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il63 [147K]
3 years ago
10

Howe Corporation calculates inventory and cost of goods sold one time at the end of every accounting period. In contrast, Kelty

Industries updates their inventory and cost of goods sold accounts multiple times in one day. What is the difference between Howe and Kelty?
Business
1 answer:
Mariana [72]3 years ago
3 0

The difference is only in the strategy the company wants to use. For some market segments calculating the cost of goods sold by the permanent or periodic method may be more advantageous and allow a better monitoring of business efficiency and profitability. Companies often choose the method that best fits their organizational strategy. The periodic method, for example, as used by Kelty Industries, can be useful for greater input and output control, process optimization, consumer behavior assessment, and other advantages. But if Howe and Kelty wanted to change the calculation method, it would not affect anything, as the result would be the same regardless of the calculation, periodic or daily.

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3 years ago
A body in the solar system has a period of 10,759.22 days and a perihelion speed of 10.18 km/s. a. Calculate the aphelion radius
anastassius [24]

Answer:

Explanation:

From the information given, by applying Kepler's 3rd law,

T^2 \alpha  a^3

where;

T = period

a = semi major axis

T = 356 days (for earth)

a = 1 AU = 1.496 \times 10^8 \ km

Therefore, T^2 = ca^3

c= \dfrac{365^2}{(1.496 \times 10^8)^3}

c = 3.9791 \times 10^{20} \ day^2/km^3

However, if the body in the solar system has a period of 10.759.22 days, then, a =?

∴

T^2 = ca^3

a3 = \dfrac{10759.22^2}{3.9791 \times 10^{-20}}

a^3 = 2.9092 \times 10^{27}

a= \sqrt[3]{2.9092 \times 10^{27}}

a = 1.4275 \times 10^9 \ km

However, the velocity for a perihelion = 10.18 km/s

Using the formula

v = \sqrt{GM ( \dfrac{2}{r}-\dfrac{1}{a})} to calculate the radius, we have:

G = 6.674 \times 10^{-11}

M = 1.989\times 10^{30} \ kg

r = perihelion

v ^2= GM ( \dfrac{2}{r}-\dfrac{1}{a})

(10.18 \times 10^3) ^2= 6.674 \times 10^{-11} \times 1.989 \times 10^{30}  ( \dfrac{2}{r}-\dfrac{1}{1.425 \times 10^{12}})

7.8068 \times 10^{-13}= \dfrac{2}{r}-\dfrac{1}{1.425 \times 10^{12}}

\dfrac{2}{r} = 1.4824 \times 10^{-12}

r = \dfrac{2}{1.4824 \times 10^{-12}}

r = 1.349 \times 10^{12}

Similarly, the perihelion is expressed by the equation,

r = a(1 - e)

where ;

e= eccentricity

∴

1.349 \times 10^{12} = 1.425 \times 10^{12} ( 1 - e)

1.349 \times 10^{12}  -  1.425 \times 10^{12}= -  1.425 \times 10^{12} (e)

-7.6\times 10^{10}= -  1.425 \times 10^{12} (e)

\dfrac{-7.6\times 10^{10}}{-  1.425 \times 10^{12}}=  (e)

e ( eccentricity) = 0.0533

Aphelion radius in natural miles, r = a( 1+ e)

r = 1.425 \times 10^{12} ( 1 + 0.0533)

r = 1.50 \times 10^{12} \ m

to nautical miles, we have:

r = 1.50 \times 10^{12} \times 0.00054  \ nautical \ mile

radius of aphelion \mathbf{r = 8.10 \times 10^8} nautical miles

In respect to the value of a( i.e 1.4275 \times  10^9 \ km)

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Answer:

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sergejj [24]

Answer:

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Explanation:

Giving the following information:

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<u>The manufacturing overhead includes all indirect costs regarding production. </u>

<u></u>

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Manufacturing overhead=  $96,000

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