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Hitman42 [59]
3 years ago
15

On March 1, 2019, Baltimore Corporation had 65,000 shares of common stock outstanding with a par value of $5 per share. On March

1, Baltimore Corporation authorized a 15% stock dividend when the market value was $18 per share. Use this information to calculate the amount either (debited) or credited to retained earnings. Enter as a negative number if retained earnings is debited and a positive number if retained earnings is credited.
Business
1 answer:
andreyandreev [35.5K]3 years ago
7 0

Answer:

retained earnings 175,500

      common stock               48,750

      paid in excess of par   126,750

Explanation:

The diivdends are 15% so we multiply this by the shares outstanding to know the amount of shares:

65,000 x 15% = 9,750 shares

Then we multiply by the market value to know the amount needed:

9,750 x $18 market value = $175,500 stock dividends

The common stock will be 9,750 at par

and the remainder will be paid in excess.

9,750 x 5 = 48,750 CS

175,500 - 48,750 = 126,750

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3 years ago
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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Answer:

1170839.28 dyn/cm^2

16.9816 psia

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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

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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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