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Julli [10]
3 years ago
5

using the following data for july, calculate the cost of goods manufactured: beginning finished goods inventory 150,475. Ending

finished goods inventory 145,750. sales 400,000. Gross Margin 120,000. The cost of goods manufactured was
Engineering
1 answer:
Alla [95]3 years ago
5 0

Answer:

The correct response is "$275,275".

Explanation:

The given values are:

Sales,

= 400,000

Gross margin,

= 120,000

Beginning Inventory goods,

= 150,475

Finished inventory goods,

= 145,750

Now,

The cost of goods sold will be:

= Sales-Gross \ margin

On substituting the values, we get

= 400,000-120,000

= 280,000

As we know,

⇒ Cost \ of \ goods \ sold=Beginning \ inventory \ goods+ cost \ of \ goods \  manufactured-Ending \ inventory \ goods

⇒ 280,000=150,475+ cost \ of \ goods \ manufactured-145750

⇒ 280,000=cost \ of \ goods \ manufactured+4,725

⇒ Cost \ of \ goods \ manufactured=280,000-4,725

⇒                                                 =275,275 ($)

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Answer:

-273.16 °C

-459.677 °F

0 °K

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Explanation:

The lowest temperature is the absolute zero.

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Celsius and Fahrenheit degrees are relative scales, these have their zeroes above the absolute zero.

Celsius scale has the same degree separation as the Kelvin scale, but the zero is separated by 273.16 degrees. Therefore the lowest temperature in the Celsius scale is -273.16 °C.

The Fahrenheit degrees have the same degree separation as the Rankine degrees, and the zero is 459.67 degrees. Therefore the lowest temperature in the Fahrenheit scale is -459.67 °F.

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3 years ago
Which type of irrigation conserves more water than other types of irrigation?
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3 years ago
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What is the maximal coefficient of performance of a refrigerator which cools down 10 kg of water (and then ice) to -6C. Upper he
inysia [295]

Given:

Temperature of water, T_{1} = -6^{\circ}C =273 +(-6) =267 K

Temperature surrounding refrigerator, T_{2} = 21^{\circ}C =273 + 21 =294 K

Specific heat given for water, C_{w} = 4.19 KJ/kg/K

Specific heat given for ice, C_{ice} = 2.1 KJ/kg/K

Latent heat of fusion,  L_{fusion} = 335KJ/kg

Solution:

Coefficient of Performance (COP) for refrigerator is given by:

Max COP_{refrigerator} = \frac{T_{2}}{T_{2} - T_{1}}

= \frac{267}{294 - 267} = 9.89

Coefficient of Performance (COP) for heat pump is given by:

Max COP_{heat pump} = \frac{T_{1}}{T_{2} - T_{1}}\frac{294}{294 - 267} = 10.89

6 0
3 years ago
Example – a 100 kW, 60 Hz, 1175 rpm motor is coupled to a flywheel through a gearbox • the kinetic energy of the revolving compo
rjkz [21]

Answer:

1200KJ

Explanation:

The heat dissipated in the rotor while coming down from its running speed to zero, is equal to three times its running kinetic energy.

P (rotor-loss) = 3 x K.E

P = 3 x 300 = 900 KJ

After coming to zero, the motor again goes back to running speed of 1175 rpm but in opposite direction. The KE in this case would be;

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Since it is in opposite direction, it will also add up to rotor loss

P ( rotor loss ) = 900 + 300 = 1200 KJ

7 0
2 years ago
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