Answer:
V=1601gal
Explanation:
Hello! This problem is solved as follows,
First we must raise the equation that defines the pressure at the bottom of the tank with the purpose of finding the height that olive oil reaches.
This is given as the sum due to the atmospheric pressure (1atm = 101.325kPa), and the pressure due to the weight of the olive oil, taking into account the above, the following equation is inferred.
P=Poil+Patm
P=total pressure or absolute pressure=26psi=179213.28Pa
Patm= the atmospheric pressure =101325Pa
Poil=pressure due to the weight of olive oil=0.86αgh
α=density of water=1000kg/m^3
g=gravity=9.81m/s^2
h= height that olive oil reaches
solving
P=Poil+Patm
P=Patm+0.86αgh
[/tex]
Now we can use the equation that defines the volume of a cylinder.
V=
D=3ft=0.9144m
h=9.23m
solving
finally we use conversion factors to find the volume in gallons
Hey, it depends what you subscribed to, or what platform you are trying to unsubscribe from?
Answer:
165 mm
Explanation:
The mass on the piston will apply a pressure on the oil. This is:
p = f / A
The force is the weight of the mass
f = m * a
Where a in the acceleration of gravity
A is the area of the piston
A = π/4 * D1^2
Then:
p = m * a / (π/4 * D1^2)
The height the oil will raise is the heignt of a colum that would create that same pressure at its base:
p = f / A
The weight of the column is:
f = m * a
The mass of the column is its volume multiplied by its specific gravity
m = V * S
The volume is the base are by the height
V = A * h
Then:
p = A * h * S * a / A
We cancel the areas:
p = h * S * a
Now we equate the pressures form the piston and the pil column:
m * a / (π/4 * D1^2) = h * S * a
We simplify the acceleration of gravity
m / (π/4 * D1^2) = h * S
Rearranging:
h = m / (π/4 * D1^2 * S)
Now, h is the heigth above the interface between the piston and the oil, this is at h1 = 42 mm. The total height is
h2 = h + h1
h2 = h1 + m / (π/4 * D1^2 * S)
h2 = 0.042 + 10 / (π/4 * 0.14^2 * 0.8) = 0.165 m = 165 mm
Answer:
We can compute the diameter of the tree T by a pruning procedure, starting at the leaves (external nodes).
- Remove all leaves of T. Let the remaining tree be T1.
-
Remove all leaves of T1. Let the remaining tree be T2.
-
Repeat the "remove" operation as follows: Remove all leaves of Ti. Let remaining tree be Ti+1.
-
When the remaining tree has only one node or two nodes, stop! Suppose now the remaining tree is Tk.
-
If Tk has only one node, that is the center of T. The diameter of T is 2k.
-
If Tk has two nodes, either can be the center of T. The diameter of T is 2k+1.
Explanation:
We can compute the diameter of the tree T by a pruning procedure, starting at the leaves (external nodes).
- Remove all leaves of T. Let the remaining tree be T1.
-
Remove all leaves of T1. Let the remaining tree be T2.
-
Repeat the "remove" operation as follows: Remove all leaves of Ti. Let remaining tree be Ti+1.
-
When the remaining tree has only one node or two nodes, stop! Suppose now the remaining tree is Tk.
-
If Tk has only one node, that is the center of T. The diameter of T is 2k.
-
If Tk has two nodes, either can be the center of T. The diameter of T is 2k+1.
