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iris [78.8K]
2 years ago
13

Company acquired a new truck at at cost of $52,000 on January 1, 2019. The truck is expected to be used for 4 years and have a s

alvage value of $8,000 at that time. The company uses the straight-line method to calculate depreciation. #1. Calculate depreciation expense for 2019 #2 Calculate depreciation expense for 2020 #3 Calculate accumulated depreciation as of 12/31/2019 #4 Calculate accumulated depreciation as of 12/31/2020 #5 Calculate the book value as of 12/31/2020
Business
1 answer:
AfilCa [17]2 years ago
3 0

Answer and Explanation:

The computation is shown below:

= (Original cost - residual value) ÷ (useful life)

= ($52,000 - $8,000) ÷ (4 years)

= ($44000) ÷ (4 years)  

= $11,000

In this method, the depreciation is same for all the remaining useful life

a. The depreciation expense for 2019 is $11,000

b.  The depreciation expense for 2020 is $11,000

c. The accumulated depreciation for year 2019 is $11,000

d. The accumulated depreciation for year 2020 is $22,000 ($11,000 + $11,000)

e. The book value is

= Original cost - accumulated depreciation

= $52,000 - $22,000

= $30,000

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Pressure with Two Liquids, Hg and Water. An open test tube at 293 K is filled at the bottom with 12.1 cm of Hg, and 5.6 cm of wa
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1170839.28 dyn/cm^2

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117.083928 kN/m^2

Explanation:

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where \rho is the respective density, g gravity and h is height.

So we have all the data required to use the above equations (P_{atm}, height and density of each column) we only need to be carefully with the units.

For simplicity we can to express all pressure contributions in mmHg ( P_{atm}, P_{w} and P_{Hg}). Note that the units "x" mmHg  means the pressure at the bottom of a column of mercury of "x" mm high. For example, in this case we have a column 12.1 cm of Hg, that is a column of 121 mmHg (passing from cm to mm only requires multiply by 10) pressure exerted by that column is 121 mmHg.

Now pressure of 5.6 cm (56 mm) of water would be 56 mm of water, but it is not the same that mmHg, since the density of water is lower, the pressure exerted by 1 mm of water is lower than the exerted by 1 mm of Hg. The conversion between mmHg and mm of water is given by the relation between the densities.

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mmHg=\frac{0.998*mmH_2O}{13.55}=0.0737 mmH_2O

And pressure of water in mmHg is

0.0737*56=4.1246 mmHg

The absolute pressure is:

P_{abs}=P_{atm}+P_G= 756 + 121 + 4.1246  = 881.1246 mmHg = 88.11246cmHg

To pass to dyn/cm^2 units we need to use the equation (*)

P= \rho g h = 13.55 \frac{g}{cm^3} * 980.665 \frac{cm}{s^2} * 88.11246 cmHg = 1170839.28 \frac{g}{cm s^2} = 1170839.28 \frac{dyn}{cm^2}

Note: We need to use cm Hg for units coherence

Now passing from dyn/cm^2 to kN/m^2 (or kPa) we need to consider that 1 dyn is 10^{-8} kN and 1 cm^2 is 10^{-4} m^2.  

1170839.28 \frac{dyn}{cm^2} * \frac{10^{-8}kN}{1 dyn}*\frac{cm^2}{10^{-4}m^2}=117.083928kN/m^2

Now passing kN/m^2 to psia. We need to consider that 1 psia is 6.89476.

117.083928kN/m^2*\frac{1psia}{6.89476kN/m^2}=16.9816 psia

 

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