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ra1l [238]
3 years ago
10

The council members of a small town have decided that an earth levee should be rebuilt and strengthen to protect against future

flooding. The town engineer estimates that the cost of the work at the end of the first year will be $ 100,000. He estimates that in subsequent years the annual repair will decline by $ 15,000, making the second-year cost $ 85,000; the third-year $ 70,000 and so forth. The council members want to know what the equivalent present worth cost is for the first 5 years of repair work if the interest rate is 6%. The value is closest to?
Engineering
1 answer:
Ket [755]3 years ago
5 0

Answer:

present cost = $302218.15

Explanation:

given data

cost of the work 1st year A1  = $100,000

cost of the work 2nd year A2 = $85,000

cost of the work 3rd year A3 = $70,000

cost of the work 4th year A4 = $55,000

cost of the work 5th year A4 = $40,000

interest rate r = 6% = 0.06

to find out

present worth cost is for the first 5 years

solution

present worth cost will be here as per given equation

present cost = \frac{A1}{(1+r)^1}+ \frac{A2}{(1+r)^2} + \frac{A3}{(1+r)^3} + \frac{A4}{(1+r)^4}+ \frac{A5}{(1+r)^5}     .........................1

here r is rate of interest and A is amount given

put here value we get

present cost = \frac{100,000}{(1+0.06)^1}+ \frac{85,000}{(1+0.06)^2} + \frac{70,000}{(1+0.06)^3} + \frac{55,000}{(1+0.06)^4}+ \frac{40,000}{(1+0.06)^5}  

present cost = $302218.15

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Answer:

Q= 4.6 × 10⁻³ m³/s

actual velocity will be equal to 8.39 m/s

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\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2

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