PHL100Y1 Lecture Notes - Lecture 3: Absolute Monarchy, Thomas Hobbes, Wrinkle

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29 Feb 2012

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We realize it is better to co-operate. Hobbes thought it was a one shot game and so a lot of things get ruled out. For a one shot game he has a much stronger argument. And because of that coalitions cannot be formed as they are formed over time. Many scholars think that what is going on with hobbes is not prisoners dilemma but assurance games. What we want is to be assured with what you do. Hobbes s question is not whether we should co-operate but whether we are assured if they will cooperate. The real issue it poses is the question why other people will do what they say and keep to their agreement. If you keep your agreements you have harmony internally and in for itself. Hobbes vs. the fool (who thought there was no god anselm) now the fool says there is no justice.

Related Questions

MAC, Inc. (the company you work for) has decided to move forwardwith a feasibility
study for a product it wishes to manufacture and sell. This productis a particle launcher that should have the capability, for everyactivation, of launching two particles into the air at twodifferent times and at two different angles of elevation, in anautomated fashion so that they both land at the same distance fromthe launch point at the same time. For example, let P1 and P2 betwo particles, let T1 and T2 be two times, and let theta1 andtheta2 be two angles of elevation. Once activated, the launchershould be able to operate so that particle P1 is launched at timeT1 with an angle of elevation theta 1, and particle P2 islaunched
at time T2 with an angle of elevation theta2.
The launcher must be configured by a technician, prior to eachactivation, so that the
initial speed is set for the launcher, and the distance from thelauncher to where both
particles will land is set. For each activation of the launcher,the initial speed will be the
same for both particles. Once the technician has set the initialspeed and desired distance,
the launcher must calculate the angles of elevation to use and howmuch time to wait in
between particle launches. Then, when the technician activates thelauncher, both
particles will be launched at the appropriate angles of elevationand at the appropriate
times so that they both land at the configured distance from thelauncher at the same time.
The launcher, based on its calculations, will change the angle ofelevation to be ready for
the second particle launch. The launcher, however, must flash awarning for the
technician if the information the technician used to configure thelauncher does not result
in a two-particle launching solution.
Certain assumptions must be made. It is assumed that that launchercan always change
the angle of elevation to get ready for the second particle launchfaster than the time
delay calculated for the second particle launch. It is assumed thatthe units for speed are
meters per second, that the units for distance are meters, and thatgravity is 9.8 meters per
second squared. It is assumed that gravity is the only force actingon the motion of the
particles after launch, and that there is no wind to be considered,so that the particles land
at a point exactly in line with the direction the launcher ispointed.
The questions that must be answered and written up in anengineering report are as
follows:
1. Given the initial speed and desired landing distance for theparticles, find the
mathematical expressions that must be used to find 1, 2, and thetime delay
between particle launches.
2. Determine under which conditions the launcher will not be ableto perform the
operation requested, and will consequently flash a warning to thetechnician.
3. Each particle launched will follow a path of motion that isparabolic. Show that
the vertex of the parabola for each particleâs path occurs at thesame distance from
the launcher.
4. Along the path of each launched particle, there will be amaximum curvature.
Find that maximum curvature for each launched particle.

3. 3 Stratosphere-troposphere exchange

The rate of exchange of air between the troposphere and thestratosphere is critical for determining the potential of variouspollutants emitted from the surface to reach the stratosphere andaffect the stratospheric ozone layer. One of the first estimates ofthis rate was made in the 1960s using measurements of strontium-90(90Sr) in the stratosphere. Strontium-90 is a radioactive isotope(half-life 28 years) produced in nuclear explosions. It has nonatural sources. Large amounts of 90Sr were injected into thestratosphere in the 1950s by above-ground nuclear tests. Thesetests were banned by international treaty in 1962. Following thetest ban the stratospheric concentrations of 90Sr began to decreaseas 90Sr was transferred to the troposphere. In the troposphere,90Sr is removed by wet deposition with a lifetime of 10 days (bycontrast there is no rain, and hence no wet deposition, in thestratosphere). An intensive stratospheric measurement network wasoperated in the 1960s to monitor the decay of 90Sr in thestratosphere. We interpret these observations here using a 2-boxmodel for stratosphere-troposphere exchange with transfer rateconstants kTS and kST (yr-1) between the tropospheric andstratospheric reservoirs. The reservoirs are assumed to beindividually well-mixed.

Let mS and mT represent the masses of 90Sr in the stratosphere andin the troposphere respectively. Observations of the decrease inthe stratospheric inventory for the period 1963-1967 can be fittedto an exponential mS(t) = mS(0)exp(-kt) where k = 0.77 yr-1.

1. Write mass balance equations for mS and mT in the 1963-1967period.

2. Assuming that transfer of 90Sr from the troposphere to thestratosphere is negligible (we will verify this assumption later),show that the residence time of air in the stratosphere is tS =1/kST = 1.3 years.

3. Let m'T and m'S represent the total masses of air in thetroposphere and the stratosphere, respectively. Show that theresidence time of air in the troposphere is tT = tS (m'T/m'S) = 7.4years. Conclude as to the validity of your assumption in question2.

4. Hydrochlorofluorocarbons (HCFCs) have been adopted asreplacement products for the chlorofluorocarbons (CFCs), which werebanned by the Montreal protocol because of their harmful effect onthe ozone layer. In contrast to the CFCs, the HCFCs can be oxidizedin the troposphere, and the oxidation products washed out byprecipitation, so that most of the HCFCs do not penetrate into thestratosphere to destroy ozone. Two common HCFCs have trade namesHCFC-123 and HCFC-124; their lifetimes against oxidation in thetroposphere are 1.4 years and 5.9 years, respectively. There are noother sinks for these species in the troposphere. Using our 2-boxmodel, determine what fractions of the emitted HCFC-123 andHCFC-124 penetrate the stratosphere.

PHL100Y1 Lecture Notes - Lecture 3: Absolute Monarchy, Thomas Hobbes, Wrinkle

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