Answer:
The stress in the rod is 39.11 psi.
Explanation:
The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:
Replacing the diameter the area results:
Therefore the the stress results:
Answer:
hello your question is incomplete attached below is the complete question
answer: There is a hydraulic jump
Explanation:
First we have to calculate the depth of flow downstream of the gate
y1 = ----------- ( 1 )
Cc ( concentration coefficient ) = 0.61 ( assumed )
Yg ( depth of gate opening ) = 0.5
hence equation 1 becomes
y1 = 0.61 * 0.5 = 0.305 m
calculate the flow per unit width q
q = Q / b ----------- ( 2 )
Q = 10 m^3 /s
b = 2 m
hence equation 2 becomes
q = 10 / 2 = 5 m^2/s
next calculate the depth before hydraulic jump y2 by using the hydraulic equation
answer : where y1 < y2 hence a hydraulic jump occurs in the lined channel
attached below is the remaining part of the solution
Given:
diameter of sphere, d = 6 inches
radius of sphere, r = = 3 inches
density, = 493 lbm/
S.G = 1.0027
g = 9.8 m/ = 386.22 inch/
Solution:
Using the formula for terminal velocity,
=
(1)
where,
V = volume of sphere
= drag coefficient
Now,
Surface area of sphere, A =
Volume of sphere, V =
Using the above formulae in eqn (1):
=
=
=
Therefore, terminal velcity is given by:
=
inch/sec