Answer:
The pump horse power required per mile of pipe 
Explanation:
The explanation is shown on the second third and fourth image
It make sense that the temperature of the oil is high even though the pipeline is in a cold environment this is so because where the oil is been tapped from is way hotter than the temperature of the pipeline environment because it is closer to the earths crust than the environment of the pipeline
Answer: a) 0.948 b) 117.5µf
Explanation:
Given the load, a total of 2.4kw and 0.8pf
V= 120V, 60 Hz
P= 2.4 kw, cos θ= 80
P= S sin θ - (p/cos θ) sin θ
= P tan θ(cos^-1 (0.8)
=2.4 tan(36.87)= 1.8KVAR
S= 2.4 + j1. 8KVA
1 load absorbs 1.5 kW at 0.707 pf lagging
P= 1.5 kW, cos θ= 0.707 and θ=45 degree
Q= Ptan θ= tan 45°
Q=P=1.5kw
S1= 1.5 +1.5j KVA
S1 + S2= S
2.4+j1.8= 1.5+1.5j + S2
S2= 0.9 + 0.3j KVA
S2= 0.949= 18.43 °
Pf= cos(18.43°) = 0.948
b.) pf to 0.9, a capacitor is needed.
Pf = 0.9
Cos θ= 0.9
θ= 25.84 °
(WC) V^2= P (tan θ1 - tan θ2)
C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2
f=60, π=22/7
C= 117.5µf
Answer:
F(x) = 0 ; x < 0
0.064 ; 0 ≤ x < 1
0.352 ; 1 ≤ x < 2
0.784 ; 2 ≤ x < 3
1 ; x ≥ 3
Explanation:
Each wafer is classified as pass or fail.
The wafers are independent.
Then, we can modelate X : ''Number of wafers that pass the test'' as a Binomial random variable.
X ~ Bi(n,p)
Where n = 3 and p = 0.6 is the success probability
The probatility function is given by :

Where
is the combinatorial number

Let's calculate f(x) :




For the cumulative distribution function that we are looking for :



The cumulative distribution function for X is :
F(x) = 0 ; x < 0
0.064 ; 0 ≤ x < 1
0.352 ; 1 ≤ x < 2
0.784 ; 2 ≤ x < 3
1 ; x ≥ 3
Answer:
the percent increase in the velocity of air is 25.65%
Explanation:
Hello!
The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.
m1=m2
Now remember that mass flow is given by the product of density, cross-sectional area and velocity
(α1)(V1)(A1)=(α2)(V2)(A2)
where
α=density
V=velocity
A=area
Now we can assume that the input and output areas are equal
(α1)(V1)=(α2)(V2)

Now we can use the equation that defines the percentage of increase, in this case for speed

Now we use the equation obtained in the previous step, and replace values

the percent increase in the velocity of air is 25.65%