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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?

User Jma
by
8.7k points

1 Answer

1 vote

Answer:

m = 1.6 x
10^(7) kg

Step-by-step 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

User Dmytro Shevchenko
by
7.7k points

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