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fenix001
13 days ago
9

A hydraulic cylinder is to be used to move a workpiece in a manufacturing operation through a distance of 50 mm in 10 s. A force

of 10 kN is required to move the workpiece. Determine the required working pressure and hydraulic liquid flow rate if a cylinder with a piston diameter of 100 mm is available.
Engineering
1 answer:
Viktor [391]13 days ago
5 0

Response:

The solution to this question is 1273885.3 ∅

Clarification:

The first step is to ascertain the required hydraulic flow rate liquid based on the working pressure if a cylinder with a piston diameter of 100 mm is utilized.

Given that,

The distance = 50mm

The time t =10 seconds

The force F = 10kN

The piston diameter = 100mm

The pressure = F/A

10 * 10^3/Δ/Δ

P = 1273885.3503 pa

Subsequently

Power = work/time = Force * distance /time

= 10 * 1000 * 0.050/10

which amounts to =50 watt

Power =∅ΔP

50 = 1273885.3 ∅

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Mrrafil [318]

Answer:

Change in length = 0.0913 in

Explanation:

Given data:

Length = 6 ft

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Solution:

We start by applying the equilibrium moment about point C, expressed as

∑M(c) = 0.............1

This can be used to find the force in AB.

10× 200 × ( 5) - (T cos(30)) × 10 = 0

Solving gives us

Tension in wire T(AB) = 1154.7 lb

We also know the modulus of elasticity for A992 is

E = 29000 ksi

And the area will be

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The change in length is expressed as

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1 month ago
A shopaholic has to buy a pair of jeans , a pair of shoes l,a skirt and a top with budgeted dollar.Given the quantity of each pr
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Answer:

you may be struggling to pinpoint the separation between your inquiry and my perspective

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1 month ago
In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle tempe
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The value obtained is 0.60. From the given conditions, we take k = 1.4 for air and r = 16. We know that for calculations involving diesel engine efficiency, we arrive at 0.60.
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19 days ago
6.15. In an attempt to conserve water and to be awarded LEED (Leadership in Energy and Environmental Design) certification, a 20
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Explanation:

At a temperature of 33^{\circ} C and relative humidity of 86%, the humidity ratio stands at 0.0223 with a specific volume of 14.289.

At a temperature of 33^{\circ} C and relative humidity of 40%, the humidity ratio is 0.0066 while the specific volume is 13.535.

To determine the mass of air, the following formula can be used:

\begin{aligned}m _{1} &=\frac{ v }{ v }(1- w ) \\&=\frac{1 \times 10^{5}}{13.535}(1-0.0066) \\&=7339.49 lb / min \\v _{ a } &=\frac{ m _{1} v }{(1- w )} \\v _{ a } &=\frac{7339.49 \times 14.289}{(1-0.0223)} \\v _{ a } &=107266.0 ft ^{3} / min\end{aligned}

Now, we will calculate the volume

\begin{aligned}m _{ w } &=\frac{ v _{ a }}{ v _{ a }} w _{ a }-\frac{ v _{ i }}{ v _{ i }} w _{ i } \\&=\frac{107266.0}{14.289} \times 0.0223-\frac{100000}{13.535} \times 0.0066 \\&=118.64 lb / min\end{aligned}

The duration required to fill the cistern can be determined with the equation:

Time \(=\frac{\text { cistern volume }}{\text { removal water perminute volume }}\)

By substituting the values into the preceding formula, we find:

\(\frac{\left(15 \times 10^{3} L\right) \times\left(0.0353147 ft ^{3} / L \right)}{(118.641 b / min ) \times\left(\frac{1}{62.41 lb / ft ^{3}}\right)}\)\\\(=279.09\) minutes\\\(=4.65\) hours.

Thus, the hours necessary to fill the cistern amount to 4.65 hours.

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The collector should bid an amount he or she is comfortable paying for the statue, to ensure they do not feel disadvantaged by the bidding process. In this auction system, the final price paid corresponds to the second-highest bid rather than the highest bid submitted.

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