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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 controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.227 μF capacitor. T
anyanavicka [17]

Answer:

R min = 28.173 ohm

R max = 1.55 × 10^{4}  ohm

Explanation:

given data

capacitor = 0.227 μF

charged to 5.03 V

potential difference across the plates =  0.833 V

handled effectively = 11.5 μs to 6.33 ms

solution

we know that resistance range of the resistor is express as

V(t) = V_o \times e^{t\RC}    ...........1

so R will be

R = \frac{t}{C\times ln(\frac{V_o}{V})}    ....................2

put here value

so for t min 11.5 μs

R = \frac{11.5}{0.227\times ln(\frac{5.03}{0.833})}

R min = 28.173 ohm

and

for t max 6.33 ms

R max = \frac{6.33}{11.5} \times 28.173  

R max = 1.55 × 10^{4}  ohm

4 0
2 years ago
Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes
loris [4]

Answer:

a). 8.67 x 10^{-3} m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L  = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X 10^{-5} Pa-s

We know density of air is ρ = 1.21 kg /m^{3}

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = \frac{\rho .U.x}{\mu }

Re = \frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = \frac{5.x}{\sqrt{Re}}

   = \frac{5\times 1}{\sqrt{332052.6}}

   = 8.67 x 10^{-3} m

a). Boundary layer thickness at x = 1 is δ = 8.67 X 10^{-3} m

b). Given Re = 100000

    Therefore the critical distance from the leading edge can be found by,

     Re = \frac{\rho .U.x}{\mu }

     100000 = \frac{1.21\times5\times x}{1.822 \times10^{-5}}

     x = 0.3011 m

c). Given x = 3 m from the leading edge

    The Reyonld no at x = 3 m from the leading edge

     Re = \frac{\rho .U.x}{\mu }

     Re = \frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }

     Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

    δ = \frac{0.38\times x}{Re^{\frac{1}{5}}}

       = \frac{0.38\times 3}{996158.06^{\frac{1}{5}}}

       = 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

  Re = \frac{\rho \times U\times x}{\mu }

5 X 10^{5} = \frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}

 x = 1.505 m

We know that the force acting on the plate is

F_{D} = \frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}

and C_{D} at x= 1.505 for a laminar flow is = \frac{1.328}{\sqrt{Re}}

                                                                         = \frac{1.328}{\sqrt{5\times10 ^{5}}}

                                                                       = 1.878 x 10^{-3}

Therefore, F_{D} =  \frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}

                                          = \frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}

                                         = 0.2137 N

e). The flow is turbulent at the end of the plate.

  Re = \frac{\rho \times U\times x}{\mu }

       = \frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }

       = 1992316

Therefore C_{D} = \frac{0.072}{Re^{\frac{1}{5}}}

                                           = \frac{0.072}{1992316^{\frac{1}{5}}}

                                           = 3.95 x 10^{-3}

Therefore F_{D} = \frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}

                                           = \frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}

                                          = 1.792 N

3 0
3 years ago
Exercises
Feliz [49]

Answer:

Rocket

Gas

Explanation:

5 0
2 years ago
9. How is the Air Delivery temperature controlled during A/C operation?
Rzqust [24]

Answer:

i believe the answer is a but i could be wrong

Explanation:

i hope it helps

6 0
3 years ago
It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 630 MPa. Calc
ratelena [41]

Answer:

The answer is below

Explanation:

Given that:

Diameter (D) = 0.03 mm = 0.00003 m, length (L) = 2.4 mm = 0.0024 m, longitudinal tensile strength (\sigma_{cd})=630\ MPa = 630*10^6\ Pa, Fracture strength

(\sigma_f)=5100\ MPa=5100*10^6\ Pa,fiber-matrix\ stres(\sigma_m)=17.5\ MPa=17.5*10^6\ Pa,matrix\ strength=\tau_c=17\ MPa=17 *10^6\ Pa

a) The critical length (L_c) is given by:

L_c=\sigma_f*(\frac{D}{2*\tau_c} )=5100*10^6*\frac{0.00003}{2*17*10^6}=0.0045\ m=4.5\ mm

The critical length (4.5 mm) is greater than the given length, hence th composite can be produced.

b) The volume fraction (Vf) is gotten from the formula:

\sigma_{cd}=\frac{L*\tau_c}{D}*V_f+\sigma_m(1-V_f)\\\\V_f=\frac{\sigma_{cd}-\sigma_{m}}{\frac{L*\tau_c}{D}-\sigma_{m}}  \\\\Substituting:\\\\V_f=\frac{630*10^6-17.5*10^6}{\frac{0.0024*17*10^6}{0.00003} -17.5*10^6} \\\\V_f=0.456

6 0
2 years ago
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