One cubic meter of water is poured off a 100-m high bridge. if the change in gravitational potential energy is converted into electricity with an efficiency of 90%, how long can this energy illuminate a standard 60-w light bulb?

the light bulb can be illuminated 14700 sec (= 245 minutes or 4 hours and 5 minutes)

Step-by-step explanation:

the change in gravitational potential energy ΔE is

ΔE= m*g*h = ρ*V*g*h = 1000 kg/m³ * 1 m³ *9.8 m/s² * 100 m = 980000 J

where

ρ= density of water

V= volume of water

g= gravity

h= height

the energy required to illuminate a light bulb is

e*ΔE = P*t

where

P= power required by the lightbulb

t= time

e= efficiency

then

t =  e*ΔE/P = 0.9 * 980000 J / 60 W = 14700 sec (= 245 minutes or 4 hours and 5 minutes)

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.