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tino4ka555 [31]
1 year ago
7

At the beginning of last year, tarind corporation budgeted $1,000,000 of fixed manufacturing overhead and chose a denominator le

vel of activity of 500,000 machine-hours. At the end of the year, tari's fixed manufacturing overhead budget variance was $14,000 favorable. Its fixed manufacturing overhead volume variance was $20,000 favorable. Actual direct labor-hours for the year were 525,000. What was tari's total standard machine-hours allowed for last year's output?.
Engineering
1 answer:
maxonik [38]1 year ago
7 0

The tari's total standard machine-hours allowed for last year's output 600,000 hours

Budgeted at start of year:  $100,000 fixed manufacturing overhead  for 500,000 machine hours

Standard = $100,000 / 500,000 hours = $0.2 fixed overhead / maching hour

At end of year, manufacturing overhead volume was $20,000 favorable which means

$20000 / $1=0.2 = 100000 additional hours.

Total Standard Machine Allowance Allowed for output = 500,000 + 100,000 = 600,000 hours.

the use of one machine running for an hour as a basis for cost estimation and operating effectiveness evaluation. In order to determine the contribution margin per machine hour for a specific product: a. Total cost per unit divided by the quantity of machine hours required to produce each unit of the target product. The manufacturing overhead cost divided by the activity driver yields the predetermined overhead rate. For instance, if machine hours were the activity driver, you would divide overhead costs by the anticipated number of machine hours.

Learn more about machine hours here:

brainly.com/question/14298561

#SPJ4

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A 10.2 mm diameter steel circular rod is subjected to a tensile load that reduces its cross- sectional area to 52.7 mm^2. Determ
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The percentage ductility is 35.5%.

Explanation:

Ductility is the ability of being deform under applied load. Ductility can measure by percentage elongation and percentage reduction in area. Here, percentage reduction in area method is taken to measure the ductility.

Step1

Given:

Diameter of shaft is 10.2 mm.

Final area of the shaft is 52.7 mm².

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Step2

Initial area is calculated as follows:

A=\frac{\pi d^{2}}{4}

A=\frac{\pi\times(10.2)^{2}}{4}

A = 81.713 mm².

Step3

Percentage ductility is calculated as follows:

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5 0
3 years ago
A one-dimensional plane wall of thickness 2L=80 mm experiences uniform thermal generation of q= 1000 W/m^3 and is convectively c
Eduardwww [97]

Answer:

h=1.99998\ W/m^2.C

k=33.333\ W/m.C

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Considering the one dimensional and steady state:

From Heat Conduction equation considering the above assumption:

\frac{\partial^2T}{\partial x^2}+\frac{\dot e_{gen}}{k}=0       Eq (1)

Where:

k is thermal Conductivity

\dot e_{gen} is uniform thermal generation

T(x) = a(L^2-x^2)+b

\frac{\partial\ T(x)}{\partial x}=\frac{\partial\ a(L^2-x^2)+b}{\partial x}=-2ax\\\frac{\partial^2\ T(x)}{\partial x^2}=\frac{\partial^2\ -2ax}{\partial x^2}=-2a

Putt in Eq (1):

-2a+\frac{\dot e_{gen}}{k}=0\\ k=\frac{\dot e_{gen}}{2a}\\ k=\frac{1000}{2*15}\\ k=33.333\ W/m.C

Energy balance is given by:

Q_{convection}=Q_{conduction}

h(T_L-T_{inf})=-k(\frac{dT}{dx}) _L     Eq  (2)

T(x) = a(L^2-x^2)+b

Putting x=L

T(L) = a(L^2-L^2)+b\\T(L)=b\\T(L)=40^oC

\frac{dT}{dx}=\frac{d(a(L^2-x^2)+b}{dx}=-2ax\\Put\ x\ =\ L\\\frac{dT}{dx}=-2aL\\(\frac{dT}{dx})_L=-2*15*0.04=-1.2

From Eq (2)

h=\frac{-k*-1.2}{(40-20)} \\h=\frac{-33.333*-1.2}{(40-20)}\\h=1.99998\ W/m^2.C

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