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Nimfa-mama [501]

3 years ago

14

XYZ Company purchased a new machine on January 1, 2020 for $160,000. The machine was assigned a 40-year life and a $3,000 residu

al value. XYZ Company will depreciate the machine using the double-declining balance depreciation method. Calculate the amount of accumulated depreciation related to the new machine that would appear in XYZ Company's December 31, 2023 balance sheet.

1 answer:

AlekseyPX3 years ago

4 0

Answer:

Accumulated depreciation at December 32, 2023 = 29679

Explanation:

machine = 160000

useful life = 40 years

residual value = 3000

straight-line depreciation rate = (160000 - 3000) ÷ 40

= 3925 per year

Annual depreciation rate = 100% ÷ 40 = 2.5%

double-declining rate = 2 × 2.5% = 5%

Depreciation for 2020 = 5% × 160000 = 8000

Depreciation for 2021 = 5% × (160000 - 8000) = 7600

Depreciation for 2022 = 5% × (160000 - 8000 - 7600) = 7220

Depreciation for 2023 = 5% × (160000 - 8000 - 7600 - 7220) = 6859

Accumulated depreciation at December 32, 2023

= 8000 + 7600 +7220 + 6859

= 29679

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Two sites are being considered for wind power generation. On the first site, the wind blows steadily at 7 m/s for 3000 hours per

kirill [66]

Solution :

Given :

$V_1 = 7 \ m/s$

Operation time, $T_1$ = 3000 hours per year

$V_2 = 10 \ m/s$

Operation time, $T_2$ = 2000 hours per year

The density, ρ = $1.25 \ kg/m^3$

The wind blows steadily. So, the K.E. = $(0.5 \dot{m} V^2)$

$= \dot{m} \times 0.5 V^2$

The power generation is the time rate of the kinetic energy which can be calculated as follows:

Power = $\Delta \ \dot{K.E.} = \dot{m} \frac{V^2}{2}$

Regarding that $\dot m \propto V$ . Then,

Power $\propto V^3$ → Power = constant x $V^3$

Since, $\rho_a$ is constant for both the sites and the area is the same as same winf turbine is used.

For the first site,

Power, $P_1= \text{const.} \times V_1^3$

$P_1 = \text{const.} \times 343 \ W$

For the second site,

Power, $P_2 = \text{const.} \times V_2^3 \ W$

$P_2 = \text{const.} \times 1000 \ W$

5 0

2 years ago

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

$V_{1}=100\ \mathrm{ft}^{3}$

First has a natural water content of 25% = $\frac{25}{100}$ = 0.25

Shrinkage limit, $w_{1}=12 \%=\frac{12}{100}=0.12$

$G_{s}=2.70$

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

$V \propto[1+e]$

$\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}$ ------> eq 1

$e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}$

The above equation is at $S_{r}=1$ ,

$e_{1}=w_{1} \times G_{s}$

Applying the given values, we get

$e_{1}=0.25 \times 2.70=0.675$

Shrinkage limit is lowest water content

$e_{2}=w_{2} \times G_{s}$

Applying the given values, we get

$e_{2}=0.12 \times 2.70=0.324$

Applying the found values in eq 1, we get

$\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904$

$V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}$

7 0

2 years ago

Multiple Choice

ra1l [238]

I need more details to your question

4 0

2 years ago

Read 2 more answers

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Alecsey [184]

Answer:

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6 0

2 years ago

Read 2 more answers

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of

bixtya [17]

Answer:

h = 287.1 m

Explanation:

the density of mercury \rho =13570 kg/m3

the atmospheric pressure at the top of the building is

$p_t = \rho gh = 13570*908*0.73 = 97.08 kPa$

the atmospheric pressure at bottom

$p_b = \rho gh = 13570*908*0.75 = 100.4 kPa$

$\frac{w_{air}}{A} =p_b -p_t$

we have also

$(\rho gh)_{air} = p_b - p_t$

1.18*9.81*h = (100.4 -97.08)*10^3

h = 287.1 m

7 0

2 years ago