Hopefully that helps you out and is this for history or science?
Answer:
80.7lbft/hr
Explanation:
Flow rate of water in the system = 3.6x10^-6
The height h = 100
1s = 1/3600h
This implies that
Q = 3.6x10^-6/[1/3600]
Q = 0.0000036/0.000278
Q = 0.01295
Then the power is given as
P = rQh
The specific weight of water = 62.3 lb/ft³
P = 62.3 x 0.01295 x 100
P = 80.675lbft/h
When approximated
P = 80.7 lbft/h
This is the average power that could be generated in a year.
This answers the question and also corresponds with the answer in the question.
Answer:
1561.84 MPa
Explanation:
L=20 cm
d1=0.21 cm
d2=0.25 cm
F=5500 N
a) σ= F/A1= 5000/(π/4×(0.0025)^2)= 1018.5916 MPa
lateral strain= Δd/d1= (0.0021-0.0025)/0.0025= -0.16
longitudinal strain (ε_l)= -lateral strain/ν = -(-0.16)/0.3
(assuming a poisson's ration of 0.3)
ε_l =0.16/0.3 = 0.5333
b) σ_true= σ(1+ ε_l)= 1018.5916( 1+0.5333)
σ_true = 1561.84 MPa
ε_true = ln( 1+ε_l)= ln(1+0.5333)
ε_true= 0.4274222
The engineering stress on the rod when it is loaded with a 5500 N weight is 1561.84 MPa.
