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.
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Umm the Water cycle sorry I’m trying
Answer:
8 mm
Explanation:
Given:
Diameter, D = 800 mm
Pressure, P = 2 N/mm²
Permissible tensile stress, σ = 100 N/mm²
Now,
for the pipes, we have the relation as:
where, t is the thickness
on substituting the respective values, we get
or
t = 8 mm
Hence, the minimum thickness of pipe is 8 mm
Answer:
0.024 m = 24.07 mm
Explanation:
1) Notation
= tensile stress = 200 Mpa
= plane strain fracture toughness= 55 Mpa
= length of a surface crack (Variable of interest)
2) Definition and Formulas
The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.
By definition we have the following formula for the tensile stress:
(1)
We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for
Multiplying both sides of equation (1) by 
(2)
Sequaring both sides of equation (2):
(3)
Dividing both sides by 
we got:
![\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B1%7D%7B%5Cpi%7D%5B%5Cfrac%7BK%7D%7BY%5Csigma_c%7D%5D%5E2)
(4)
Replacing the values into equation (4) we got:
![\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7B1%7D%7B%5Cpi%7D%5B%5Cfrac%7B55%20Mpa%5Csqrt%7Bm%7D%7D%7B1.0%28200Mpa%29%7D%5D%5E2%20%3D0.02407m)
3) Final solution
So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.
The change in annual cost when Q is increased from 340 to 341 is -1.23 and the instantaneous rate of change when Q = 340 is -1.25
<h3>How to find the Instantaneous rate of change?</h3>
The annual inventory cost C for a manufacturer is given as;
C = (1012000/Q) + 7.5Q
where Q is the order size when the inventory is replenished.
Now, the change in C can be calculated by evaluating the cost function at Q = 340 and Q = 341
Change in C = [1,012,000/341 + 7.5*341] - [1,012,000/340 + 7.5*340] ≈ -1.23
Instantaneous rate of change in C is first order derivative C':
C'(Q) = -1,012,000/(Q²) + 7.5
C'(340) = -1,012,000/(340²) + 7.5 ≈ -1.25
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