Answer: hello the diagram related to your question is missing please the third image is the missing part of the question
Fx = 977.76 Ib/ft
Explanation:
<u>Estimate the force that water exerts on the pier </u>
V = 12 ft/s
D( diameter ) = 6 ft
first express the force on the first half of the cylinder as
Fx1 = -
---------------- ( 1 )
where ; Fy = 0
Ps = Po + 1/2 Pv^2 ( 1 - 4 sin^2β ) ------------- ( 2 )
Input equation (2) into equation ( 1 ) (note : assuming Po = 0 )
attached below is the remaining part of the solution
The modulus of elasticity is 28.6 X 10³ ksi
<u>Explanation:</u>
Given -
Length, l = 5in
Force, P = 8000lb
Area, A = 0.7in²
δ = 0.002in
Modulus of elasticity, E = ?
We know,
Modulus of elasticity, E = σ / ε
Where,
σ is normal stress
ε is normal strain
Normal stress can be calculated as:
σ = P/A
Where,
P is the force applied
A is the area of cross-section
By plugging in the values, we get
σ = 
σ = 11.43ksi
To calculate the normal strain we use the formula,
ε = δ / L
By plugging in the values we get,
ε = 
ε = 0.0004 in/in
Therefore, modulus of elasticity would be:

Thus, modulus of elasticity is 28.6 X 10³ ksi
U mean how to connect them to your phone?
Answer:

Explanation:
First, we will find actual properties at given inlet and outlet states by the use of steam tables:
AT INLET:
At 4MPa and 350°C, from the superheated table:
h₁ = 3093.3 KJ/kg
s₁ = 6.5843 KJ/kg.K
AT OUTLET:
At P₂ = 125 KPa and steam is saturated in vapor state:
h₂ =
= 2684.9 KJ/kg
Now, for the isentropic enthalpy, we have:
P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K
Since s₂ is less than
and greater than
at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

Now, we will find
(enthalpy at the outlet for the isentropic process):

Now, the isentropic efficiency of the turbine can be given as follows:

Express it in standard form and apply the basic indices laws to simplify