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2 years ago

Hello how are you all bye everyone have a great day ahead

Engineering

2 answers:

Step2247 [10]2 years ago

5 0

Answer:

Bye, have a great rest of your day. ;D

Explanation:

yKpoI14uk [10]2 years ago

3 0

Answer:

days good

Explanation:

because why not right

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Given the following phasors and the information related to the frequency of that phasor, provide the corresponding time-domain r

Evgesh-ka [11]

Answer:

Explanation:

In Engineering and Physics a Phasor That is a portmanteau of phase vector, is a complex number that represents a sinusoidal function whose Amplitude (A), Angular Frequency (ω), and Initial Phase (θ) are Time-invariant.

For the step by step solution to the question you asked, go through the attached documents.

4 0

2 years ago

A 650-kN column load is supported on a 1.5 m square, 0.5 m deep spread footing. The soil below is a well-graded, normally consol

insens350 [35]

<u>Explanation:</u>

Determine the weight of footing

$W_{f}=\gamma(L)(B)(D)$

Where $W_{f}$ is the weight of footing, γ is the unit weight of concrete, L is the length of footing is the width of footing, and D is the depth of footing

Substitute $2 m \text { for } L, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 23.6 kN / m ^{3}$ for γ in the equation

$\begin{aligned}W_{f} &=\left(23.6 kN / m ^{3}\right)(2 m )(1.5 m )(0.5 m ) \\&=35.4 kN\end{aligned}$

Therefore, the weight of the footing is 35.4 kN

Determine the initial vertical effective stress.

$\sigma_{z p}^{\prime}=\gamma(D+B)-u$

Here, $\sigma_{z^{p}}^{\prime}$ is initial vertical stress at a depth below ground surface γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute $18 kN / m ^{3} \text { for } \gamma, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 0$ for u in the equation

$\begin{aligned}\sigma_{z p}^{\prime} &=\left(18 kN / m ^{3}\right)(1.5+0.5) m -0 \\&=36 kPa\end{aligned}$

Therefore, the initial vertical stress is 36 kPa

Determine the vertical effective stress.

$\sigma_{z D}^{\prime}=\gamma D$

Here, $\sigma_{z^{p}}^{\prime}$ is initial vertical stress at a depth below ground surface γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute $18 kN / m ^{3}\) for $\gamma, 0.5 m$ for $D$ and 0 for \(u$ in the equation.

$\begin{aligned}\sigma_{z b}^{\prime} &=\left(18 kN / m ^{3}\right)(0.5 m )-0 \\&=9 kPa\end{aligned}$

Therefore, the vertical stress at a depth below the ground surface is

9 kPa

Determine the influence factor at the midpoint of soil layer,

$I_{e p}=0.5+0.1 \sqrt{\frac{q-\sigma_{s 0}^{\prime}}{\sigma_{z p}^{\prime}}}$

Here $I_{e p}$ is the influence factor at the midpoint of soil layer $\sigma_{z^{p}}^{\prime}$ is initial vertical stress, $\sigma_{z^{p}}^{\prime}$ is vertical effective stress, and $Q$ is bearing pressure

Substitute $36 kPa for $\sigma_{z p}^{\prime}, 228.47$ kPa for $q,$ and 9 kPa for $\sigma_{z D}^{\prime}$$ in the equation $\begin{aligned}I_{\epsilon P} &=0.5+0.1 \sqrt{\frac{228.47 kPa -9 kPa }{36 kPa }} \\&=0.75\end{aligned}$

Therefore the influence factor at the midpoint of the soil layer is 0.693

6 0

3 years ago

Vai trò của chủ đầu tư

White raven [17]

Answer:

can't understand the writting

4 0

2 years ago

Read 2 more answers

What are primary and secondary super-heaters?

Maru [420]

Explanation:

Superheater has two types of parts which are:

The primary super-heater
The secondary super-heater

Primary super-heater is first heater which is passed by the steam after steam comes out of steam drum.

After steam is heated on super primary heater, then the steam is passed on secondary super-heater so to be heated again. Thus, on secondary super-heater, the steam formed is hottest steam among others.

Steam from secondary super-heater which becomes the superheated steam, flow to rotate the High-Pressure Turbine.

3 0

2 years ago

Find the derivative of y = sin(ln(5x2 − 2x))

pickupchik [31]

Answer:

$y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)$

Explanation:

Let $y = \sin[\ln(5\cdot x^{2}-2\cdot x)]$ and we proceed to find the derivative by the following steps:

1) $y = \sin[\ln(5\cdot x^{2}-2\cdot x)]$ Given

2) $y = \sin [\ln[x\cdot (5\cdot x - 2)]]$ Distributive property

3) $y = \sin[\ln x + \ln (5\cdot x - 2 )]$ $\ln (a\cdot b) = \ln a + \ln b$

4) $y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)$ $\frac{d}{dx} (\sin x) = \cos x$ / $\frac{d}{dx}(\ln x) = \frac{1}{x}$ / $\frac{d}{dx}(c\cdot x^{n}) = n\cdot c\cdot x^{n-1}$ /Rule of chain/Result

3 0

2 years ago

Hello how are you all bye everyone have a great day ahead​

Hello how are you all bye everyone have a great day ahead