The manufacturer of a 9[V] flashlight battery says that the battery will deliver 20[mA] for 80 continuous hours. However, during that time the voltage will drop from 9[V] to 6[V]. Assume the voltage change with time is linear in time; in other words, the plot of voltage versus time is a straight line. Assume that the current is constant during these 80 hours.

Your question should includes how much energy does the battery deliver in 80 hours interval.

The battery delivers 17,280 J

Explanation:

Starting voltage v1 = 9V

Final voltage v2 = 6V = voltage drop after 80hrs

v = v1 - v2 = 9 - 6 = 3V

Electrical current (i) = 20[mA] = 20*10^-3 A

Discharging time = 80 Hrs = 80*(60*60) sec = 288,000 sec

Work done = energy (W) = p.t

Where p = v*i

W = v*i*t

W = 3*(20*10^-3)*288,000

W = 17,280 J

The battery delivers 17,280 J

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

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