81 students took the exam.  The number of students getting a question right
is for the whole class on that question.

Quest.	Version A		Version B		Version C
	ans.	# right		ans.	# right		ans.	# right
1.	A	62		C	64		B	51
2.	A	71		B	72		D	66
3.	C	70		B	51		C	69
4.	C	80		C	71		B	62
5.	B	51		D	66		C	48
6.	B	62		C	63		B	63
7.	A	61		C	70		C	63
8.	B	74		D	74		C	80
9.	D	65		A	58		A	61
10.	A	80		A	61		A	58
11.	C	69		B	62		B	68
12.	B	63		A	80		B	62
13.	B	62		C	69		C	69
14.	D	74		C	80		B	53
15.	A	58		B	68		D	74
16.	C	64		C	71		C	70
17.	C	63		C	69		C	71
18.	C	71		A	62		A	80
19.	B	53		A	71		A	71
20.	B	68		B	62		B	74
21.	C	69		B	74		D	65
22.	D	66		D	65		D	69
23.	C	48		B	63		A	62
24.	B	72		D	69		C	64
25.	D	69		B	53		B	72

26.	T	60		F	75		F	47
27.	F	58		F	54		T	73
28.	F	75		T	73		T	71
29.	T	73		F	58		F	75
30.	T	65		T	65		T	65
31.	T	73		T	78		F	58
32.	T	71		T	71		T	78
33.	F	54		F	47		F	54
34.	F	47		T	60		T	60
35.	T	78		T	32		T	73

36. (a): 1 m^2 area times 10 m (amount of air that passes in a second) = 10 m^3
    (b): density 1.3 kg/m^3 times a volume of 10 m^3 = 13 kg
    (c): in each second, the kinetic energy passing through the windmill is
         0.5*m*v^2 = 0.5*(13 kg)*(10 m/s)^2 = 650 J
         so the power available is 650 J in each second, or 650 W
    (d): only 40% (0.4) of the above is extractable by the windmill, so the
         practical power is 0.4*650 W = 260 W

36. (8 points): The shower uses 75 kg of water whose temperature has been
    raised by 25 C, requiring an energy of (4184 J/kg/C)*(75 kg)*(25 C)
    = 7.8 MJ.  For a 1500 W (1500 J/s) space heater to blow through 7.8 MJ
    would take (7.8x10^6 J)/(1500 J/s) = 5230 seconds, or about 1.5 hours.
    Let this sink in: a 10 minute shower uses as much energy as 1.5 hours of
    personal space heater.

38. (a): 0.15 kW * 720 h = 108 kWh
    (b): uwave: 1 kW * 0.2 h = 0.2 kWh every day
         stove: 2 kW * 0.3 h = 0.6 kWh every day
         total is 0.8 kWh each day, or 0.8*30 = 24 kWh in a month
    (c): 0.075 kW * 720 h = 54 kWh (same technique as part a)
    (d): each load has 5 kW*1 h = 5 kWh from dryer, 0.2 kWh from washer
         so 5.2 kWh per load.  At 12 loads per month, this is 62.4 kWh
    (e): 248 kWh in a month means 8.3 kWh per day
         (note: at $0.15/kWh, this is $37/month)
         note the relative importance of the above contributions:
         an always-on computer plus random household devices might dominate
    (f): If the unknown part uses 2 kWh of energy every 24 hours, the
         average power associated with this is 2kWh/24h = 1/12 kW = 83 W
    (g): If the missing 2 kWh is being used in a 5 hr period, the
         power associated with this source is 2 kWh/5h = 0.4 kW = 400 W