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4 years ago

How are isometric drawings and orthographic drawings similar?

Engineering

1 answer:

sashaice [31]4 years ago

5 0

Answer:

The horizontal lines of the orthographic drawing

are NOT horizontal in the isometric drawing but are projected at 30-degree and 60. ... degree angles, and the length of the lines remain the same in the isometric as they were in the orthographic.

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Javier’s class visited a power plant near his city, and they learned how it produced electricity. What does this form of power d

adelina 88 [10]

The electricity is generated by the burning of fuel in the power plant that increases the kinetic energy of water vapor which rotates the turbine and shaft of the generator.

<h3>What is a power plant?</h3>

A power plant is a place where heavy machinery objects are installed to produce electricity.

The power is generated in the power plant with the help of the burning of fuel that increases the temperature of the boiler. There are so many tubes in which water is present. Water absorbs heat and increases its kinetic energy. And this kinetic energy help to rotate the turbine and the shaft of the generator is attached to the turbine as result electricity is generated.

More about the power plant link is given below.

brainly.com/question/7670779

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2 years ago

Problem 3: Soil Classification using the AASHTO and USCS Systems

nataly862011 [7]

<u>Solution:</u>

$Given\\ $\quad W=3000 Ib , \quad m=\frac{W}{g}=\frac{3000}{322} \ slug =93.1677 slug$\\K_{e q}=2160 lbs / wp =2100 \frac{ lbs }{10} \frac{ x 12}{1 ft }=(2160 \times 12) lb / ft$$$

a) The natural frequency

$\begin{aligned}&\left(\omega_{n}\right)=\sqrt{\frac{K_{e q}}{m}}\\&=\sqrt{\frac{2160 \times 12}{93.1677}}\\&\omega_{n}=16.68 \text { rad } | s\\&\omega_{n}=\frac{2 \pi}{T}\\&16.68=\frac{2 \pi}{T}\\&T=0.3767 s\end{aligned}$

$Given, $t=10 s , \quad y(t)=6 in = A$\\$y(t)=A \cos \left(\omega_{n} t+\phi\right) \rightarrow 0$\\$6=6 \cos (16.68 \times 10+\phi)$\\$1=\cos (166.8+\phi)$\\$166.8+\phi=0$\\\phi=-166.8\)\\At $t=0, \quad y(0)=6 \cos (16.68 \times 0-166.8)$ {y(0)}=-5.74 in$

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3 years ago

What is the difference between a single-model production line and a mixed-model production line?

Nat2105 [25]

The unique model production line is responsible for producing identical pieces. For this purpose the balancing of the assembly line is only responsible for assembling a model throughout the line.

This is a considerable difference compared to the mixed model assembly line where many models are assembled during the same production line, that is, it produces parts or products that have slight changes accommodated in them, with slight variations in their model or products of soft variety

The choice of the type of production depends on the type of company and its own demand, always prioritizing the efficiency in the operation. Generally, the mixed model tends to be chosen when demand is very large and customer demand is required to be met. In others it is considered a plant model in which half of the line is mixed and the other one is the only model in order to keep the efficiency balanced.

6 0

3 years ago

Water exiting the condenser of a power plant at 45 Centers a cooling tower with a mas flow rate of 15,000 kg/s. A stream of cool

MariettaO [177]

Answer: hello your question is incomplete below is the missing part

question :Determine the temperature of the cooled water exiting the cooling tower

answer : T = 43.477° C

Explanation:

Temp of water at exit = 45°C

mass flow rate of cooling tower = 15,000 kg/s

Temp of makeup water = 20°C

Assuming an atmospheric pressure of = 101.3 kPa

<u>Determine temperature of the cooled water exiting the cooling tower</u>

Water entering cooling tower at 45°C

Given that Latent heat of water at 45°C = 43.13 KJ/mol

Cp(wet air) = 1.005+ 1.884(y1)

where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273

Hence : Cp(wet air) = 29.145 + (0.01273) (33.94) = 29.577 KJ/kmol°C

<u>First step : calculate the value of Q </u>

Q = m*Cp*(ΔT) + W(latent heat)

Q = 321.6968 (29.577) (40-30) + 43.13 (18.26089)

Q = 95935.8547 KJ/s

Given that mass rate of water = 15000 kg/s

<u>Hence the temperature of the cooled water can be calculated using the equation below</u>

Q = m*Cp*∆T

Cp(water) = 4.2 KJ/Kg°C

95935.8547 = (15000)*(4.2)*(45 - T)

( 45 - T ) = 95935.8547/ 63000. ∴ T = 43.477° C

5 0

3 years ago

Does somebody know how to do this?

maksim [4K]

No I don’t sorry, I hope you do well

4 0

3 years ago