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prohojiy [21]
3 years ago
8

Please answer the questions !

Engineering
1 answer:
gizmo_the_mogwai [7]3 years ago
6 0

Answer:

120

Explanation:

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Consider a fully-clamped circular diaphragm poly-Si with a radius of 250 μm and a thickness of 4 μm. Assume that Young’s modulus
Alina [70]

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Explanation:

find attached the solution to the question

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3 years ago
What are some quality assurance systems
Aloiza [94]
Examples of quality assurance activities include process checklists, process standards, process documentation and project audit. Examples of quality control activities include inspection, deliverable peer reviews and the software testing process. You may like to read more about the quality assurance vs quality control.
7 0
2 years ago
It is true about Metals and alloys: a)-They are good electrical and thermal conductors b)-They can be used as semi-conductors c)
ycow [4]

Answer:

(d) a and c are correct

Explanation:

METALS : Metal are those materials which has very high ductility, high modulus of elasticity, good thermal and electrical conductivity

for example : iron, gold ,silver, copper

ALLOYS: Alloys are those materials which are made up of combining of two or more than two metals these also have good thermal and electrical conductivity and me liable property

for example ; bronze and brass

so from above discussion it is clear that option (d) will be the correct option

8 0
3 years ago
Read 2 more answers
Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)
vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

7 0
3 years ago
Hareem and Shahad are on a road trip to Florida. They pull over to get gas, and as you may know it is sold by the gallon in the
Alona [7]

Answer:

53.3

Explanation:

4×14=56

56÷1.05=53.3

4 0
2 years ago
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