Unit 3 Conclusion
Completion requirements
Unit 3
Energy Flow in Technological Systems
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.
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.
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.
- 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.
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.
- 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.
- 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.
- 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.
- 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»
- 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»
- Use the following graph to answer the questions.
©Explore Learning
C5.24 graph of a runner with described motion
- Describe the motion of the object from point 0 s to point 1 s.A positive slope indicates the object is moving forward.
- 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.
- Describe the motion of the object from poins 3 s to point 4 s.A positive slope indicates the object is moving forward.
- 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. - 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». - 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»
- Describe the motion of the object from point 0 s to point 1 s.
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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) - 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.
nullStep 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» - 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. - 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) - 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.