PHIL 1000Y Study Guide - Robert Merrihew Adams, Divine Command Theory, God Game

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olivemacaw308

15 Nov 2011

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To play the god game, one must believe in god. If you play, the happiness you get if you win is infinite. If there is no proof of god"s existence, then the odds of his existence and heaven must be set at 0. Happiness (infinite) x odds (0) = expected utility (0) Since the expected utility is 0, a rational person would only play the god game if there were no costs. However, there are costs to playing (ex. no drinking, smoking, premarital sex, etc. ), so it is not rational to believe in god. Even if the costs are viewed as benefits to leading a healthy lifestyle, they have nothing to do with going to heaven. Response to the argument against logic, philosophy, science. Firstly, it is only important to believe in god is god"s existence is actually proven.

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Imagine you are looking down from a great height onto a pool ofcross sectional area A. Throughout the water are suspended N blackballs of cross section sigma.

a) Find the expected fraction of the bottom of the pool you cansee, as a function of the number N of balls in the pool.

b) Now let the ball sizes be distributed as a power-law in sigmaover some wide range of sigma. Find the value of that power suchthat the small balls are just as important for blocking the pool asis the large balls.

Here's what I've got
For part a:
I assume the pool is round (can I make this assumption?)
Let N = total number of balls
Let R = radius of pool
Let A = cross-sectional area of pool = 4piR^2
Let r = radius of one ball
let a = cross-sectional area of one ball = 4pir^2

Then the total cross-sectional area occupied by all the balls is =Na = N(4pir^2)
Then the area of the pool that can be seen is =
[A - Na]/A = [(4piR^2) - N(4pir^2)]/(4piR^2) = 1 -[N(r^2)]/R^2
I believe this is correct for part a.

For part b I have no idea how to do a power-law so if you couldshow as many details as possible I would appreciate it.

(1) Using Equation delta.d = 2do+ n glass d glass - d1 and Figure 1-3 and 1-4 located in theExperimental
Procedure, determine the speed of light assuming that the distancebetween
the transmitter-side mirror and the face of the corner-cube do is1.7m and the
distance between the two mirrors d1 is 4 em. (Note: The corner-cubeis
designed so that light always travels 6.7 em within the corner-cube- i.e.
dglass)). nglass=1.5
(2) Calculate the time-of-flight you expect for light to propagatea distance of 20
m in (a) vacuum and (b) a medium with an index of refraction n =1.5.
Express your answer in nanoseconds (ns).
(3) In telephone calls using a satellite, you sometimes can hear adelay in the
response of the other person. Satellites are usually either inlow-Earth orbit
Uust above the atmosphere) or much higher in geosynchronous orbit.The
latter is at an altitude of 35,785 kilometers (22,236 miles), wherethe satellite
completes one orbit in exactly one day. Because the orbitalvelocity matches
the spin rate of the Earth, a satellite in a circular equatorialgeosynchronous
orbit appears to hover motionless over a single location on theequator. The
total delay in hearing your correspondent's reply, due to thefinite speed of
light, is due to two round trips. If you and your correspondentwere located
nearby, and on the equator, the only distance involved is thedistance from
you to the satellite, the total distance corresponding to the delayis 4 x 35,785
km. Calculate this delay, assuming the radio signal travels at thespeed of
light in a vacuum. Report your answer in units of seconds, anddiscuss in one
sentence whether this delay is long enough for you to notice.Identify at least
one factor that could cause the actual delay to be longer than whatyou
calculated.

Consider three identical spherical conductors. One conductor,labeled A, is given a
charge of 7mC of type Nell. The charge state of the second andthird conductors, B and C, are unknown. The conductors undergo asequence of 3 contacts that occur ONLY between pairs of conductors.After the contacts, the net charge on each conductor is A: (0.25mCof type Nell), B: (1.875mC of type Nell) and C: (1.875mC of typeNell). Letâs make the assumption that the type Paul charge statecorresponds to an excess of electrons.

Because these three spheres are ______________________ the_______________ can be
freely exchanged between any pair of conductors in contact. Sincethese conductive
spheres all have the same size, when the spheres are placed incontact the ______
_____________ will evenly distribute between the two spheres.Because charge is a
property of particles and the particles can not be destroyed ANDthere is no outside
source of charge, the _______________ charge of the ______________must remain
constant at 4mC of type Nell. Based on this information, we candetermine that the
charge of at least one of B or C must be of type_______________.

To determine the order of contacts we must start from the lastcontact and work our way to the first contact. It is important tonote that because we have no initial charge state information forspheres B and C that we have the ability to make an arbitrarydecision about the initial charge state of these two spheres. Inother words, the charge states of sphere B and C areinterchangeable. Third and last Contact: Because spheres ________and _______ have the same charge we can determine that they were______ in contact. If we look at spheres B and C as the system, wefind that the total charge of the sub system consisting of B and Cwas ______________ of type _________.

Second contact: We can safely assume that A touched either B or Cprior to B and C
being brought into contact. Here we can make the arbitrary choicethat B touched A in the second contact. If this is the case, weknow that the charge of B ___________ it touched C must have been______________ of type __________ because it must have the sameamount and type of charge as the final state of _______. Therefore,at this instant we know that the charge on sphere C was________________ of type ____________________ it touched B.

First contact: Now we have an interesting situation. We find thatthe charge state of
sphere C would be consistent with sphere C being initially____________ and being
placed in contact with _______________ since the charge state of Cis exactly
__________ that of ____________ initially. This would mean that thecharge state of
sphere A after contact with ______ would be _____________ of type__________.
However, we already determined that the __________ charge of thesystem containing sphere _______ and sphere _______ in the secondcontact was ____________ of type __________. The only way this ispossible is if sphere B has a charge of __________ of type____________.

Final Results
The initial charge state of sphere Ais:_________________________________________
The initial charge state of sphere Bis:_________________________________________
The initial charge state of sphere Cis:_________________________________________
In the first contact of sphere ___________ to sphere_________________ how many
electrons were transferred from one sphere to the other?
In the second contact of sphere ___________ to sphere_________________ how many
electrons were transferred from one sphere to the other?
In the third contact of sphere ___________ to sphere_________________ how many
electrons were transferred from one sphere to the other?

PHIL 1000Y Study Guide - Robert Merrihew Adams, Divine Command Theory, God Game

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