Philosophy 2070E Lecture Notes - Thought Experiment, Infor, Fiduciary

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ceruleanmouse560

9 Nov 2011

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Wrote during the time of the english civil war conflict between royal and parliamentary forces. The political strife was a cause for great concern. This is a prescriptive claim; a prescription of reason (a moral claim towards peace). 1st law: seek peace when possible - use all the advantages of war when peace is not possible. 3rd law: perform covenants made: mutual contract. Almost there, but we have come to an impasse: hobbes"s solution: mutually institute a political body with the power, via the fear of punishment, to collapse inforo interno obligation into infor externo obligation. Sovereign is not party to the contract to institute sovereignty: not possible for the. Sovereign to breach a covenant, and more importantly, to perpetrate an injustice. Not party to the multitude because there is no sovereign before the mutual contract . Not party to contract with each individual insipient subject .

Related Questions

Learning Goal: Understand how to determine the constants in thegeneral equation for simple harmonic motion, in terms of giveninitial conditions.
A common problem in physics is to match the particular initialconditions - generally given as an initial position x_0 andvelocity v_) at t=0 - once you have obtained the general solution.You have dealt with this problem in kinematics, where theformula

1. x(t) = x_0 + v_0t + at^2

has two arbitrary constants (technically constants of integrationthat arise when finding the position given that the acceleration isa constant). The constants in this case are the initial positionand velocity, so "fitting" the general solution to the initialconditions is very simple.

For simple harmonic motion, it is more difficult to fit the initialconditions, which we take to be

x_0, the position of the oscillator at t = 0, and
v_0, the velocity of the oscillator at t = 0.

There are two common forms for the general solution for theposition of a harmonic oscillator as a function of time t:

2. x(t) = A\cos(omega t + phi) and
3. x(t) = C\cos(omega t) + S\sin(omega t),

where A, phi, C, and S are constants, omega is the oscillationfrequency, and t is time.

Although both expressions have two arbitrary constants--parametersthat can be adjusted to fit the solution to the initialconditions--Equation 3 is much easier to use to accommodate x_0 andv_0. (Equation 2 would be appropriate if the initial conditionswere specified as the total energy and the time of the first zerocrossing, for example.)

Find C and S in terms of the initial position and velocity of theoscillator.
Give your answers in terms of x_0, v_0, and omega. Separate youranswers with a comma.

Materials fracture when the force per unit area, called STRESS,exceeds a critical value. In general, the critical stress dependson the direction as well as the magnitude of the force. Forexample, bones break differently when subjected to torsional(twisting) stress than when subjected to compressional (orsqueezing) stress. Let us consider compressional stress. Wet humanbone for 20-39 year olds has an ultimate compressive strength of ~16x104Ncm2 [8]. Consider a person who falls from a height to theground. Upon landing, the person's kinetic energy is converted towork. If the person decelerates over a distance h, we can estimatethe average force during the collision by

If a person lands stiff-legged on a hard surface and doesn'tbounce, the deceleration occurs over a very short distance, h ~ 1cm. Since the force is transmitted up the leg, the stress isgreatest where the cross-sectional area is least, i.e., the tibiajust above the ankle [8].
The following lists some different conditions that may occur duringlanding and physical explanations for why they make bone fracturefrom a fall more or less likely [8]:

(a) bounce [more likely. The impulse is twice as large for anelastic collision as for an inelastic collision.]
(b) bent knees [less likely. Bending the knees increases the timeand distance over which deceleration occurs, which reduces theforce.]
(c) landing in loose earth or sand [less likely. This increases hwhich reduces F.]
(d) turning the body to land on the side [less likely. Thisincreases h. It also distributes the force widely over the body(increases A), thereby reducing the stress.]

So, why can a properly executed karate kick break a concrete blockwithout fracturing bones [16]? First, bone is a very strongmaterial. Its ultimate compressive strength is approximately 40times larger than concrete. Second, contact is made with the edgeof the foot. This concentrates the force into a small area of thetarget and reduces the likelihood of bending a bone to the point offracture. Third, the collision with the target is essentiallyinelastic and extends over several milliseconds, so the peak force,though large, does not exceed the strength of the bone. If A(cross-sectional area of the tibia) ~ 3.3 cm2, compute how far a 69kg person can fall and land stiff-legged on both legs withoutbreaking a bone. Assume F is split evenly between two legs. (Hint:F<or =2(theta)A )

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.

Philosophy 2070E Lecture Notes - Thought Experiment, Infor, Fiduciary

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