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Ray Of Light [21]

3 years ago

12

A company sells a plant asset that originally cost $450000 for $200000 on December 31, 2022. The accumulated depreciation accoun

t had a balance of $180000 after the current year's depreciation of $45000 had been recorded. The company should recognize a ________.

1 answer:

STALIN [3.7K]3 years ago

5 0

Answer:

$90,000 loss on disposal

Explanation:

If the current year's depreciation of $45,000 is recorded, the loss on disposal will be $45,000 multiplied by 2 which is $90,000

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2 years ago

The classified Balance Sheet will subsection the assets section as follows a. Current Assets and Property, Plant, and Equipment

anyanavicka [17]

Answer:

a. Current Assets and Property, Plant, and Equipment

Explanation:

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3 years ago

Vesnalui [34]

Answer:

d. willingness to pay of all buyers in the market.

Explanation:

The demand curve shows the relationship between the price of a good or service and the quantity demanded at a particular time.

Therefore, a demand curve reflects:

a. highest price buyers are willing to pay for each quantity.

b.quantity that each buyer will ultimately purchase.

c. value each buyer in the market places on the good.

With this in mind, what the demand curve does not reflect, with these in mind is a willingness to pay of all buyers in the market.

8 0

2 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

Radda [10]

Answer:

$1170839.28 dyn/cm^2$

16.9816 psia

$117.083928 kN/m^2$

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.

$mmHg=\frac{\rho_w*mmH_2O}{\rho_{Hg}}$

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

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.

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

8 0

3 years ago