PHL232H1 Lecture : causation and hume

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purplezebra711

26 Oct 2011

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Hume"s version of the problem of causation: all mental content is derived from sense experience, ex. Billiard balls: the idea of a necessary connection causes a problem: hume says we don"t experience a necessary connection in the world, ex. Switching off your lights and the lighthouse coming on: you have (a->b every time but it is not causation). The lights do not come on because you switch the light in your living room. Just seeing a->b once wont make you think its causation but seeing it many times will show causation: expectation of b is the passing".

Related Questions

I already did the first part of the problem, which is:

One pound of air undergoes a thermodynamic cycle consisting ofthree processes.

~ Process 1 -> 2: Constant volume from P1=20 lbf/in^2, T1= 500 Rto T2=820 R.
~ Process 2 -> 3: Adiabatic expansion to v3=1.4(v2).
~ Process 3 -> 1: Constant-Pressure compression.

Employing the ideal gas equation of state:
1) sketch the cycle on P-v coordinates.
2)Determine the pressure at state 2, in lbf/in^2.
3) Determine the temprature at state 3, in R.
4) Is this power cycle?

*****The scecond part that i need help in is thefollowing:

Solve the problem assuming constant specificheats, which solution is more accurate anwhy?

How do you solve for the current in when the system isoverconstrained. We know Q1+Q2 must equal Qtotal (=C1V0). We alsoknow that dQtotal/dt=0 and therefore d(Q1+Q2)/dt must equal 0 whichimplies that dQ1/dt=-dQ2/dt and we have equations for those changesin charge. This doesn't work though because the expression at lefthas no variables (other than time). The remainder of the problem issimple we can derive the current equation.

gravity force probelm
1. The problem statement, all variables and given/known data
When the Earth, Moon, and Sun form a right triangle, with the Moonlocated at the right angle, as shown in the figure , the Moon is inits third-quarter phase. (The Earth is viewed here from above itsNorth Pole.)

Find the magnitude of the net force exerted on the Sun.
Find the direction of the net force exerted on the Sun. Give thedirection relative to the line connecting the Sun and theMoon.

2. Relevant equations
Gm1m2/R2
Fnet=vF12+F22
LaTeX Code: \\theta = inverse trig (whatever necessary)
Mass sun=1.99E30
Mass earth= 5.98E24
mass moon= 7.35E22
radius sun to moon=1.50E11
radius moon to earth=3.85E8
3. The attempt at a solution
F1(Sun-Moon)=4.336E20 found by using gravity force eq.
F2(sun earth)= 3.53E22, however this is wrong because it should be3.58E22, so that means the distance between the sun and the earthat this time would have to be 1.49E11, which doesn't make sincebecause this should be the hypotenuse, right???. So when using thesencond value for F2, and using hte pythagorean theory I getFnet=3.57E22, which is the right answer.
NOW to find the direction, i thought you can use the cos inverse*F1/F2, but this turns out to be the wrong answer I have no ideawhat else to do. Please help!

PHL232H1 Lecture : causation and hume

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