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ehidna [41]
3 years ago
14

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

is 130 GPa and Poisson’s ratio is 0.3. A uniform pressure of 20 kPa is applied.
(a) What is the stress at the center of the diaphragm?
(b) What is the radial stress at the fully-clamped edge of the diaphragm?
(c) What is the transverse stress at the fully-clamped edge of the diaphragm?
(d) What is the amount of deflection at the center of the diaphragm?

Engineering
1 answer:
Alina [70]3 years ago
4 0

Answer:

Explanation:

find attached the solution to the question

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What is the angular velocity (in rad/s) of a body rotating at N r.p.m.?
Darina [25.2K]

Answer:

0.1047N

Explanation:

To solve this problem we must remember the conversion factors, remembering that 1 revolution equals 2π radians and 1min equals 60s

N\frac{rev}{min} \frac{2\pi }{1rev} \frac{1min}{60} =N\frac{2\pi }{60} =0.1047N

in conclusion, to know how many rad / s an element rotates which is expressed in Rev / min we must only multiply by 0.1047

3 0
3 years ago
R-744 refrigerant is bad why
mars1129 [50]

Answer:

Explanation:

R-744 is seen as the 'perfect' natural refrigerant as it is climate neutral and there is not a flammability or toxicity risk. It is rated as an A1 from ASHRAE. While it is non-toxic there is still risk if a leak occurs in an enclosed area as R-744 will displace the oxygen in the room and could cause asphyxiation

6 0
3 years ago
The 10 foot wide circle quarter gate AB is articulated at A. Determine the contact force between the gate and the smooth surface
slamgirl [31]

Answer:

F = 641,771.52 \dfrac{lb-ft}{s^2}

Explanation:

Given that

R=8 ft

Width= 10 ft

We know that hydro statics force given as

  F=ρ g A X

ρ is the density of fluid

A projected area on vertical plane

X is distance of center mass of projected plane from free surface of water.

Here

X=8/2  ⇒X=4 ft

A=8 x 10=80  ft^2

So now putting the values

F=ρ g A X

F=62.4(32.14)(80)(4)

F = 641,771.52 \dfrac{lb-ft}{s^2}

   

4 0
3 years ago
A sample of sand weighs 490 g in stock and 475 in Oven Dry (OD) condition, respectively. If absorption capability of the sand is
Ivahew [28]

The weight of the specimen in SSD condition is 373.3 cc

<u>Explanation</u>:

a) Apparent specific gravity = \frac{A}{A-C}

Where,

A = mass of oven dried test sample in air = 1034 g

B = saturated surface test sample in air = 1048.9 g

C = apparent mass of saturated test sample in water = 975.6 g

apparent specific gravity = \frac{A}{A-C}

                                         = \frac{1034}{1034-675 \cdot 6}

Apparent specific gravity = 2.88

b) Bulk specific gravity G_{B}^{O D}=\frac{A}{B-C}

G_{B}^{O D}=\frac{1034}{1048.9-675 \cdot 6}

       =  2.76

c) Bulk specific gravity (SSD):

G_{B}^{S S D}=\frac{B}{B-C}

=\frac{1048 \cdot 9}{1048 \cdot 9-675 \cdot 6}

G_{B}^{S S D} = 2.80

d) Absorption% :

=\frac{B-A}{A} \times 100 \%

=\frac{1048 \cdot 9-1034}{1034} \times 100

Absorption = 1.44 %

e) Bulk Volume :

v_{b}=\frac{\text { weight of dispaced water }}{P \omega t}

=\frac{1048 \cdot 9-675 \cdot 6}{1}

= 373.3 cc

5 0
2 years ago
Which of the following is a possible consequence of poor measurement in construction as stated in the segment?
Kamila [148]

Answer:

Decreased risk of structure failure

6 0
2 years ago
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