COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 6, Problem 102QAP
To determine

(a)

The average output power per hydroelectric plant.

Expert Solution
Check Mark

Answer to Problem 102QAP

The average output power per hydroelectric plant is 7.77MW/dam.

Explanation of Solution

Given:

  E=282×109kWhnumberofdams=4138Elecrticalenergy=90%h=50.0m1.00kWh=3.60×106J

Concept Used:

Theory of power for transfer of energy.

Calculation:

According to theory of power for transfer of energy:

  P=W(Amountofworkdone)t(Time)P=282×109kWh(1yr)(4138dams)×3.60×106J1.00kWh×1yr365.25days×1day24hr×1hr3600secP=7.77×106W/dam×1MW106WP=7.77MW/dam

Conclusion:

The average output power per hydroelectric plant is 7.77MW/dam.

To determine

(b)

The total mass of water flowed over the dams during 2006.

Expert Solution
Check Mark

Answer to Problem 102QAP

The total mass of water flowed over the dams during 2006 is 2.30×1015kg.

Explanation of Solution

Given:

  E=282×109kWhnumberofdams=4138Elecrticalenergy=90%h=50.0mg=9.80m/sec21.00kWh=3.60×106J

Concept Used:

Interpreting potential energy.

Calculation:

Potential energy U associated with the gravity during its motion:

  Ugrav=mgh

As per given condition each plant is efficient of converting 90% mechanical energy to electrical energy.

  Ugrav=282×109×100%mechanicalenergy90%electricalenergyUgrav=3.13×1011kWhm=Ugravgh

On replacing the values, we get

  g=9.8m/sec2h=50.0m

  Ugrav=3.13×1011kWh1.00kWh=3.60×106Jm=3.13×1011kWh9.8×50.0×3.60×106J1.00kWhm=2.30×1015kg

Conclusion:

The total mass of water flowed over the dams during 2006 is 2.30×1015kg.

To determine

(c)

The average mass of water per dam and average volume of water per dam that provides the mechanical energy to generate the electricity.

Expert Solution
Check Mark

Answer to Problem 102QAP

The average mass of water per dam is 5.56×1011kg/dam and average volume of water per dam that provides the mechanical energy to generate the electricity is 5.56×108m3/dam.

Explanation of Solution

Given:

  E=282×109kWhnumberofdams=4138Elecrticalenergy=90%h=50.0mg=9.80m/sec21.00kWh=3.60×106JDensity=1000kg/m3

Concept Used:

Interpreting potential energy.

.

Calculation:

Potential energy U associated with the gravity during its motion:

  Ugrav=mgh

As per given condition each plant is efficient of converting 90% mechanical energy to electrical energy.

  Ugrav=282×109×100%mechanicalenergy90%electricalenergyUgrav=3.13×1011kWhm=Ugravgh

On replacing the values, we get

  g=9.8m/sec2h=50.0m

  Ugrav=3.13×1011kWh1.00kWh=3.60×106Jm=3.13×1011kWh9.8×50.0×3.60×106J1.00kWhm=2.30×1015kg

  Average mass of water per dam=MassofwaterNo.ofdams

On replacing the value, we get

  Average mass of water per dam=2.30×1015kg4138damsAverage mass of water per dam=5.56×1011kg/dam

The average volume of water per dam that provides the mechanical energy to generate the electricity is:

  Volume=Average mass of water per damDensityofwaterVolume=5.56×1011103=5.56×108m3/dam

Conclusion:

The average mass of water per dam is 5.56×1011kg/dam and average volume of water per dam that provides the mechanical energy to generate the electricity is 5.56×108m3/dam.

To determine

(d)

The gallons that the gasoline for the 4138 dams save.

Expert Solution
Check Mark

Answer to Problem 102QAP

The gallons that the gasoline for the 4138 dams save is 22.6×109gallons.

Explanation of Solution

Given:

  E=282×109kWhnumberofdams=4138Elecrticalenergy=90%h=50.0mg=9.80m/sec21.00kWh=3.60×106JDensity=1000kg/m3

A gallon of gasoline contains 45.0×106J of energy.

Concept Used:

Interpreting potential energy.

Calculation:

We know that energy is given as:

  E=282×109kWh1.00kWh=3.60×106J

A gallon of gasoline contains 45.0×106J of energy

So, the gallons that the gasoline for the 4138 dams save is:

  =282×109kWh×3.60×106J1.00kWh×1gal45.0×106J=22.6×109gal

Conclusion:

The gallons that the gasoline for the 4138 dams save is 22.6×109gallons.

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Chapter 6 Solutions

COLLEGE PHYSICS

Ch. 6 - Prob. 11QAPCh. 6 - Prob. 12QAPCh. 6 - Prob. 13QAPCh. 6 - Prob. 14QAPCh. 6 - Prob. 15QAPCh. 6 - Prob. 16QAPCh. 6 - Prob. 17QAPCh. 6 - Prob. 18QAPCh. 6 - Prob. 19QAPCh. 6 - Prob. 20QAPCh. 6 - Prob. 21QAPCh. 6 - Prob. 22QAPCh. 6 - Prob. 23QAPCh. 6 - Prob. 24QAPCh. 6 - Prob. 25QAPCh. 6 - Prob. 26QAPCh. 6 - Prob. 27QAPCh. 6 - Prob. 28QAPCh. 6 - Prob. 29QAPCh. 6 - Prob. 30QAPCh. 6 - Prob. 31QAPCh. 6 - Prob. 32QAPCh. 6 - Prob. 33QAPCh. 6 - Prob. 34QAPCh. 6 - Prob. 35QAPCh. 6 - Prob. 36QAPCh. 6 - Prob. 37QAPCh. 6 - Prob. 38QAPCh. 6 - Prob. 39QAPCh. 6 - Prob. 40QAPCh. 6 - Prob. 41QAPCh. 6 - Prob. 42QAPCh. 6 - Prob. 43QAPCh. 6 - Prob. 44QAPCh. 6 - Prob. 45QAPCh. 6 - Prob. 46QAPCh. 6 - Prob. 47QAPCh. 6 - Prob. 48QAPCh. 6 - Prob. 49QAPCh. 6 - Prob. 50QAPCh. 6 - Prob. 51QAPCh. 6 - Prob. 52QAPCh. 6 - Prob. 53QAPCh. 6 - Prob. 54QAPCh. 6 - Prob. 55QAPCh. 6 - Prob. 56QAPCh. 6 - Prob. 57QAPCh. 6 - Prob. 58QAPCh. 6 - Prob. 59QAPCh. 6 - Prob. 60QAPCh. 6 - Prob. 61QAPCh. 6 - Prob. 62QAPCh. 6 - Prob. 63QAPCh. 6 - Prob. 64QAPCh. 6 - Prob. 65QAPCh. 6 - Prob. 66QAPCh. 6 - Prob. 67QAPCh. 6 - Prob. 68QAPCh. 6 - Prob. 69QAPCh. 6 - Prob. 70QAPCh. 6 - Prob. 71QAPCh. 6 - Prob. 72QAPCh. 6 - Prob. 73QAPCh. 6 - Prob. 74QAPCh. 6 - Prob. 75QAPCh. 6 - Prob. 76QAPCh. 6 - Prob. 77QAPCh. 6 - Prob. 78QAPCh. 6 - Prob. 79QAPCh. 6 - Prob. 80QAPCh. 6 - Prob. 81QAPCh. 6 - Prob. 82QAPCh. 6 - Prob. 83QAPCh. 6 - Prob. 84QAPCh. 6 - Prob. 85QAPCh. 6 - Prob. 86QAPCh. 6 - Prob. 87QAPCh. 6 - Prob. 88QAPCh. 6 - Prob. 89QAPCh. 6 - Prob. 90QAPCh. 6 - Prob. 91QAPCh. 6 - Prob. 92QAPCh. 6 - Prob. 93QAPCh. 6 - Prob. 94QAPCh. 6 - Prob. 95QAPCh. 6 - Prob. 96QAPCh. 6 - Prob. 97QAPCh. 6 - Prob. 98QAPCh. 6 - Prob. 99QAPCh. 6 - Prob. 100QAPCh. 6 - Prob. 101QAPCh. 6 - Prob. 102QAPCh. 6 - Prob. 103QAPCh. 6 - Prob. 104QAPCh. 6 - Prob. 105QAPCh. 6 - Prob. 106QAPCh. 6 - Prob. 107QAP
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