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Nadya [2.5K]
3 years ago
15

Paint can shaker mechanisms are common in paint and hardware stores. While they do a good job of mixing the paint, they are also

noisy and transmit their vibrations to the shelves and counters. A better design of the paint can shaker is possible using a balanced fourbar linkage. Design such a portable device to sit on the floor (not bolted down) and minimize the shaking forces and vibrations while still effectively mixing the paint.
Engineering
1 answer:
Ymorist [56]3 years ago
6 0

Answer:

A good design for a portable device to mix paint minimizing the shaking forces and vibrations while still effectively mixing the paint. Is:

The best design is one with centripetal movement. Instead of vertical or horizontal movement. With a container and system of holding structures made of materials that could absorb the vibration effectively.

Explanation:

First of all centripetal movement would be friendlier to our objective as it would not shake the can or the machine itself with disruptive vibrations. Also, we would have to use materials with a good grade of force absorption to eradicate the transmission of the movement to the rest of the structure. Allowing the reduction of the shaking forces while maintaining it effective in the process of mixing.

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This question allows you to practice proving a language is non-regular via the Pumping Lemma. Using the Pumping Lemma (Theorem 1
Ulleksa [173]

Answer:

<em>L is not a regular language with formal proofs  </em>

Explanation:

<em>(a) To prove that L is not a regular language, we will use a proof by contradiction. the assumption entails  that L is a regular language. Then by the Pumping Lemma for Regular Languages, </em>

<em>there exists a pumping length p for L such that for any string s ∈ L where |s| ≥ p, </em>

<em>s = xyz subject to the following conditions: </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L</em>

<em />

<em>(b) To determine that L is not a regular language, we mke use of proof by contradiction.  lets assume, that L is regular. Then by the Pumping Lemma for Regular Languages, it states also,</em>

<em>The pumping length, p for L such that for any string s ∈ L where |s| ≥ p, s = xyz subject  to the condtions as follows : </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L. </em>

<em>Choose s = 0p10p </em>

<em>. Clearly, |s| ≥ p and s ∈ L. By condition (b) above, it follows is shown. by the first condition x and y are zeros.</em>

<em>for some  k > 0. Per (c), we can take i = 0 and the resulting string will still be in L. Thus,  xy0 </em>

<em>z should be in L. xy0 </em>

<em>z = xz = 0(p−k)10p </em>

<em>It is shown that is is  not in L. This is a  contraption with the pumping lemma.  our assumption that L is regular is  incorrect, and L is not a regular language</em>

6 0
3 years ago
A sample of wastewater is diluted 10 times. The diluted solution has an ultimate biochemical oxygen demand (BOD), Lo, of 30 mg/L
zzz [600]

Answer:

474.59 mg/L

Explanation:

Given that

BOD = 30 mg/L

Original BOD  = 30 mg/L × dilution factor

Original BOD  = 30 mg/L  × 10 = 300 mg/L

L_o = \frac{BOD}{1-e^{-5t}}

here L_o is the ultimate BOD ; BOD is the  biochemical oxygen demand ;  t = 0.20 /day

L_o = \frac{300}{1-e^{-5(0.20)}}

L_o = 474.59 \ mg/L

3 0
3 years ago
A fluid of specific gravity 0.96 flows steadily in a long, vertical 0.71-in.-diameter pipe with an average velocity of 0.90 ft/s
KengaRu [80]

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∅ ) D^{2} / 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

8 0
3 years ago
Read 2 more answers
Exercises
Feliz [49]

Answer:

Rocket

Gas

Explanation:

5 0
2 years ago
Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb
guajiro [1.7K]

Answer:

\eta_{turbine} = 0.603 = 60.3\%

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in  vapor state:

h₂ = h_{g\ at\ 125KPa} = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than s_g and greater than s_f at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88

Now, we will find h_{2s}(enthalpy at the outlet for the isentropic process):

h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg

Now, the isentropic efficiency of the turbine can be given as follows:

\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%

3 0
3 years ago
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