phys171HW4

# After a long time the voltages at point a and point b

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After a long time the voltages at point A and point B will be the same d. (a.) yes, the current would flow until the capacitor reached it’s maximum  charge (b.) no, there would be no current (flowing electrons) because they  wouldn’t be able to “go uphill” any more towards areas of higher voltage aka  the capacitor would have reached a maximum charge. e. after a long period of time, the voltages at D and E will have the same voltage  (0V) and point C will have a voltage of 9 V 3. P5.1: 0.03kW= 30W x 3600 seconds =108,000J * kg sooo… 108,000J x 20kg = 216000J are contained in a 20 kg battery 4. P5.4:      current (A)= voltage (V)/ resistance (Omega) a. A = 9/50= 0.18A b. A = 9/1000000=9x10     -6     A     5. P3.14: a. 1 Kw = 1000W x 3600 hours = 3,600,000 joules of energy b. 60W x 5= 300W x 24 hours = ( 7,200 joules)  7.2 Kw * hours x 3 cents * Kw  (hour)= 21.6 cents is the cost of leaving those 5 lights on c. 2 Kw * hours x 720 hours x \$0.03= \$43.2 electrical bill

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d. 100 MW = 100,000 Kw/ 2Kw = Power plant could serve 50,000 homes 6.) P3.15: 24 x 365 = 8760 hours/year 15W x 8760 hours = 131.4 Kw * hours x \$0.09 =\$11.82 60W x 8760 hours = 525.6 Kw * hours x \$0.09 = \$47.30 \$47.30 - \$11.82 = \$35.48 savings per year 7.)P3.16: a. 50 MW= 5000000 W/200W = 250,000 servers b. 50 MW = 50000 KW x 8760 hours x \$0.08 = 35,040,000 50 MW = 50000 KW x 8760 hours x \$0.02 = \$8,760,000 35,040,000 – 8,760,000 = \$26,280,000 saved in 1 year

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