Question

At Hoover Dam, the distance the water effectively falls before encountering the electric generators depends on the water levels in Lake Mead. Assume that the water intakes are effectively 170 m above the electric generators. How much water must pass through the generators to power 2.2 million 35-W Las Vegas lightbulbs for 6.0 minutes?

Answer 1

Answer:

m = 1.6 x $10^{7}$ kg

Explanation:

It is given :

Height, h = 170 m

Number of light bulbs = 2.2 million

Power of one bulb = 35 W

Time = 6 minutes = 6 x 60 s

                             = 360 s

Now power required for 2.2 million bulbs = 2.2 million x 35 W

                                                                     = 2200000 x 35

                                                                     = 77000000 W

Now we know that Power = energy / time

Therefore, Energy = Power x time

                               = 77000000 x 360

                                = 27720000000

                                = 2.772 x $10^{10}$ J

For Dam, we know

Energy = m.g.h

where g = acceleration due to gravity

              =  9.8 m/s2

          m = mass of water required

          h = height from which water falls

              = 175 m

Therefore,

E = m.g.h

m = E / (g.h)

   = 2.772 x $10^{10}$ / ( 9.8 x 175 )

   = 2.772 x $10^{10}$ / 1715

   = 1.6 x 10$10^{7}$ kg

Therefore mass of water = 1.6 x $10^{7}$ kg