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Lesson 11 Sustainable Energy

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Course: Science 10 [5 cr] - AB Ed copy 1
Book: Lesson 11 Sustainable Energy
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Date: Sunday, 7 September 2025, 6:46 PM

Table of contents

  Introduction

Which energy sources seem to have an infinite supply with minimal environmental effects?



C11.1 electric power lines in Banff National Park
Technologies that promote sustainable energy are those that use renewable energy sources and those that are designed to improve energy efficiency.

Sustainable energy is energy that is used at a much slower rate than is required to replenish it, so its supply seems endless. It is also an energy source that has products and effects on the environment that are manageable and minimal.

  Targets

By the end of this lesson, you will be able to

  • compare the energy content of fuels used in thermal power plants in Alberta in terms of costs, benefits, efficiency, and sustainability
  • explain the need for efficient energy conversions to protect our environment and to make judicious use of natural resources

  Watch This

Which Power Source Is Most Efficient? © YouTube Seeker  


This video provides you with an extensive overview of the different power sources in creating electricity. It focuses primarily on the efficiencies of the different power sources and also discusses the advantages and disadvantages of them. This will help get you in the right mindset for this lesson in this section.

  Energy Content of Energy Sources Used in Alberta

How do the energy sources used in power plants compare?


C11.2 power lines in a wheat field      
It is important that we look at many factors when evaluating which is the best energy source to be used in a power plant in Alberta. We can compare energy content of fuels used in power plants in Alberta, in terms of costs, benefits, efficiency, and sustainability.

All types of power plants used in Alberta will be compared, not just thermal power plants.  

Type of Generating Station
Coal Natural Gas
Biomass Biogas Waste Heat Recovery
Hydroelectric Wind
Energy Content (MJ/kg)
24–35
 55.5
 16–21
 55-55.7
 - - -
Cost* ($/kWh) 0.11–0.12
0.053–0.11
0.098
 0.03–0.05
 0.048
 0.064
0.044–0.20
Benefits
  • abundant in Alberta
  • by-product of combustion (ash) that can be used for concrete and roads
    • 70% less CO2 compared to coal
    • has high heating value
    • Alberta still has large reserve still in Alberta
    • renewable
    • abundant in Alberta
    • carbon-neutral
    • can produce heat and electricity together
    • diverts waste from agriculture to a usable source
    • renewable
    • produces fewer emissions than combustion energies
    • Has a smaller effect on the enhanced greenhouse effect
    • diverts waste from agriculture to a usable source
    • renewable
    • removes waste heat from environment is beneficial
    • reuse of waste heat that is normally lost
    • helps to conserve resources used in operation of buildings
    • renewable
    • clean source of energy (no emissions)
    • constant supply of water from river
    • renewable
    • clean source of energy (no emissions)
    • multiple uses for land
    Efficicency (%) 32–42 32–38 22–34 40–45 20–25 85–90 35–45
    Sustainability Low Low High High High High High
    Learn More About Energy Source
    		Learn More
    		Learn More
    		Learn More
    		Learn More
    		Learn More
    		Learn More
    		Learn More
    *The levelized energy costs (LEC) reflects all the costs including initial capital, return on investment, continuous operation, fuel, and maintenance, as well as the time required to build a plant and its expected lifetime.

      Digging Deeper

    C11.3 solar-powered construction traffic light
    Did you ever notice that most of the construction signs use solar panels to power them? What about the roadside help phones as you drive down major highways? There is a large increase in the use of solar energy to power devices such as these.

    To read more about the solar energy developments in Alberta, click on this link. https://solaralberta.ca/

    Learn More

      Read This

    Please read pages 221 to 223, and 227 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on the different energy sources used in power generation, with a specific focus on sustainability and the province of Alberta. Remember, if you have any questions, or do not understand something, ask your teacher!

      Practice Questions

    Complete the following practice questions to check your understanding of the concept you just learned. Make sure you write complete answers to the practice questions in your notes. After you have checked your answers, make corrections to your responses (where necessary) to study from.

    1. What connection does there appear to be between the sustainability of renewable vs. non-renewable energy sources? Explain.

      Your answer should be a variation of the following. Non-renewable resources are not as sustainable as renewable resources. This is because sustainable energy is a form of energy that meets today's demands for energy without putting them in danger of running out and can be used over and over again.

    2. Why is hydroelectricity so much more efficient than any of the other sources of energy?

      Your answer should be a variation of the following. The laws of thermodynamics have taught us that waste energy in the form of heat is produced in every energy conversion. Hydroelectricity has few energy conversions, so more useful energy is available. Also, many other sources require the combustion of the fuel, which releases large amounts of “waste” energy in the form of heat.

    3. Why is wind energy so much less efficient than hydroelectricity, when it too does not require many inefficient energy conversions?

      Your answer should be a variation of the following. Wind is not always available, and wind speeds below around 50 km/h produce very little energy because the blades are too heavy to be turned by a slower wind.

      Drawbacks of Energy Sources Used in Alberta

    What are the drawbacks to the energy sources used in power plants?


    C11.4 aerial view of pit mining project in Alberta’s oilsands near Fort McMurray
    It is equally important to consider the drawbacks when evaluating energy sources for use in Alberta power plants. Wise energy use must involve consideration of environmental impacts.

    Energy Source
    		Drawbacks
    Coal
    • It contributes to the enhanced greenhouse effect and releases CO2(g) when combusted.
    • It is non-renewable, so it is not sustainable and supplies will be depleted.
    • It contributes to environmental effects such as acid deposition and thermal pollution.
    Natural Gas
    • It contributes to the enhanced greenhouse effect and releases CO2(g) when combusted, though less than coal.
    • It is non-renewable, so it is not sustainable and supplies will be depleted.
    • It contributes to environmental effects such as acid deposition and thermal pollution.
    Biomass
    • It is quite an expensive process due to the cost of gathering and transporting the biomass to the power plants.
    • The land area needed for biomass power plants is very large.
    Biogas
    • When combusted, some of the products are corrosive to metal machine components.
    • Since it is combusted, it still adds to the greenhouse effect and thermal pollution.
    • It is expensive when used on a large industrial scale.
    Waste Heat Recovery
    • It is quite an expensive process and equipment is large.
    • Waste heat is often not very warm, so a large amount is needed to make it an effective energy source
    Hydroelectricity
    • It cannot produce the quantities of energy that fossil fuels can.
    • It is dependent upon the weather for its source of power (precipitation).
    • It can only be used where there is a large amount of water to be harnessed.
    Wind
    • The best wind sites are often located in remote locations, far from where electricity is needed.
    • It can only be used in areas of high wind.
    • Turbines cause noise and visual pollution.
    • Poorly-placed wind turbines can be dangerous to birds and bats.

      Digging Deeper

    ©Wikimedia Commons
    C11.5 hoary bat

    Wind turbines in Alberta have affected certain bat species. The fall migration paths of hoary and silver-haired bats have led to high fatality rates. Fatalities vary from site to site, but increased fatality have been associated with taller wind turbines.

    Both species are currently listed as “sensitive” as of 2005 – they make up more than 80% of all the bats killed by wind turbines in North America.



      Read This

    Please read pages 221 to 223, and 225 to 227 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on the different energy sources used in power generation, with a specific focus on drawbacks and the province of Alberta. Remember, if you have any questions, or do not understand something, ask your teacher!

      Practice Questions

    Complete the following practice questions to check your understanding of the concept you just learned. Make sure you write complete answers to the practice questions in your notes. After you have checked your answers, make corrections to your responses (where necessary) to study from.

    1. Why must a balance between the pros and cons be considered when evaluating an energy source for power generation? Explain.

      Your answer should be a variation of the following. Each energy source has strong positives and drawbacks. When evaluating an energy source for power generation, not just one or two positive factors can be considered while ignoring strong negative factors. We need to find a balance between our energy demands and saving the environment for future generations.


      Sustainable Energy


    C11.6 wood pellets for biomass boiler
    This last lesson of the unit looked at comparing the benefits and drawbacks of different energy sources for power generation in Alberta. We compared energy content of fuels used in power plants in Alberta, in terms of costs, benefits, efficiency, and sustainability. Then, we considered the drawbacks in terms of the importance of preserving our environment and using energy sources wisely.

    All types of energy have pros and cons. There is no perfect source of energy, so we need to weigh the pros and cons of each. Is the key strategy to reducing the effects of power production to reduce the amount of power being used, or to make the production of power more efficient?

      Watch This

    Bill Gates and the Quest for Sustainable Energy © YouTube The Atlantic 


    Watch this video for a detailed explanation of sustainable energy. It reviews the concept of sustainable energy, and the many types of alternative energy technologies.


      Conclusion

    There are many types of energy and ways that energy can be used.


    CC.1 kicking a soccer ball
    The movement of an object can be described using the concepts of work, force, displacement and distance, speed and velocity, and acceleration.

    The energy of an object can be explained through gravitational potential, kinetic, and mechanical energies. Modern and efficient energy conversion devices have been developed based upon the first and second laws of thermodynamics. These laws are used to explain energy conservation during conversions, efficiency and how thermal energy flows from hot to cold.

    In this unit, we investigated mechanical energy conversions and transfers in systems. We learned that while energy is conserved, useful energy becomes less with each conversion, because waste energy is released, often in the form of heat. We looked at how useful energy can be observed when it is being transferred, and that mechanical energy can be measured.

    In the next unit, you will study energy flow in global systems.
    CC.2 walking trail

      Review Questions

    Complete the following review questions to check your understanding of the concepts you have learned. Make sure you write complete answers to the review questions in your notes. After you have checked your answers, make corrections to your responses (where necessary) to study from.

    1. Identify at least one natural system and one technological system that uses solar energy as its primary source.
    Your answer should be a variation of the following. A natural system that uses the energy from the sun is plants (through photosynthesis). Also, the greenhouse effect uses the conversion of the sun’s solar energy to heat Earth.

    A technological system that uses the sun’s energy is solar panels that can be used for heat traps or for electrical generation. Solar energy can also be used in devices for cooking or heating.
    1. With a single pulley, you lift a crate. If you exerted a force of 455 N and did 3 276 J of work, how far did you lift the crate?

      Step 1: List the variables.

      «math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»W«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mn»276«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»F«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»455«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»N«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»d«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«/mtable»«/math»
      Step 2: Identify the correct formula, and rearrange if necessary.

      W = Fd
      To isolate d, you must divide each side by F. To move F to the other side, you must use the opposite operation. Division is opposite to multiplication.

      «math» «mfrac» «mi»W«/mi» «mi»F«/mi» «/mfrac» «mo»=«/mo» «mfrac» «mrow» «mi»F«/mi» «mi»d«/mi» «/mrow» «mi»F«/mi» «/mfrac» «/math»

      Now, cancel the like terms.

      «math» «mfrac» «mi»W«/mi» «mi»F«/mi» «/mfrac» «mo»=«/mo» «mi»d«/mi» «mo»§#160;«/mo» «mi»or«/mi» «mo»§#160;«/mo» «mi»d«/mi» «mo»=«/mo» «mfrac» «mi»W«/mi» «mi»F«/mi» «/mfrac» «/math»
      Step 3: Substitute the values into the formula.

      «math»«mi»d«/mi»«mo»=«/mo»«mfrac»«mrow»«mn»3«/mn»«mo»§#160;«/mo»«mn»276«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»455«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»N«/mi»«/mrow»«/mfrac»«/math»
      Step 4: Calculate the answer.

      d = 7.2 m

      The answer must be expressed to three significant digits.

      The height that the delivery clerk carried the package was 7.20 m.
    2. A 4 325 g chandelier hangs from the ceiling of a large ballroom. If the chandelier is 12.3 m above the floor, what is its gravitational potential energy relative to the floor?

      Step 1: List the variables.

      «math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»4«/mn»«mo»§#160;«/mo»«mn»325«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»=«/mo»«mn»4«/mn»«mo».«/mo»«mn»325«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»g«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mo»§#160;«/mo»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mi»h«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»12«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mtd»«/mtr»«/mtable»«/math»
      Step 2: Identify the correct formula, and rearrange if necessary.

      Ep = mgh
      Step 3: Substitute the values into the formula.

      «math»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«mo»=«/mo»«mfenced»«mrow»«mn»4«/mn»«mo».«/mo»«mn»325«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mrow»«/mfenced»«mfenced»«mrow»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mfrac»«/mrow»«/mfenced»«mfenced»«mrow»«mn»12«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mrow»«/mfenced»«/math»
      Step 4: Calculate the answer.

      Ep = 521.867...J

      The answer must be rounded to three significant digits.

      The gravitational potential energy of the chandelier is 522 J.
    3. A 5.32 kg bowling ball is rolling and has a kinetic energy of 8.62 J. What is the speed that the ball is rolling at?

      Step 1: List the variables.

      «math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»8«/mn»«mo».«/mo»«mn»62«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»5«/mn»«mo».«/mo»«mn»32«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«/mtable»«/math»
      Step 2: Identify the correct formula, and rearrange if necessary.

      «math»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/math»

      To isolate v2, you must divide each side by «math»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«/math». To move «math» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «/math»and m to the other side you need to use the opposite operation.

      «math» «mfrac» «msub» «mi»E«/mi» «mi»k«/mi» «/msub» «mstyle displaystyle=¨true¨» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «mi»m«/mi» «/mstyle» «/mfrac» «mo»=«/mo» «mfrac» «mstyle displaystyle=¨true¨» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «mi»m«/mi» «msup» «mi»v«/mi» «mn»2«/mn» «/msup» «/mstyle» «mstyle displaystyle=¨true¨» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «mi»m«/mi» «/mstyle» «/mfrac» «/math»

      Now, cancel the like terms.

      «math» «mfrac» «mrow» «mn»2«/mn» «msub» «mi»E«/mi» «mi»k«/mi» «/msub» «/mrow» «mi»m«/mi» «/mfrac» «mo»=«/mo» «msup» «mi»v«/mi» «mrow» «mn»2«/mn» «mo»§#160;«/mo» «/mrow» «/msup» «mo»§#160;«/mo» «mi»or«/mi» «mo»§#160;«/mo» «mo»§#160;«/mo» «msup» «mi»v«/mi» «mn»2«/mn» «/msup» «mo»=«/mo» «mfrac» «mrow» «mn»2«/mn» «msub» «mi»E«/mi» «mi»k«/mi» «/msub» «/mrow» «mi»m«/mi» «/mfrac» «/math»

      Remember that square root is the opposite of squaring, so to remove the square we must square root the other side.

      «math» «mi»so«/mi» «mo»,«/mo» «mo»§#160;«/mo» «mi»v«/mi» «mo»=«/mo» «msqrt» «mfrac» «mrow» «mn»2«/mn» «msub» «mi»E«/mi» «mi»k«/mi» «/msub» «/mrow» «mi»m«/mi» «/mfrac» «/msqrt» «/math»
      Step 3: Substitute the values into the formula.

      «math» «mi»v«/mi» «mo»=«/mo» «msqrt» «mfrac» «mrow» «mn»2«/mn» «mfenced» «mrow» «mn»8«/mn» «mo».«/mo» «mn»62«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»J«/mi» «/mrow» «/mfenced» «/mrow» «mrow» «mn»5«/mn» «mo».«/mo» «mn»32«/mn» «mo»§#160;«/mo» «mi»kg«/mi» «/mrow» «/mfrac» «/msqrt» «/math»
      Step 4: Calculate the answer.

      v = 1.800 1...m/s

      The answer must be rounded to three significant digits.

      The speed of the ball is 1.80 m/s.
    4. A 204.3 kg roller-coaster car is sitting motionless at a point on its track, 15.0 m above the ground. If the car starts to roll down the track, what will its speed be when it reaches a point 6.0 m above the ground?

      As the roller-coaster car moves, the gravitational potential energy gets converted into kinetic energy. Ep= Ek due to the conservation of energy.

      Step 1: List the variables.

      «math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi mathvariant=¨normal¨»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»204«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»g«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mi»h«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»15«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»-«/mo»«mn»6«/mn»«mo».«/mo»«mn»1«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»=«/mo»«mn»9«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mtd»«/mtr»«/mtable»«/math»
      Step 2: Identify the correct formula, and rearrange if necessary.

      «math» «mtable columnspacing=¨0px¨ columnalign=¨right center left¨» «mtr» «mtd» «msub» «mi»E«/mi» «mi»p«/mi» «/msub» «/mtd» «mtd» «mo»=«/mo» «/mtd» «mtd» «msub» «mi»E«/mi» «mi»k«/mi» «/msub» «/mtd» «/mtr» «mtr» «mtd» «mi»m«/mi» «mi»g«/mi» «mi»h«/mi» «/mtd» «mtd» «mo»=«/mo» «/mtd» «mtd» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «mi»m«/mi» «msup» «mi»v«/mi» «mn»2«/mn» «/msup» «/mtd» «/mtr» «/mtable» «/math»

      To move «math» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «/math»and m to the other side, you need to use the opposite operation.

      «math» «msup» «mi»v«/mi» «mn»2«/mn» «/msup» «mo»=«/mo» «mfrac» «mrow» «mn»2«/mn» «mi»m«/mi» «mi»g«/mi» «mi»h«/mi» «/mrow» «mi»m«/mi» «/mfrac» «/math»

      cancel like terms of m

      «math» «msup» «mi»v«/mi» «mn»2«/mn» «/msup» «mo»=«/mo» «mn»2«/mn» «mi»g«/mi» «mi»h«/mi» «/math»
      Step 3: Substitute the values into the formula.

      «math»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«mo»=«/mo»«mfenced»«mn»2«/mn»«/mfenced»«mfenced»«mrow»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mfrac»«/mrow»«/mfenced»«mfenced»«mrow»«mn»9«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mrow»«/mfenced»«mo»=«/mo»«mn»176«/mn»«mo».«/mo»«mn»58«/mn»«mfrac»«msup»«mi mathvariant=¨normal¨»m«/mi»«mn»2«/mn»«/msup»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mfrac»«/math»
      Step 4: Calculate the answer.

      «math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»176«/mn»«mo».«/mo»«mn»58«/mn»«mfrac»«msup»«mi mathvariant=¨normal¨»m«/mi»«mn»2«/mn»«/msup»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msqrt»«mn»176«/mn»«mo».«/mo»«mn»58«/mn»«mfrac»«msup»«mi mathvariant=¨normal¨»m«/mi»«mn»2«/mn»«/msup»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mfrac»«/msqrt»«mo»=«/mo»«mn»13«/mn»«mo».«/mo»«mn»288«/mn»«mo»§#8230;«/mo»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«mi mathvariant=¨normal¨»s«/mi»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«mtd»«/mtd»«mtd»«/mtd»«/mtr»«/mtable»«/math»

      The answer must be rounded to two significant digits.
      The girls speed is «math»«mn»13«/mn»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«mi mathvariant=¨normal¨»s«/mi»«/mfrac»«mo».«/mo»«/math»

      Alternative way to solve this question:

      «math» «mi»m«/mi» «mi»g«/mi» «mi»h«/mi» «mo»=«/mo» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «mi»m«/mi» «msup» «mi»v«/mi» «mn»2«/mn» «/msup» «/math»

      Since mass is found on both sides of the equation, you can cancel the common factor. To move «math» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «/math»to the other side, you need to use the opposite operation.

      «math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»g«/mi»«mi»h«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«mfenced»«mrow»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mfrac»«/mrow»«/mfenced»«mfenced»«mrow»«mn»9«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mrow»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mfenced»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mfenced»«/mtd»«/mtr»«mtr»«mtd»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfenced»«mn»2«/mn»«/mfenced»«mfenced»«mrow»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mfrac»«/mrow»«/mfenced»«mfenced»«mrow»«mn»9«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mrow»«/mfenced»«mo»=«/mo»«mn»176«/mn»«mo».«/mo»«mn»58«/mn»«mfrac»«msup»«mi mathvariant=¨normal¨»m«/mi»«mn»2«/mn»«/msup»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msqrt»«mn»176«/mn»«mo».«/mo»«mn»58«/mn»«mfrac»«msup»«mi mathvariant=¨normal¨»m«/mi»«mn»2«/mn»«/msup»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mfrac»«/msqrt»«mo»=«/mo»«mn»13«/mn»«mo».«/mo»«mn»288«/mn»«mo»§#8230;«/mo»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«mi mathvariant=¨normal¨»s«/mi»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«mtd»«/mtd»«mtd»«/mtd»«/mtr»«/mtable»«/math»
      The girls speed is «math»«mn»13«/mn»«mfrac»«mi mathvariant=¨normal¨»m«/mi»«mi mathvariant=¨normal¨»s«/mi»«/mfrac»«mo».«/mo»«/math»
    5. A boy rides his bicycle 823 m north and stops to talk to a friend. He then rides 382 m north but realizes that he needs to return a book to another friend’s house. He rides 540 m south. After dropping off the book, he rides 1 450 m north. What distance did the boy ride? What was his displacement?

      To calculate the distance, you would perform the calculation Δd = 823 m + 382 m + 540 m + 1 450 m = 3 195 m.

      To calculate the displacement, you would add the four values, taking into account that north is a positive direction and south is a negative direction.

      «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»§#8710;«/mo»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mfenced»«mrow»«mo»+«/mo»«mn»823«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mrow»«/mfenced»«mo»+«/mo»«mfenced»«mrow»«mo»+«/mo»«mn»382«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mrow»«/mfenced»«mo»+«/mo»«mfenced»«mrow»«mo»-«/mo»«mn»540«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mrow»«/mfenced»«mo»+«/mo»«mfenced»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»450«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mrow»«/mfenced»«mo»=«/mo»«mo»+«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mn»115«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math»
       
      Because north is positive, an answer of +2 115 m means that you could express it as «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»§#8710;«/mo»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mo»+«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mn»115«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»o«/mi»«mi»r«/mi»«mo»§#160;«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mn»115«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mfenced open=¨[¨ close=¨]¨»«mi mathvariant=¨normal¨»N«/mi»«/mfenced»«mo».«/mo»«/math»

    6. Use the following graph to answer the questions.


      ©Explore Learning
      C5.24 graph of a runner with described motion

      1. Describe the motion of the object from point 0 s to point 1 s.

        A positive slope indicates the object is moving forward.
      2. Describe the motion of the object from point 1 s to point 3 s.

        The flat portion of the graph indicates the object is not moving.
      3. Describe the motion of the object from poins 3 s to point 4 s.

        A positive slope indicates the object is moving forward.
      4. What is the total distance travelled by the object?

        Read from the graph that the distance from 0 s to 1 s is 10 m.
        Read from the graph that the distance from 1 s to 3 s is 0 m.
        Read from the graph that the distance from 3 s to 4 s is 10 m.
        To calculate the distance, you would perform the calculation Δd = (10 m) + (0 m) + (10 m) = 20 m.
      5. What is the total displacement of the object?

        To calculate the displacement, you would add the three values, taking into account that positive slope is a positive direction and negative slope is a negative direction.

        «math»«mo»§#8710;«/mo»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mfenced»«mrow»«mo»+«/mo»«mn»10«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mrow»«/mfenced»«mo»+«/mo»«mfenced»«mrow»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mrow»«/mfenced»«mo»+«/mo»«mfenced»«mrow»«mo»+«/mo»«mn»10«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mrow»«/mfenced»«mo»=«/mo»«mo»+«/mo»«mn»20«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/math»

        Because positive slope is positive, then an answer of +20 m means that you could express it as «math»«mo»§#8710;«/mo»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mo»+«/mo»«mn»20«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«mo»§#160;«/mo»«mi»or«/mi»«mo»§#160;«/mo»«mo»§#160;«/mo»«mn»20«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«mfenced open=¨[¨ close=¨]¨»«mi»forward«/mi»«/mfenced»«/math».
      6. What is the average velocity of the object from 3 s to 4 s?

        «math»«mi»s«/mi»«mi»l«/mi»«mi»o«/mi»«mi»p«/mi»«mi»e«/mi»«mo»=«/mo»«mfrac»«mrow»«mi»r«/mi»«mi»i«/mi»«mi»s«/mi»«mi»e«/mi»«/mrow»«mrow»«mi»r«/mi»«mi»u«/mi»«mi»n«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mfrac»«mrow»«msub»«mi»y«/mi»«mn»1«/mn»«/msub»«mo»-«/mo»«msub»«mi»y«/mi»«mn»2«/mn»«/msub»«/mrow»«mrow»«msub»«mi»x«/mi»«mn»1«/mn»«/msub»«mo»-«/mo»«msub»«mi»x«/mi»«mn»2«/mn»«/msub»«/mrow»«/mfrac»«mspace linebreak=¨newline¨»«/mspace»«mi»s«/mi»«mi»l«/mi»«mi»o«/mi»«mi»p«/mi»«mi»e«/mi»«mo»=«/mo»«mfrac»«mrow»«mn»20«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»-«/mo»«mn»10«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mrow»«mrow»«mn»4«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»-«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mfrac»«mrow»«mn»10«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mrow»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mrow»«/mfrac»«mo»=«/mo»«mn»10«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/math»
    7. If you walk an average velocity of 1.25 m/s [S], how much time will it take for you to go 2.37 km [S]?

      Step 1: List the variables.

      v = 1.25 m/s [S]

      Remember that the displacement needs to be in meters.
      So, we need to convert kilometers to meters. There are 1 000 m in 1 km.

      «math»«mo»§#8710;«/mo»«mi»d«/mi»«mo»=«/mo»«mn»2«/mn»«mo».«/mo»«mn»37«/mn»«mo»§#160;«/mo»«mi»km«/mi»«mo»§#160;«/mo»«mfenced open=¨[¨ close=¨]¨»«mi mathvariant=¨normal¨»S«/mi»«/mfenced»«mo»=«/mo»«mfenced»«mrow»«mn»2«/mn»«mo».«/mo»«mn»37«/mn»«mo»§#160;«/mo»«mi»km«/mi»«mo»§#160;«/mo»«mfenced open=¨[¨ close=¨]¨»«mi mathvariant=¨normal¨»S«/mi»«/mfenced»«/mrow»«/mfenced»«mfenced»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mrow»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/mrow»«/mfrac»«/mfenced»«mo»=«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mn»370«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«mfenced open=¨[¨ close=¨]¨»«mi mathvariant=¨normal¨»S«/mi»«/mfenced»«mspace linebreak=¨newline¨»«/mspace»«mo»§#8710;«/mo»«mi»t«/mi»«mo»=«/mo»«mo»?«/mo»«/math»
      Step 2: Identify the correct formula, and rearrange if necessary.

      «math»«mi»v«/mi»«mo»=«/mo»«mfrac»«mrow»«mo»§#8710;«/mo»«mi»d«/mi»«/mrow»«mrow»«mo»§#8710;«/mo»«mi»t«/mi»«/mrow»«/mfrac»«/math»

      Re-arrange the formula to solve for Δt.

      To isolate Δt, you must multiple each side by Δt. To move Δt to the other side, you must use the opposite operation. Multiplication is opposite to division.

      Then you would have the equation «math»«mover»«mi»v«/mi»«mo»§#8594;«/mo»«/mover»«mo»§#8710;«/mo»«mi»t«/mi»«mo»=«/mo»«mo»§#8710;«/mo»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«/math».

      To isolate Δt, you must divide each side by «math»«mover»«mi»v«/mi»«mo»§#8594;«/mo»«/mover»«/math». To move «math»«mover»«mi»v«/mi»«mo»§#8594;«/mo»«/mover»«/math» to the other side, you must use the opposite operation. Division is opposite to multiplication.

      «math»«mo»§#8710;«/mo»«mi»t«/mi»«mo»=«/mo»«mfrac»«mrow»«mo»§#8710;«/mo»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«/mrow»«mover»«mi»v«/mi»«mo»§#8594;«/mo»«/mover»«/mfrac»«/math»
      Step 3: Substitute the values into the formula.

      «math»«mo»§#8710;«/mo»«mi»t«/mi»«mo»=«/mo»«mfrac»«mrow»«mn»2«/mn»«mo»§#160;«/mo»«mn»370«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«mfenced open=¨[¨ close=¨]¨»«mi mathvariant=¨normal¨»S«/mi»«/mfenced»«/mrow»«mrow»«mn»1«/mn»«mo».«/mo»«mn»25«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»§#160;«/mo»«mfenced open=¨[¨ close=¨]¨»«mi mathvariant=¨normal¨»S«/mi»«/mfenced»«/mrow»«/mfrac»«/math»
      Step 4: Calculate the answer.

      Δt = 1 896 s = 1.90 × 103 s (to 2 sig digs)
    8. A car slows from 27 m/s [W] to 10.0 m/s [W] before reaching a highway exit. If it took the car 6.5 s to reach the exit after starting to slow down, what was the car’s acceleration?

      Step 1: List the variables.

      null
      Step 2: Identify the correct formula, and rearrange if necessary.

      «math»«mover»«mi»a«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mfrac»«mrow»«mo»§#8710;«/mo»«mover»«mi»v«/mi»«mo»§#8594;«/mo»«/mover»«/mrow»«mrow»«mo»§#8710;«/mo»«mi»t«/mi»«/mrow»«/mfrac»«/math»
      Step 3: Substitute the values into the formula.

      «math»«mover»«mi»a«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mfrac»«mrow»«mo»-«/mo»«mn»17«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«mo»§#160;«/mo»«mfenced open=¨[¨ close=¨]¨»«mi mathvariant=¨normal¨»W«/mi»«/mfenced»«/mrow»«mrow»«mn»6«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mrow»«/mfrac»«/math»
      Step 4: Calculate the answer.

      «math»«mover»«mi»a«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mo»-«/mo»«mn»2«/mn»«mo».«/mo»«mn»625«/mn»«mo»§#8230;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«mo»=«/mo»«mo»-«/mo»«mn»2«/mn»«mo».«/mo»«mn»6«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«mo»§#160;«/mo»«mo»§#160;«/mo»«mi»or«/mi»«mo»§#160;«/mo»«mo»§#160;«/mo»«mn»2«/mn»«mo».«/mo»«mn»6«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«mo»§#160;«/mo»«mfenced open=¨[¨ close=¨]¨»«mi mathvariant=¨normal¨»W«/mi»«/mfenced»«mo»§#160;«/mo»«mfenced»«mrow»«mi»to«/mi»«mo»§#160;«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi»sig«/mi»«mo»§#160;«/mo»«mi»digs«/mi»«/mrow»«/mfenced»«/math»
    9. Compare the energy conversions that occur in a hydroelectric power generation plant and a wind power generator.

      Hydroelectric power:
      sun (nuclear energy) → gravitational potential energy → kinetic energy → mechanical energy → electrical energy

      Wind power:
      sun (nuclear energy) →  kinetic energy → mechanical energy → electrical energy

      The energy conversions are the same, except for the gravitational potential energy of the water behind the dam.
    10. If a light bulb is 6.39% efficient and it emits a total of 6.18 x 103 J of light energy, how much electrical energy did it use?

      Step 1: List the variables.

      «math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»6«/mn»«mo».«/mo»«mn»39«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»6«/mn»«mo».«/mo»«mn»18«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mn»3«/mn»«/msup»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«/mtable»«/math»
      Step 2: Identify the correct formula, and rearrange if necessary.

      «math»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«mo»=«/mo»«mfrac»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/math»

      To isolate total input energy, you must multiple each side by total input energy. To move total input energy to the other side, you must use the opposite operation. Multiplication is opposite to division.

      Then you would have the equation:
      (total input energy)(percent efficiency)=useful output energy × 100%

      To isolate total input energy, you must divide each side by percent efficiency. To move percent efficiency to the other side, you must use the opposite operation. Division is opposite to multiplication.

      Then you would have the equation «math»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»=«/mo»«mfrac»«mrow»«mi»ueful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/math».
      Step 3: Substitute the values into the formula.

      «math»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»=«/mo»«mfrac»«mrow»«mn»6«/mn»«mo».«/mo»«mn»18«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mrow»«mn»3«/mn»«mo»§#160;«/mo»«/mrow»«/msup»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»6«/mn»«mo».«/mo»«mn»39«/mn»«mo»%«/mo»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/math»
      Step 4: Calculate the answer.

      total input energy = 96 713.615...J = 9.67 × 104 J (to 3 sig digs)
    11. A company wants to build a hydroelectric dam for power generation on a river that runs through the city where you live. What information would you want to know before permission is granted to the company to build the dam? What suggestions might you make to the company to reduce the impact of the dam on the environment? Hint: You may need to do a bit of research.

      Your answer should be a variation of the following.

      Information required: Will the walls and floor of the river valley support the water pressure of a reservoir? Is there a section with a deep enough canyon to make an efficient dam? Has the company researched the environment in the area and how the dam and flooding of the area will impact organisms native to the area?

      Suggestions: Make sure the dam does not interfere with mating patterns, seasons, and breeding areas of the water animals. Make sure that there is a large enough area available for the dam and reservoir. Materials used in the construction must be of high quality materials. Talk to neighbouring communities that may be impacted by a change in river flow.

    3.6 Assignment

    Unit 3 Assignment Lessons 9-11

    It is now time to complete the Lesson 11 portion of 3.6 Assignment. This assignment has two parts.

    1. Part 1 Written Portion: Select the preferred document type from the options below. Download and save the assignment on your desktop (or documents folder).

      PDF Document       
    2. Open and print this saved document.
    3. Record your responses in the appropriate textboxes.
    4. When you have completed the assignment, scan it and save it on your desktop (or documents folder).
    5. Once you have completed the written portion of your assignment, click on the button below to go to the submission page.

      Written Portion Submission Page
    6. Part 2 Online Portion: It is now time to complete the online portion of this assignment. Click on the button below to go to the online questions of this assignment.

      Online Questions

    This assignment is worth ___% of your final grade.