Answer:
The answer is below
Explanation:
The practical considerations you might encounter when you increase the moment of inertia (I) while keeping the cross-sectional area fixed are:
1. Shapes of moment of inertia: Engineers should consider or know the different shapes of moment of inertia for different shape
2. Understanding the orientation of the beam: this will allow engineers to either increase or decrease the moment of inertia of a beam without increasing its cross sectional area.
Answer:
0.00650 Ib s /ft^2
Explanation:
diameter ( D ) = 0.71 inches = 0.0591 ft
velocity = 0.90 ft/s ( V )
fluid specific gravity = 0.96 (62.4 ) ( x )
change in pressure ( P ) = 0 because pressure was constant
viscosity = (change in p - X sin∅ ) 
/ 32 V
= ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2 / 32 * 0.90
= - 59.904 sin (-90) * 0.0035 / 28.8
= 0.1874 / 28.8
viscosity = 0.00650 Ib s /ft^2
We can actually deduce here that making a airtight seal will take different format. You can:
- Use an epoxy-resin to create an airtight seal
- Create a glass-metal airtight seal
- Make a ceramic-metal airtight seal.
<h3>What is an airtight seal?</h3>
An airtight seal is actually known to be a seal or sealing that doesn't permit air or gas to pass through. Airtight seal are usually known as hermetic seal. They are usually applied to airtight glass containers but the advancement in technology has helped to broaden the materials.
We can see that epoxy-resin can used to create an airtight seal. They create airtight seals to copper, plastics, stainless steels, etc.
When making glass-metal airtight seal, the metal should compress round the solidified glass when it cools.
Learn more about airtight seal on brainly.com/question/14977167
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In order to understand a monomer let´s first see the structure of a polymer. As an example, in the first figure polyethylene (or polyethene) is shown. This polymer, like every other one, is composed of many repeated subunits, these subunits are called monomer. In the second figure, polyethylene's monomer is shown.
Explanation:
Sum of forces in the x direction:
∑Fx = ma
Rx − 250 N = 0
Rx = 250 N
Sum of forces in the y direction:
∑Fy = ma
Ry − 120 N − 300 N = 0
Ry = 420 N
Sum of forces in the z direction:
∑Fz = ma
Rz − 50 N = 0
Rz = 50 N
Sum of moments about the x axis:
∑τx = Iα
Mx + (-50 N)(0.2 m) + (-120 N)(0.1 m) = 0
Mx = 22 Nm
Sum of moments about the y axis:
∑τy = Iα
My = 0 Nm
Sum of moments about the z axis:
∑τz = Iα
Mz + (250 N)(0.2 m) + (-120 N)(0.16 m) = 0
Mz = -30.8 Nm
