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3 years ago

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QS 7-13 Note receivable interest and maturity LO P4 On December 1, Daw Co. accepts a $36,000, 45-day, 10% note from a customer.

(1) Prepare the year-end adjusting entry to record accrued interest revenue on December 31. (2) Prepare the entry required on the note's maturity date assuming it is honored. (Use 360 days a year.)

1 answer:

asambeis [7]3 years ago

6 0

Answer and Explanation:

The journal entries are shown below:

1. Interest Receivable $300($36,000 × 10% x 30 ÷ 360)

To Interest Revenue $300

(Being accrued interest revenue is recorded)

2. Cash $36,450

To Interest Receivable A/c $300

To Interest Revenue A/c $150 ($36,000 × 10% x 15 ÷ 360)

To Notes Receivable A/c $36000

(Being note maturity date it is honoured is recorded)

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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:

$1170839.28 dyn/cm^2$

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.

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

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hoa [83]

Answer:

The annualized rate of return to the Swiss investor is -7.93%.

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This is an instance of foreign currency bond.

Using the exchange rate of $1 = 1.420, purchase price of the bond is calculated as $9,708.74 x 1.420 = 13,786.4108 Swiss Francs

Using the exchange rate of $1 = 1.324, maturity value is $10,000 x 1.324 = 13,240 Swiss Francs

Holding period is 6 months.

So, annualized rate of return is: (Maturity amount - Purchase price)/Purchase price x 12 / No of months

Annualized rate of return is: (13,240 - 13,786.4108)/13,786.4108 x 12/6 = -0.079268028.

Annualized rate of return is -7.93% approximately.

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