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3 years ago

You can change lanes during a turn long as there’s no traffic and you driving slowly

Engineering

1 answer:

Vanyuwa [196]3 years ago

5 0

Your allowed to switch lanes as long as the road is clear and you use signals.

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What is a smooth flow of air over a surface is called?

pochemuha

Answer:

laminar flow

Explanation:

In fluid dynamics, laminar flow is characterized by fluid particles following smooth paths in layers, with each layer moving smoothly past the adjacent layers with little or no mixing.

4 0

3 years ago

System Integration summary

kenny6666 [7]

Answer:

System integration can be defined as the progressive linking and testing of system components to merge their functional and technical characteristics into a comprehensive interoperable system.

Explanation:

....

6 0

3 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

Jelena is 26 and has worked in the service industry for most of her work history. She is willing to advance her education and tr

katrin2010 [14]

Customer service

Since she has reckless driving and her record, anything involving driving of some sorts mechanics as well won’t work well with her!! Hope this helps

8 0

3 years ago

(a) The reverse-saturation current of a pn junction diode is IS = 10−11 A. Determine the diode voltage to produce currents of (i

kirill115 [55]

Answer:

The equation used to solve a diode is

$i_d = I_se^\frac{V_d}{V_T}-1$

$i_d$ is the current going through the diode
$I_s$ is your saturation current
$V_D$ is the voltage across your diode
$V_T$ is the voltage of the diode at a certain room temperature. by default, you always use $V_T=25.9mV$ for room temperature.

If you look at the equation, $i_d = I_se^\frac{V_d}{V_T}-1$ , you'd notice that the $e^\frac{V_d}{V_T}$ grow exponentially fast, so we can ignore the -1 in the equation because it's so small compared to the exponential.