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Bezzdna [24]
1 year ago
12

The annual inventory cost C for a manufacturer is given below, where Q is the order size when the inventory is replenished. Find

the change in annual cost when Q is increased from 340 to 341, and compare this with the instantaneous rate of change when Q = 340. (Round your answer to two decimal places.
Engineering
1 answer:
Nataly_w [17]1 year ago
8 0

The change in annual cost when Q is increased from 340 to 341 is -1.23 and the instantaneous rate of change when Q = 340 is -1.25

<h3>How to find the Instantaneous rate of change?</h3>

The annual inventory cost C for a manufacturer is given as;

C = (1012000/Q) + 7.5Q

where Q is the order size when the inventory is replenished.

Now, the change in C can be calculated by evaluating the cost function at Q = 340 and Q = 341

Change in C = [1,012,000/341 + 7.5*341] - [1,012,000/340 + 7.5*340] ≈ -1.23

Instantaneous rate of change in C is first order derivative C':

C'(Q) = -1,012,000/(Q²) + 7.5

C'(340) = -1,012,000/(340²) + 7.5 ≈ -1.25

Read more about Instantaneous rate of change at; brainly.com/question/14666106

#SPJ1

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If a pilot-operated check valve (POC) does not check flow, you will see a. erratic actuator movement b, no actuator movement c.
Volgvan

If a pilot-operated check valve (POC) does not check flow, you will see a. erratic actuator movement.

<h3>What is a pilot-operated check valve (POC)?</h3>

Pilot operated test valves paintings through permitting loose float from the inlet port via the opening port. Supplying a pilot strain to the pilot port permits float withinside the contrary direction. Air strain on pinnacle of the poppet meeting opens the seal permitting air to float freely.

An actuator fault is a form of failure affecting the machine inputs. Due to strange operation or fabric aging, actuator faults might also additionally arise withinside the machine. An actuator may be represented through additive and/or multiplicative fault.

Read more about the pilot-operated check valve:

brainly.com/question/13001928

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7 0
2 years ago
A work element in a manual assembly task consists of the following MTM-1 elements: (1) R16C, (2) G4A, (3) M10B5, (4) RL1, (5) R1
ella [17]

Answer:

a)

1) R16C ; Tn = 17 TMU

2) G4A ; Tn = 7.3 TMU

3) M10B5 ; Tn = 15.1 TMU

4) RL1 ; Tn = 2 TMU

5) R14B ; Tn = 14.4 TMU

6) G1B ; Tn = 3.5 TMU

7) M8C3 ; Tn = 14.7 TMU

8) P1NSE ; Tn = 10.4 TMU

9) RL1 ; Tn = 2 TMU

b) 3.1 secs

Explanation:

a) Determine the normal times in TMUs for these motion elements

1) R16C ; Tn = 17 TMU

2) G4A ; Tn = 7.3 TMU

3) M10B5 ; Tn = 15.1 TMU

4) RL1 ; Tn = 2 TMU

5) R14B ; Tn = 14.4 TMU

6) G1B ; Tn = 3.5 TMU

7) M8C3 ; Tn = 14.7 TMU

8) P1NSE ; Tn = 10.4 TMU

9) RL1 ; Tn = 2 TMU

b ) Determine the total time for this work element in seconds

first we have to determine the total TMU = ∑ TMU = 86.4 TMU

note ; 1 TMU = 0.036 seconds

hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds

7 0
2 years ago
Calculate the maximum load that a 7075 series aluminum alloy bar (with a T6 temper heat treatment) can support without permanent
Aleksandr [31]

Answer:

The maximum load the bar can withstand = 35.43 KN

Explanation:

Ultimate tensile strength of the given aluminium bar \sigma = 540 M pa

Cross section area of the bar = 8.1^{2}  = 65.61 mm^{2}

We know that the ultimate strength of the bar is calculated from

\sigma = \frac{P_{max} }{A}

540 = \frac{P_{max} }{65.61}

P_{max} = 540 × 65.61

P_{max} =  35.43 KN

Therefore the maximum load the bar can withstand = 35.43 KN

6 0
4 years ago
What is the first thing a person must do before anything can be done today corroded area
melamori03 [73]

Answer:

Explanation:

We have learned that three things are required for the anodic and cathodic steps of corrosion to occur: an electrolyte, an exposed metal surface, and an electron acceptor. It follows, then, that we can prevent corrosion by removing one of these essential conditions.

8 0
2 years ago
A mass of 1.9 kg of air at 120 kPa and 24°C is contained in a gas-tight, frictionless piston–cylinder device. The air is now com
emmainna [20.7K]

Answer:

W=-260.66 kJ (negative answer means, that the work was done on the gas)

Explanation:

1) Convert temperature from C to K- T=24+273=297K- all temperature in the gas problems should be used in Kelvins;

2) We need to analyse type of the process- it is given, that the temperature is constant, so it is an Isothermal process, which means, that the equation of the process is: pV=const (constant);

3) Work, done on the system, should be calculated using the following equation: W=\int\limits^{Vb}_{Va} {p} \, dV

4) To calculate initical and final volumes (Va and Vb), we can use the following equation: pV=mRT, so V=mRT/p. Note, that the pressure is changing, thus we can calculate volumes for the both cases- initial and final, using initial (120kPa) and final (600kPa) pressures, in addition, we can find equation for the pressure, as function of the volume, which we need to use for the integration in step 3: p=mRT/V;

5) Now we can calculate the integral, given in the step 3: W=mRT ln(\frac{Vb}{Va}). As we have pressure as a known values, we can re-write the equation, using pressures: W=mRT ln(\frac{pa}{pb})=1.9*0.287*279*ln(\frac{120}{160})=-260.66 kJ

Note, that natural logarithm (ln) yields negative answer, which supports the question, that the work was done on the gas, not by the gas.

6 0
3 years ago
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