PHI 210 Lecture 7: A Positive and Negative Sorites Paradox

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maroonraven450

8 Feb 2017

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A positive and negative sorites paradox: 10000 grains of sand is a heap, if 10000 grains of sand is a heap, then 9999 grains of sand is a heap, therefore, 9999 grains of sand is a heap. Therefore, 1 grain of sand is a heap: 1 grain of sand isn"t a heap, if 1 grain of sand isn"t a heap, then 2 grains isn"t a heap, therefore, 2 grains of sand isn"t a heap. Therefore, 10000 grains of sand isn"t a heap. Possible responses: claim that any amount of sand is a heap. Accept the conclusion of the positive and deny the first premise of the negative: claim that no amount of sand is a heap. Accept the conclusion of the negative and deny the first premise of the positive: deny the validity of the reasoning. Explain why modus ponens is only valid when used with terms that have a precise definition: reject toleration.

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1. The USS Farragut (DDG-99) Guided Missle Destroyer) displaces9200 long tons (1 long ton= 2240 lbs) of seawater. Assuming theaverage density of seawater surrounding the ship is 1.024 g/mL,calculate the total volume of seawater(in m^3) displaced by theship's hull.

2. A cargo ship of 455 ft. length, 62 ft. beam, and 38 ft. depth(deack to keel) has been loaded so much that is displaces 15,036.75m^3 of seawater (assume a uniform density of 1.024 g/m). What isthe total weight of the cargo ship (in kg) with its present load?If you first calculate the ship's overall volume, what is theship's overall density (in g/mL)? If there was a hull breach, howmuch seawater (in m^3) could she take on before sinking?

3. Your ship is in the middle of the Pacific Ocean, in waters 4500meters deep. While enjoying the day on deck, a breeze picks up atiny grain of quartz sand (0.3 mm diameter, 2.65 g/mL) in your hairand carries it out to sea. If the sand grain is deposited on theocean surface, how long will it take to sink to the ocean floor(assuming a perfectly still ocean with a uniform density=1.024g/mL)?
v (setting)=ï»¿ï»¿ ï»¿(2r^2g(d(particle)-d(fluid))/9u

r=particleradius (m)

g= 9.8m/s^2

d=density (kg*m^-3)

u=(1.17*10^-6 m^2 s^-1)* d(fluid)

4. If the particle in Question 3 were a flake of dandruffinstead (1.035 g/mL), at what depth would the dandruff flake ceaseits descent in a layered ocean like the one dipicted here? How longwill it take for the dandruff flake to reach its final destination?Make sure you choose a meaingful unit of time; a simple algebraicmethod is fine.

1.017g/mL 500m

1.025g/mL 2000m

1.028g/mL 2600m

1.032g/mL 3175m

1.037g/mL 3800m

1.042g/mL 4500m

A relatively new method of treating cancers is to bombard the tumorwith protons (The nuclei of hydrogen). The idea is that the protonswill eject electrons from the molecules of the cell, and thenelectrostatic forces (discussed in physics 1202W) will then pullthe molecule apart. If this happens in the cell DNA, the cell willbe sterile and thus will not reproduce but will disappear after itsnatural lifetime.

Electrons are however bound to their individual atoms, which meanthat relative to the world they have negative potential energy. Itis therefore necessary to transfer energy to them in order torelease them.

Consider the most favorable case. A proton collides head on with anelectron, so that after the collision, they both leave in the samedirection as the proton was incident.. The collision is elastic,i.e. no loss of energy, and the average binding (potential) energyof the electrons in the tumor cells is -5 electron volts.

What is the minimum energy of the incoming proton which will ejectan average electron? If instead of it being an elastic collision,the electron sticks to the proton to form a hydrogen atom, whatthen would be the minimum energy of the incoming proton?

( 1 electron volt = 6 x 10-18 Joules.

Mass of a proton 1..6 x 10- 27 Kg,

Mass of an electron 9.1 x 10-31 Kg).

PHI 210 Lecture 7: A Positive and Negative Sorites Paradox

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