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Bezzdna [24]
2 years 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]2 years 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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An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate is found to be 1% pe
LenaWriter [7]

Answer:

a) Q = 251.758 kJ/mol

b) creep rate is    = 1.751 \times 10^{-5} \% per hr

Explanation:

we know Arrhenius expression is given as

\dot \epsilon =Ce^{\frac{-Q}{RT}

where

Q is activation energy

C is pre- exponential constant

At 700 degree C creep rate is\dot \epsilon = 5.5\times 10^{-2}% per hr

At 800 degree C  creep rate is\dot \epsilon = 1% per hr

activation energy for creep is \frac{\epsilon_{800}}{\epsilon_{700}} = = \frac{C\times e^{\frac{-Q}{R(800+273)}}}{C\times e^{\frac{-Q}{R(700+273)}}}

\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}

\frac{0.01}{5.5\times 10^{-4}} = ln [e^{\frac{Q}{8.314}[\frac{1}{1073} - \frac{1}{973}]}]

solving for Q we get

Q = 251.758 kJ/mol

b) creep rate at 500 degree C

we know

C = \epsilon e^{\frac{Q}{RT}}

    =- 1\% e{\frac{251758}{8.314(500+273}} = 1.804 \times 10^{12} \% per hr

\epsilon_{500} = C e^{\frac{Q}{RT}}

                         = 1.804 \times 10^{12}  e{\frac{251758}{8.314(500+273}}

                         = 1.751 \times 10^{-5} \% per hr

4 0
3 years ago
In digital communication technologies, what is an internal network also known as?
garri49 [273]
Intranet is a network that is internal and use internet technologies. It makes information of any company accessible to its employees and hence facilitates collaboration. Same methods can be used to get information, use resources, and update the data as that of the internet.
Hopefully this helped.
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Technician A says test lights are great for performing simple tests. Technician B says you can use a test light to check SRS cir
adoni [48]

The technician that is correct about either testing lights for simple tests or to check SRS Circuits is; Technician A.

<h3>Which Technician is Correct?</h3>

First of all it is pertinent to note that test lights are generally small bulbs that are turned on by the voltage and current flowing through the circuit in analog circuits.

Now, the  two values ​​of voltage and current are high and sufficient to light up the bulb. However, in digital circuits, the current is very small in the order of milliamps, and as a result there is not enough power to turn on the lights.

Thus, we can conclude that Technician A is correct.

Read more about Correct Technician at; brainly.com/question/14449935

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Answer:

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Explanation:

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List one advantage and one disadvantage of the use of the commutator?
Mariulka [41]

Answer and  Explanation:

Commutator are used in DC machine commutator is mainly used for the reversing the direction of the current .It is connected to the armature of the DC generator or motor

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DISADVANTAGE OF COMMUTATOR : The main disadvantage is due to the friction between the commutator and brushes there is a friction loss.

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