Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

410 Homework Ch06 17 wer: At a tim e when the population of the U.S. was about 2

ID: 778114 • Letter: 4

Question

410 Homework Ch06 17 wer: At a tim e when the population of the U.S. was about 260 mill 17) Po electrical energy in the U.S. was about 1.0x10 joules per ve dectrical energy ta vrage rate of electal energy consumption per person (also in watty? b) What was the average rate of consumption c) On average, the earth's surface receives about 1000 wat al energy consumption in watt year. Think units. s? ut 1000 watts per square meter from the Sun, and of this, about 40% can be collected with reasonably efficient solar panels. would be required to collect all the energy used by the U.S. at this time. How large an area (in 18)(ex) In a 10.0-km race, a 65.0-kg athlete generates an a body mass,(fa) Hu ge power of 3.50 watts per kilogram o

Explanation / Answer

rate of energy consumption P = E/t

E = 1.0*10^19


time t = 365days = 365*24 hours = 365*2483600 s

rate of energy consumption P = (1.0*10^19)/(365*2483600) = 1.1*10^10 W


-------------------------


part(b)


rate of energy consumption Per person = P/N = 1.1*10^10/(260*10^6) = 42.3 W

=========================================

part(c)

rate of energy fron sun = 1000*A

rate of energy collected by solar panel = 0.4*1000*A   W

rate of energy consumption P = 1.1*10^10 W


0.4*1000*A = 1.1*10^10

area of surface A = 27500000 m^2 = 27500000*10^-6 (Km)^2 = 27.5 (km)^2


area of surface A = 27.5 square kilometers

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote