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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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3 years ago
A piston–cylinder device contains 0.78 kg of nitrogen gas at 140 kPa and 37°C. The gas is now compressed slowly in a polytropic
DiKsa [7]

Answer:

The entropy change of nitrogen during this process. is - 0.32628 kJ/K.

Explanation:

Solution

Given that:

A piston cylinder device contains =0.78 kg of nitrogen gas

Temperature = 37°C

The  nitrogen gas constant of R = 0.2968 kJ/kg.K

At room temperature cv = 0.743 kJ/kg.K

Now,

We assume that at specific condition the nitrogen can be treated as an ideal gas

Nitrogen has a constant volume specific heat at room temperature.

Thus,

From the polytropic relation, we have the following below:

T₂/T₁ =(V₁/V₂)^ n-1 which is,

T₂ = T₁ ((V₁/V₂)^ n-1

= (310 K) (2)^1.3-1 = 381.7 K

So,

The entropy change of nitrogen is computed as follows:

ΔSN₂ = m (cv₁ avg ln T₂/T₁ + ln V₂/V₁)

= (0.78) ((0.743 kJ/kg .K) ln 381.7 K/310K + (0.2968 kJ/kg. K) ln (0.5))

= 0.57954 * 0.2080 +  (-0.2057)

= 0.12058 + (-0.2057) = -0.32628

Therefore the entropy change of nitrogen during this process. is - 0.32628 kJ/K.

6 0
3 years ago
The beam is supported by a pin at A and a roller at B which has negligible weight and a radius of 15 mm. If the coefficient of s
Anettt [7]

Answer:

33.4

Explanation:

Step 1:

\sumMo=0 (moment about the origin)

Fb(15)-Fc(15)=0

Fb=Fc

Step 2:

\sumFx=0

-Fb-Fccos\theta+Ncsin\theta=0

Fc=0.3Nc=Fb

-0.3Nc-0.3Nccos\theta+Ncsin\theta=0

(-0.3-cos\theta+sin\theta)Nc=0----(1)

\sumFy=0

Nccos\theta+Fcsin\theta-Nb=0

Nccos\theta+0.3Ncsin\theta-Nc=0

Nc[cos\theta+0.3sin\theta-1]=0--------(2)

Solving eq (1) and eq (2)

\theta=33.4

Step 3:

As the roller is a two force member

2(90-\phi)+\theta=180

\phi=\theta/2

\phi=Tan(\muN/N)-1

\phi=16.7

\theta=2x16.7=33.4

5 0
3 years ago
For a 20 ohm resistor R, the current i = 2 A. What is the voltage V? Submit your answer as a number without units. ​
svetoff [14.1K]

Answer:

20*2=40

Explanation:

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3 years ago
Write what you already know about college majors. What are they? Can you think of any examples? When do you have to pick one? Ca
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3 years ago
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