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3 years ago

To revise a monthly budget, changes in which categories might need to be addressed? Check all that apply.

Engineering

1 answer:

GREYUIT [131]3 years ago

5 0

Income

amount budgeted

expenses

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Two thousand pieces will flow through from the first machine A to the final machine F based on the given sequence of operations.

Vlad1618 [11]

The total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.

"Your question is not complete, it seems to be missing the following information;"

The sequence of operation is A - E - D - C - B - A - F

The given parameters;

number of pieces that will flow from the first machine A to machine F, = 2,000 pieces
initial unit load specified in the first machine, L₁ = 50
final unit load, L₂ = 100
the capacity of the vehicle = 1 unit load

The given sequence of operation of the vehicle;

A - E - D - C - B - A - F

the vehicle makes 6 trips for 100 unit loads

The total number of trips that the vehicle has to make, in order to transport the 2000 pieces of the load given, is calculated as follows.

100 unit loads ----------------- 6 trips

2000 unit loads --------------- ?

$= \frac{2000}{100} \times 6\\\\= 120 \ trips$

Thus, the total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.

Learn more here:brainly.com/question/21468592

6 0

2 years ago

Determine the constant speed at which the cable at A must be drawn in by the motor in order to hoist the load 6 m in 1.5s

zlopas [31]

Answer:

4m/s

Explanation:

We know that power supplied by the motor should be equal to the rate at which energy is increased of the mass that is to be hoisted

Mathematically

$Power_{motor} } =\frac{Energy }{time}$ \

We also know that Power = force x velocity ..................(i)

The force supplied by the motor should be equal to the weight (mg) of the block since we lift the against a force equal to weight of load

=> power = mg x Velocity........(ii)

While hoisting the load at at constant speed only the potential energy of the mass increases

Thus Potential energy = Mass x g x H...................(iii)

where

g = accleration due to gravity (9.81m/s2)

H = Height to which the load is hoisted

Equating equations (ii) and (iii) we get

m x g x v = $\frac{mgh}{t}$

thus we get v = H/t

Applying values we get

v = 6/1.5 = 4m/s

5 0

3 years ago

Water is flowing at a rate of 0.15 ft3/s in a 6 inch diameter pipe. The water then goes through a sudden contraction to a 2 inch

Georgia [21]

Answer:

Head loss=0.00366 ft

Explanation:

Given :Water flow rate Q=0.15 $\frac{ft^{3}}{sec}$

$D_{1}$ = 6 inch=0.5 ft

$D_{2}$ =2 inch=0.1667 ft

As we know that Q=AV

$A_{1}\times V_{1}=A_{2}\times V_{2}$

So $V_{2}=\frac{Q}{A_2}$

$V_{2}=\dfrac{.015}{\frac{3.14}{4}\times 0.1667^{2}}$

$V_{2$ =0.687 ft/sec

We know that Head loss due to sudden contraction

$h_{l}=K\frac{V_{2}^2}{2g}$

If nothing is given then take K=0.5

So head loss $h_{l}=(0.5)\frac{{0.687}^2}{2\times 32.18}$

=0.00366 ft

So head loss=0.00366 ft

4 0

3 years ago

While discussing the testing of a Hall-effect switch:

sergejj [24]

Answer:

your answer is technician A

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]

Explanation:

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