Consider two adjacent states, S1 and S2, that wish to control particulate emissions from power plants and cement plants; New Jer
sey and Pennsylvania, for example. Each state wants to remove all particulates larger than 100 nanometres from the airspace below 1000 feet. Suppose this requires 50 kilotons of abatement in each state. The aggregate cost of abatement when the marginal abatement costs of all industries within the state are equalized is 3A21 for state S1 and it is A2 for state S2, where Ai is measured in kilotons.(a) If the states do not cooperate, what is the cost of abatement in state S1?(b) Would a permit market with 50 kilotons of emissions permits distributed to the firms by means of an ascending clock auction equate the marginal abatement costs of all power and cement firms within state S1? You may assume that no firms are bankrupted.(c) If the states do not cooperate, what is the cost of abatement in state S2?(d) If the states do cooperate by means of a 100 kiloton cap and a permit market that covers both states with permits allocated by an ascending clock auction, what is the total cost of abatement? You may assume that no firms are bankrupted. You may also assume that a joint auction will not change the aggregate cost functions 3A21 for state S1 and A2 for state S2.(e) Why does the sum of the costs of abatement from question 1a and 1c not equal the cost of abatement in question 1d?
1 answer:
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Answer:
Yes
Explanation:
The given parameters are;
The speed with which the fastball is hit, u = 49.1 m/s (109.9 mph)
The angle in which the fastball is hit, θ = 22°
The distance of the field = 96 m (315 ft)
The range of the projectile motion of the fastball is given by the following formula
Where;
g = The acceleration due to gravity = 9.81 m/s², we have;
Yes, given that the ball's range is larger than the extent of the field, the batter is able to safely reach home.
Answer:
134.77 mm
Explanation:
Wave length of light λ = 599 x 10⁻⁹ m
Slit separation d = 20 x 10⁻⁶ m
Screen distance D = 3 m
Distance of second dark fringe from centre
= 1.5 x λ D / d
Putting the values given above
distance =
= 134.77 x 10⁻³ m
= 134.77 mm.