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2 years ago

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Technician A says a solenoid is not used on modern automobiles. Technician B says fuel injectors and starter motor solenoids are

two of many solenoids used on automobiles. Who is correct?
Select one
A Technician A
B. Technician B
C Both Techs A and B

1 answer:

julia-pushkina [17]2 years ago

5 0

Answer:

ON THE BOTTOM

Explanation:

red cord plus black cord

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What does it mean to say that PEER is a data-driven, consumer-centric, and comprehensive system?

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Answer:

have you heard of gnoogle?

Explanation:have you heard of goongle?

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3 years ago

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A series of concrete pillars have been built on the border between Kuwait and Iraq. They are there to __________ the border.

Gala2k [10]

A series of concrete pillars have been built on the border between Kuwait and Iraq. They are there to demarcate the border.

<h3>What is a border?</h3>

A border is a geographical boundary that separate<em> countries, states, provinces, counties, cities, and towns.</em>

A series of concrete pillars have been built on the border between Kuwait and Iraq. They are there to demarcate the border.

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1 year ago

Analyse what effect the building of an airport may have on the decision of how to use an area of land nearby. (6)

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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

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3 years ago

Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces app

stiks02 [169]

Answer:

a) Fb= 275.77 lb Fc= 142.75 lb

b) M = -779.97 lb.ft (i.e. 779.97 lb.ft in clockwise direction)

c) Fax = 195 lb

Fay = 337.75 lb

Fbx = 195 lb

Fby = 195 lb

Explanation:

Question: Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces applied should be couples. The tugboat at point A applies a 390 lb force.

(a) Determine FB and FC so that only couples are applied.

(b) Using your answers to Part (a), determine the resultant couple moment that is produced.

(c) Resolve the forces at A and B into x and y components, and identify the pairs of forces that constitute couples.

Solution:

<u>For this problem Right hand side is positive X direction and Upwards is positive Y direction. Couples and moments will be considered positive in counterclockwise direction.</u>

<u />

a) For no translation condition

∑ $F_{x} = 0$ & ∑ $F_{y} = 0$

Hence,

$F_{A}cos(30) - F_{B}cos(45) - F_{C} = 0$

$F_{A}sin(30) - F_{B} sin(45) = 0$

and

$F_{A} = 390 lb$

Inserting the value of $F_{A}$ and solving the remaining equations simultaneously yields (magnitudes),

$F_{B} = 275.77 lb\\F_{C} = 142.75 lb$

b) Summing up moments

$M=45 ( -F_{Ay}-F_{Cy}) +5 (-F_{By})+22(-F_{Ax}-F_{Bx})\\ =45(-390cos(30)-142.75)+5(-275.77cos(45))+22 (-390sin(30)-275.77sin(45))$

$M = -779.97 lb.ft$ (i.e. 779.97 lb.ft clockwise)

c)

$F_{Ax} = 390 sin(30) = 195 lb$

$F_{Ay} = 390 cos(30) = 337.75lb\\$

$F_{Bx} = 275.77 sin(45) = 195lb\\F_{By} = 275.77 cos(45) = 195 lb$

8 0

3 years ago