Philosophy 1020 Lecture Notes - Coherence Theory, Correspondence Theory Of Truth, Phantosmia

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8 Dec 2011

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The coherence theory of truth (october 27, 2011: biographical details: f. h. When he died, the intellectual movement kind of stopped. He wrote 8 books: review: problems with the correspondence theory, trouble over propositions & states of affairs. We have seen that the correspondence theory of truth: The picture seemed to become very complicated as we thought about exactly what was in correspodence -- not beliefs and things, but propositions and states of affairs: apparently useless for checking truth. It is a useless theory for checking on truth, for verification. We can"t check for correspondence between beliefs and reality because, while we can check our beliefs, we have no access to reality except via our beliefs about it. Those beliefs may alter, but they are still our beliefs. The led us to ask how we do in fact check for truth. The answer appears to be that we check for the coherence of our various beliefs with one another.

Related Questions

(a) Calculate the frequency of revolution and the orbit radius of the electron in the Bohr model of hydrogen for n = 100 (b) Calculate the photon frequency for transitions from the n to n − 1 states for the same value of n as in part (a) and compare with the revolution frequencies found in part (a). (c) Explain how your results verify the correspondence principle.

chocolatesnail205

(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.

4. 3.5/6.5 points All Submissions Notes Question: SerEssen11.AE.06. .Question part Points Submissions 1 2 3 4 5 6 7 8
.5 .5 .5 1 1 0/1 0/1 0/1
1/15 1/15 1/15 1/15 1/15 5/15 0/15 0/15
Total 3.5/6.5 ... Example 1.6 Cartesian and Polar Coordinates
Goal Understand how to convert from plane rectangular coordinatesto plan polar coordinates and vice versa.

Problem (a) The Cartesian coordinates of a point in the xy-planeare (x, y) = (-3.37, -2.73) m, as shown in Active Figure 1.5 Findthe polar coordinates of this point. (b) Convert (r, ?) = (4.61 m,36.5Â°) to rectangular coordinates.

Strategy Apply the trigonometric functions and their inverses,together with the Pythagorean theorem.
Figure 1.5 Converting from Cartesian coordinates to polarcoordinates.

Solution
(a) Cartesian to Polar
Take the square root of both sides of Equation 1.2 to find theradial coordinate. r =
= 1 m

Use Equation 1.1 for the tangent function to find the angle withthe inverse tangent, adding 180Â° because the angle is actually inthe third quadrant: tan ? = = = 0.810
? = tan-1 (0.810) = 39.0Â° + 180Â°
? = 2Â°

(b) Polar to Cartesian
Use the trigonometric definitions, Equation 1.1. x = r cos? = (4.61m) cos 36.5Â°
x = 3 m

y = r sin? = (4.61 m) sin 36.5Â°
y = 4 m

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Remarks When we take up vectors in two dimensions in Chapter 3, wewill routinely use a similar process to find the direction andmagnitude of a given vector from its components, or, conversely, tofind the components from the vector's magnitude anddirection.

Exercise 1.6 Hints: Getting Started | I'm Stuck

(a) Find the polar coordinates corresponding to (x, y) = (3.11,1.28) m.
r = 5 m
? = 6Â°
Your response differs significantly from the correct answer. Reworkyour solution from the beginning and check each stepcarefully.

(b) Find the Cartesian coordinates corresponding to (r, ?) = (4.07m, 52.5Â°)
x = 7 m
y = 8 m

Anything you could do would help, but please explain slowly. Thanksagain. It is for my technical physics 115 class.

Philosophy 1020 Lecture Notes - Coherence Theory, Correspondence Theory Of Truth, Phantosmia

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