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2 years ago

12

In Process Costing, the journal entry to record the cost of goods transferred out from the WIP shaping department to the WIP pac

kaging department would be a debit to WIP - shaping department and a credit to WIP - packaging department. True False

1 answer:

ddd [48]2 years ago

4 0

Answer:

Journal entry :

Date Account and explanation Debit credit

WIP-Packing department XXX

WIP-Shaping department XXX

(To record cost of goods transferred out from the WIP shaping department to the WIP packaging department )

So above statement is False

Explanation:

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Answer:

no

Explanation:

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An air compressor of mass 120 kg is mounted on an elastic foundation. It has been observed that, when a harmonic force of amplit

kupik [55]

Answer:

equivalent stiffness is 136906.78 N/m

damping constant is 718.96 N.s/m

Explanation:

given data

mass = 120 kg

amplitude = 120 N

frequency = 320 r/min

displacement = 5 mm

to find out

equivalent stiffness and damping

solution

we will apply here frequency formula that is

frequency ω = ω(n) √(1-∈ ²) ......................1

here ω(n) is natural frequency i.e = √(k/m)

so from equation 1

320×2π/60 = √(k/120) × √(1-2∈²)

k × ( 1 - 2∈²) = 33.51² ×120

k × ( 1 - 2∈²) = 134752.99 .....................2

and here amplitude ( max ) of displacement is express as

displacement = force / k × ( $\frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}}$ )

put here value

0.005 = 120/k × ( $\frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}}$ )

k ×∈ × √(1-2∈²) = 1200 ......................3

so by equation 3 and 2

$\frac{k\varepsilon \sqrt{1-\varepsilon^2})}{k(1-2\varepsilon^2)} = \frac{12000}{134752.99}$

$\varepsilon^{2} - \varepsilon^{4} = 7.929 * 10^{-3} - 0.01585 * \varepsilon^{2}$

solve it and we get

∈ = 1.00396

and

∈ = 0.08869

here small value we will consider so

by equation 2 we get

k × ( 1 - 2(0.08869)²) = 134752.99

k = 136906.78 N/m

so equivalent stiffness is 136906.78 N/m

and

damping is express as

damping = 2∈ √(mk)

put here all value

damping = 2(0.08869) √(120×136906.78)

so damping constant is 718.96 N.s/m

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3 years ago

Wet steam at 15 bar is throttled adiabatically in a steady-flow process to 2 bar. The resulting stream has a temperature of 130°

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Answer:

$\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}$

Explanation:

The adiabatic throttling process is modelled after the First Law of Thermodynamics:

$m\cdot (h_{in} - h_{out}) = 0$

$h_{in} = h_{out}$

Properties of water at inlet and outlet are obtained from steam tables:

State 1 - Inlet (Liquid-Vapor Mixture)

$P = 1500\,kPa$

$T = 198.29\,^{\textdegree}C$

$h = 2726.9\,\frac{kJ}{kg}$

$s = 6.3068\,\frac{kJ}{kg\cdot K}$

$x = 0.967$

State 2 - Outlet (Superheated Vapor)

$P = 200\,kPa$

$T = 130\,^{\textdegree}C$

$h = 2726.9\,\frac{kJ}{kg}$

$s = 7.1776\,\frac{kJ}{kg\cdot K}$

The change of entropy of the steam is derived of the Second Law of Thermodynamics:

$\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}$

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Answer:

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35 mi/h = 0.464hp

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For the step by step explanation of how we arrived at these answers, please go through the attached files.

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