Two players find themselves in a legal battle over a patent. The patent is worth 20 for each player, so the winner would receive
20 and the loser 0. Given the norms of the country they are in, it is common to bribe the judge of a case. Each player can secretly oer a bribe of 0, 9 or 20, and the one whose bribe is the largest is awarded the patent. If both choose not to bribe, or if the bribes are the same amount, then each has an equal chance of being awarded the patent. (If a player decides to bribe then the judge pockets it regardless of who gets the patent).
(a) Derive the game matrix.
(b) Is the game dominance solvable? If so, findnd the strategy prole surviving IDSDS.
(c) Now consider the case in which the allowed bribe amounts are instead 0, 9 and 15. Is the game dominance solvable? Find the best responses of each player to each of the pure strategies of the opponent.
(a) Derive the game matrix.
(b) Is the game dominance solvable? If so, findnd the strategy prole surviving IDSDS.
(c) Now consider the case in which the allowed bribe amounts are instead 0, 9 and 15. Is the game dominance solvable? Find the best responses of each player to each of the pure strategies of the opponent.
1 answer:
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Answer:
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Answer:
heat loss per 1-m length of this insulation is 4368.145 W
Explanation:
given data
inside radius r1 = 6 cm
outside radius r2 = 8 cm
thermal conductivity k = 0.5 W/m°C
inside temperature t1 = 430°C
outside temperature t2 = 30°C
to find out
Determine the heat loss per 1-m length of this insulation
solution
we know thermal resistance formula for cylinder that is express as
Rth = .................1
here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity
so
heat loss is change in temperature divide thermal resistance
Q =
Q =
Q = 4368.145 W
so heat loss per 1-m length of this insulation is 4368.145 W
Answer:
Explanation:
In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.
We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.
The weigth depends on the size and specific gravity.
W = 1/2 * b * h * L * SG
Then
Teq = 1/2 * b * h * L * SG * b / 2
Teq = 1/4 * b^2 * h * L * SG
The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:
The term sin(30) is because of the slope of the wall
The pressure of water is:
p(y) = SGw * (h - y)
Then:
T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)
T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)
T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)
T1 = 1/3 * SGw * sin(30) * L * h^3
To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)
1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3
In an equilateral triangle h = b * cos(30)
1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3
SG > SGw * 4/3* sin(30) * (cos(30))^2
SG > 1/2 * SGw
For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.
This is avergae specific gravity, including holes.