pedal
Explanation: mark brainlest plz