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liberstina [14]
3 years ago
9

Assume the following property and equipment footnote was obtained from the Deere & Company's 2016 10-K. Property and Depreci

ation A summary of property and equipment at October 31 follows: ($ millions) Average Useful Lives (Years) 2016 2015 Equipment Operations Land $79 $75 Buildings and building equipment 25 1,490 1,419 Machinery and equipment 10 2,961 2,870 Dies, patterns, tools, etc 7 1,039 987 All other 5 589 571 Construction in progress 232 156 Total at cost 6,390 6,078 Less accumulated depreciation 4,113 3,966 Total $2,277 $2,112 During 2016, the company reported $636.5 million of depreciation expense (this expense also includes amortization expense relating to computer software that is included with property and equipment). Estimate the percent used up of Deere's depreciable assets.
Business
1 answer:
JulijaS [17]3 years ago
5 0

Answer:

Explanation:rerererererere

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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
Radda [10]

Answer:

1170839.28 dyn/cm^2

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Explanation:

To calculate the absolute pressure in the bottom of tube we need to sum the atmosferic and gauge pressure.

P_{abs}=P_{atm}+P_G

And the gauge pressure is given by the contributions of columns of water (P_{w}) and mercury(P_{Hg}), we can calculate the contribution of each column as:

P= \rho g h (*)

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

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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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Typically, the<u> laissez-faire</u> style of a leader avoids dominating groups.

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