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Water flows over a section of Niagara Falls at a rate of 1.2 × 106 kg/s and falls 71...

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ispyy | Student, Undergraduate | eNotes Newbie

Posted November 3, 2009 at 11:07 AM via web

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Water flows over a section of Niagara Falls at a rate of 1.2 × 106 kg/s and falls 71 m. The acceleration of gravity is 9.8 m/s2 .

How many 70 W bulbs can be lit with this power?

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neela | High School Teacher | Valedictorian

Posted November 3, 2009 at 2:16 PM (Answer #1)

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The mass, m of water falling in one second =m=1.2*106kg=127.2kg.

The acceleration of falling = g=9.8m/s^2.

The force generated by water due to falling = mg Newton =127.2*9.8 N=1246.56Newton.

The work done by the force = force* distance = force * height of fall=1246.56*71m=88505.76 watts

This energy is equal to (88505.76/70)=1264.368 or 1264.368 times the energy required of a 70 w bulb.

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krishna-agrawala | College Teacher | Valedictorian

Posted November 3, 2009 at 9:18 PM (Answer #2)

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I understand that the rate of flow of water was meant to be 1.2 x 10^6 kg/s rather than 1.2 x 106 kg/s. The question is solves with this rate.

As the water comes down the fall, its potential energy is converted into kinetic energy that may be used for generating electric power.

If we assume that this whole energy is converted into electricity with 100 percent efficiency, the electric energy generated is same as potential energy of water converted into kinetic energy. The rate of generation of this electric energy ids given by the formula:

Power (in watts) = Energy (in joule)/s.

and

Energy/s = m*g*h

Where m = mass of water per second = 1.2 x 10^6 kg/s

g = acceleration due to gravity = 9.8 m/s^2, and

h = Height of fall = 71 m.

Therefore electric power generated

= m*g*h = (1.2 x 10^6)*9.8*71 = 834.96 x 10^6 W

number of 70 watts bulb that can be lit by this power is given by the formula:

Number of bulbs = (Total power generated)/(Power used per bulb)

= (834.96 x 10^6)/70 = 11 982 000 bulbs.

Answer: 11 982 000 bulbs can be lit.

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