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AnnyKZ [126]
3 years ago
11

Suppose that the president of a small island nation has decided to increase government spending by constructing three beach reso

rts to attract tourists. The legislation was enacted without any delay. From here, planning will take 6 months and construction will take 2 months. Which of the following is true? Choose one: A. This policy is contractionary. B. The Laffer curve would be used to recommend this policy. C. The planning and construction of the resorts represent a recognition lag of this policy. D. This policy shows an example of automatic stabilizers taking effect. E. The planning and construction of the resorts represent an impact lag of this policy.
Engineering
1 answer:
Mila [183]3 years ago
3 0

Answer:

Option E

The planning and construction of the resorts represent an impact lag of this policy.

Explanation:

Whereas the legislation was enacted without any delay but planning takes six months and construction taking 2 months, it means the policy has a lag. Therefore, option E, the planning and construction of the resorts represent an impact lag of this policy.

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Can the United States defeat Iranian forces
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Two variables, num_boys and num_girls, hold the number of boys and girls that have registered for an elementary school. The vari
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Using python

num_boys = int(input("Enter number of boys :"))

num_girls = int(input("Enter number of girls :"))

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3 years ago
A gasoline engine has a piston/cylinder with 0.1 kg air at 4 MPa, 1527◦C after combustion, and this is expanded in a polytropic
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The expansion work is 71.24 kJ and heat transfer is -16.89 kJ

Explanation:

From ideal gas law,

Initial volume (V1) = nRT/P

n is the number of moles of air in the cylinder = mass/MW = 0.1/29 = 0.00345 kgmol

R is gas constant = 8314.34 J/kgmol.K

T is initial temperature = 1527 °C = 1527+273 = 1800 K

P is initial pressure = 4 MPa = 4×10^6 Pa

V1 = 0.00345×8314.34×1800/(4×10^6) = 0.013 m^3

V2 = 10×V1 = 10×0.013 = 0.13 m^3

The process is a polytropic expansion process

polytropic exponent (n) = 1.5

P2 = P1(V1/V2)^n = 4×10^6(0.013/0.13)^1.5 = 1.26×10^5 Pa

Expansion work = (P1V1 - P2V2) ÷ (n - 1) = (4×10^6 × 0.013 - 1.26×10^5 × 0.13) ÷ (1.5 - 1) = 35620 ÷ 0.5 = 71240 J = 71240/1000 = 71.24 kJ

Heat transfer = change in internal energy + expansion work

change in internal energy (∆U) = Cv(T2 - T1)

T2 = PV/nR = 1.26×10^5 × 0.13/0.00345×8314.34 = 571 K

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∆U = 20.785(571 - 1800) = -25544.765 kJ/kgmol × 0.00345 kgmol = -88.13 kJ

Heat transfer = -88.13 + 71.24 = -16.89 kJ

5 0
3 years ago
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