Answer: That'd be about 6400km upwards, i belive.
Explanation:
Answer:
C. The user will not be able to see the junction object records or the field values.
Explanation:
For the profile of the user to give access permission such as create and read to the job without granting access permission to the production facility object, the value of the field or records of the junction object will not be seen by the user. This is one of the necessary criteria or principle for the universal container with a junction object.
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:
So % increment in tool life is equal to 4640 %.
Explanation:
Initially n=0.12 ,V=130 m/min
Finally C increased by 10% , V=90 m/min
Let's take the tool life initial condition is 
and when C is increased it become 
.
As we know that tool life equation for tool
At initial condition 
------(1)
At final condition 
-----(2)
From above equation
So increment in tool life =
=
So % increment in tool life is equal to 4640 %.
