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)

the answer is 28

step-by-step explanation:

d. bc = 12 and ef = 16

step-by-step explanation:

we are given that δabc : δdef such that .

also, it is given that

so, from the options, we need that .

a.

b.

c.  

d.

as, we can see that only option d gives the same ratio as i.e. .

hence, the possible lengths are bc = 12 and ef = 16.