On January 1 st 2012, Everhart Corporation, a calendar year company issues $100,000, 5%, 5-year bonds dated January 1, 2012. The
bond pays interest semiannually on January 1 and July 1 . The bonds are issued to yield 6%. 2.50% 3.00% 5.00% 6.00% Present value of a single sum for 5 periods 0.88385 0.86261 0.78353 0.74726 Present value of a single sum for 10 periods 0.78120 0.74409 0.61391 0.55839 Present value of an annuity for 5 periods 4.64583 4.57971 4.32948 4.21236 Present value of an annuity for 10 periods 8.75206 8.53020 7.72173 7.36009 If Everhart Corporation uses the effective interest method to amortize any premiums or discounts on their
outstanding bonds, what will be the journal entries to record interest expense for calendar year 2013?
outstanding bonds, what will be the journal entries to record interest expense for calendar year 2013?
1 answer:
Answer:
Interest expense 2894.7 debit
discount on Bonds Payable 394.7 credit
cash 2500 credit
Interest expense 2906.55 debit
discount on Bonds Payable 406.55 credit
interest payable 2500 credit
Explanation:
We have to solve for the 2013 year which is one year after the issuance ofthe bonds.
We solve for the bond issuance price and then, we construct the bonds schedule and take the numbers from period 3 and 4.
Issuance proceeds: present value fo the coupon payment and maturity at market rate:
C 2,500.000
time 10
rate 0.03
PV $21,325.5071
Maturity 100,000.00
time 10.00
rate 0.03
PV 74,409.39
PV c $21,325.5071
PV m $74,409.3915
Total $95,734.8986
Now we will calcautlethe interest expense by multiplying carrying value by the market value and sutract from the cash outlay to determinate the amortization on the bonds.
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Answer:
16.9816 psia
Explanation:
To calculate the absolute pressure in the bottom of tube we need to sum the atmosferic and gauge pressure.
And the gauge pressure is given by the contributions of columns of water () and mercury(
), we can calculate the contribution of each column as:
(*)
where is the respective density, g gravity and h is height.
So we have all the data required to use the above equations (, 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 ( ,
and
). 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.
And pressure of water in mmHg is
The absolute pressure is:
To pass to dyn/cm^2 units we need to use the equation (*)
Note: We need to use cm Hg for units coherence
Now passing from dyn/ to kN/
(or kPa) we need to consider that 1 dyn is
kN and 1
is
.
Now passing kN/ to psia. We need to consider that 1 psia is 6.89476.