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telo118 [61]
3 years ago
13

A company buys a machine for $12,000, which it agrees to pay for in five equal annual payments, beginning one year after the dat

e of purchase, at an annual interest rate of 4%. Immediately after the second payment, the terms are changed such that the company can pay off the loan immediately in a lump sum during the following year. What is the final single payment?
Engineering
1 answer:
Yuki888 [10]3 years ago
4 0

Answer:

$7,778.35

Explanation:

At year 3, the final payment of the remaining balance is equal to the present worth P of the last three payments.

First, calculate the uniform payments A:

A = 12000(A/P, 4%, 5)

= 12000(0.2246) = 2695.2  (from the calculator)

Then take the last three payments as its own cash flow.

To calculate the new P:

P = 2695.2 + 2695.2(P/A, 4%, 2) = 2695.2 + 2695.2(1.886) = 7778.35

Therefore, the final payment is $7,778.35

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D. Chlorofluorocarbon

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2 years ago
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(a) Calculate the heat flux through a sheet of steel that is 10 mm thick when the temperatures oneither side of the sheet are he
aleksandr82 [10.1K]

Answer:

do the wam wam

Explanation:

6 0
3 years ago
The path taken to move items from their origin to their destination is known as which of the following?
mrs_skeptik [129]

Answer:d

Explanation:

7 0
2 years ago
(a) If 5 x 10^17 phosphorus atoms per cm3 are add to silicon as a substitutional impurity, determine the percentage of silicon a
Y_Kistochka [10]

Answer:

The percentage of silicon atoms per unit volume that are displaced in the single crystal lattice = 0.001 %

The percentage of silicon atoms per unit volume that are displaced in the single crystal lattice with boron atoms = 0.4 ×10^{-5} %

Explanation:

No. of phosphorus atoms = 5 × 10^{17} \ cm^{-3}

The volume occupied by a single Si atom

V_{si} = \frac{a^{3} }{8}

V_{si} = \frac{5.43^{3}(10^{-8} )^{3}  }{8}

V_{si} = 2 × 10^{-23} \frac{cm^{3} }{atom}

n_{si} = \frac{1}{V_{si} }

n_{si} = 5 × 10^{22} \frac{atoms}{cm^{3} }

PCT = \frac{N_p}{N_{si}}   100

Put the values in above equation we get

PCT = \frac{5 (10^{17} )}{5 (10^{22}) } 100

PCT = 10^{-3} = 0.001 %

These are the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.

(b).

No. of boron atoms = 2 × 10^{15} \ cm^{-3}

The volume occupied by a single Si atom

V_{si} = \frac{a^{3} }{8}

V_{si} = \frac{5.43^{3}(10^{-8} )^{3}  }{8}

V_{si} = 2 × 10^{-23} \frac{cm^{3} }{atom}

n_{si} = \frac{1}{V_{si} }

n_{si} = 5 × 10^{22} \frac{atoms}{cm^{3} }

PCT = \frac{N_p}{N_{si}}   100

Put the values in above equation we get

PCT = \frac{2 (10^{15} )}{5 (10^{22}) } 100

PCT = 0.4 ×10^{-5} %

These are the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.

7 0
3 years ago
A sedimentation basin in a water treatment plant has a length = 48 m, width = 12 m, and depth = 3 m. The flow rate = 4 m 3 /s; p
Slav-nsk [51]

Answer:

The minimum particle diameter that is removed at 85% is 1.474 * 10 ^⁻4 meters.

Solution

Given:

Length = 48 m

Width = 12 m

Depth = 3m

Flow rate = 4 m 3 /s

Water density = 10 3 kg/m 3

Dynamic viscosity = 1.30710 -3 N.sec/m

Now,

At the minimum particular diameter it is stated as follows:

The Reynolds number= 0.1

Thus,

0.1 =ρVTD/μ

VT = Dp² ( ρp- ρ) g/ 10μ²

Where

gn = The case/issue of sedimentation

VT = Terminal velocity

So,

0.1 = Dp³ ( ρp- ρ) g/ 10μ²

This becomes,

0.1 = 1000 * dp³ (1100-1000) g 0.1/ 10 *(1.307 * 10 ^⁻3)²

= 3.074 * 10 ^⁻6 = dp³ (.g01 * 10^6)

dp³=3.1343 * 10 ^⁻12

Dp minimum= 1.474 * 10 ^⁻4 meters.

8 0
3 years ago
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