~EUvV֊{ Student Name: assignment #002
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14:50:51 `q001. Note that there are 14 questions in this assignment.
If an object moves 12 meters in 4 seconds, then at what average rate is the object moving? Explain how you obtained your result in terms of commonsense images.......!!!!!!!!...................................
RESPONSE --> 12 m / 4 s = 3 m/s
If you had 4 marks to represent the time traveled you would need to cover an equal amount between each mark, and this resulted in 3 units traveled for each mark..................................................
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14:51:04 Moving 12 meters in 4 seconds, we move an average of 3 meters every second. We can imagine dividing up the 12 meters into four equal parts, one for each second. Each part will have 3 meters, corresponding to the distance moved in 1 second, on the average.
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RESPONSE --> OK.
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14:52:23 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> It is related to rate because both determine a quantity compared to a unit.
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14:52:51 A rate is obtained by dividing the change in a quantity by the change in another quantity on which is dependent. In this case we divided the change in position by the time during which that change occurred.
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RESPONSE --> OK.
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14:53:28 `q003. Is object position dependent on time or is time dependent on object position?
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RESPONSE --> Object position is dependent of time.
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14:53:38 Object position is dependent on time--the clock runs whether the object is moving or not so time is independent of position. Clock time is pretty much independent of anything else.
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RESPONSE --> OK.
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14:54:33 `q004. So the rate here is the average rate at which position is changing with respect to clock time. Explain what concepts, if any, you missed in your explanations.
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RESPONSE --> I undertood these concepts.
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14:54:46 You should always self-critique your work in this manner. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
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RESPONSE --> OK.
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15:01:24 `q005. If an object is displaced -6 meters in three seconds, then what is the average speed of the object what is its average velocity? Explain how you obtained your result in terms of commonsense images and ideas.
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RESPONSE --> Average Speed: 6 m / 3 sec = 2 m/s
Average Velocity: -6 m / 3 sec = -2 m/s The time is divided into 3 marks and then the distance is divided equally between those marks, and this results in 3 meters per second. For velocity, the negative sign indicates the object moving to the left on the x axis..................................................
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15:01:40 Speed is the average rate at which distance changes, and distance cannot be negative. Therefore speed cannot be negative. Velocity is the average rate at which position changes, and position changes can be positive or negative.
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RESPONSE --> OK.
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15:02:49 `q006. If `ds stands for the change in the position of an object and `dt for the time interval during which the position changes, then what expression stands for the average velocity vAve of the object during this time interval?
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RESPONSE --> vAve = ds / dt
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15:04:23 Average velocity is rate of change of position. Change in position is `ds and change in clock tim is `dt, so vAve = `ds / `dt.
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RESPONSE --> Are the ` marks needed? (ex. `ds or ds)
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15:06:08 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> (triangle) x / (triangle) t
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15:07:07 You use the Greek capital Delta symbol Delta. `d is often used here because the symbol for Delta is not interpreted correctly by some Internet browsers. You should get in the habit of thinking and writing Delta when you see `d. You may use either `d or Delta when submitting work and answering questions.
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RESPONSE --> OK.
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15:09:06 `q008. If an object changes position at an average rate of 5 meters/second for 10 seconds, then how far does it move?
How is this problem related to the concept of a rate?......!!!!!!!!...................................
RESPONSE --> 5 m/s * 10 s = 50 m
This is related in that both use a rate multiplied by a quantity to find the answer..................................................
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15:10:48 In this problem you are given the rate at which position changes with respect to time, and you are given the time interval during which to calculate the change in position. Given the rate at which one quantity changes with respect to another, and the change in the second quantity, how do we obtain the resulting change in the first?
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RESPONSE --> Multiply the two together.
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15:14:33 `q009. If vAve stands for the rate at which the position of the object changes (also called velocity) and `dt for the time interval during which the change in position is to be calculated, then how to we write the expression for the change `ds in the position?
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RESPONSE --> `ds = vAve * `dt
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15:15:11 To find the change in a quantity we multiply the rate by the time interval during which the change occurs. We therefore obtain the change in position by multiplying the velocity by the time interval: `ds = vAve * `dt. The units of this calculation pretty much tell us what to do: Just as when we multiply pay rate by time (dollar / hr * hours of work) or automobile velocity by the time interval (miles / hour * hour), when we multiply vAve, in cm / sec or meters / sec or whatever, by `dt in seconds, we get displacement in cm or meters, or whatever, depending on the units of distance used.
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RESPONSE --> OK.
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15:21:33 `q010. Explain how the quantities average velocity vAve, time interval `dt and displacement `ds are related by the definition of a rate, and how this relationship can be used to solve the current problem problem.
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RESPONSE --> vAve is equivalent to rate `dt is equivalent to the change in time `ds is equivalent to the change in quantity
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15:21:51 vAve is the average rate at which position changes. The change in position is the displacement `ds, the change in clock time is `dt, so vAve = `ds / `dt.
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RESPONSE --> OK.
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15:27:59 `q011. The basic rate relationship vAve = `ds / `dt expresses the definition of average velocity vAve as the rate at which position s changes with respect to clock time t. What algebraic steps do we use to solve this equation for `ds, and what is our result?
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RESPONSE --> To solve for 'ds we multiply both sides by 'dt. This results in the equation 'ds = vAve * 'dt.
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15:28:24 To solve vAve = `ds / `dt for `ds, we multiply both sides by `dt. The steps:
vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds . Switching sides we have `ds = vAve * `dt.......!!!!!!!!...................................
RESPONSE --> OK.
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15:30:52 `q012. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
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RESPONSE --> Displacement is equal to average velocity times clock time.
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15:31:11 Our most direct intuition about velocity probably comes from watching an automobile speedometer. We know that if we multiply our average velocity in mph by the duration `dt of the time interval during which we travel, we get the distance traveled in miles. From this we easily extend the idea. Whenever we multiply our average velocity by the duration of the time interval, we expect to obtain the displacement, or change in position, during that time interval.
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RESPONSE --> OK.
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15:34:33 `q013. What algebraic steps do we use to solve the equation vAve = `ds / `dt for `dt, and what is our result?
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RESPONSE --> We multiply both sides by 'dt and then we divide both sides by vAve.
vAve = `ds / `dt (Multiply both sides by 'dt) 'dt * vAve = 'ds (Divide both sides by vAve) 'dt = 'ds / vAve.................................................
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15:34:51 To solve vAve = `ds / `dt for `dt, we must get `dt out of the denominator. Thus we first multiply both sides by the denominator `dt. Then we can see where we are and takes the appropriate next that. The steps:
vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds. We can now divide both sides by vAve to get `dt = `ds / vAve.......!!!!!!!!...................................
RESPONSE --> OK.
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15:35:47 `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
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RESPONSE --> Clock time equals displacement divided by average velocity.
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15:35:58 If we want to know how long it will take to make a trip at a certain speed, we know to divide the distance in miles by the speed in mph. If we divide the number of miles we need to travel by the number of miles we travel in hour, we get the number of hours required. We extend this to the general concept of dividing the displacement by the velocity to get the duration of the time interval.
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RESPONSE --> OK.
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ΓQxyzj뵶} Student Name: assignment #002 002. Volumes
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22:07:35 `q001. There are 9 questions and 4 summary questions in this assignment.
What is the volume of a rectangular solid whose dimensions are exactly 3 cm by 5 cm by 7 cm?......!!!!!!!!...................................
RESPONSE --> V = lwh V = (3 cm)(5 cm)(7 cm) V = 105 cm^3
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22:11:08 If we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2.
Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3. The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3. This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore V = A * h, where A is the area of the base and h the altitude. This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more important.......!!!!!!!!...................................
RESPONSE --> OK. But you answer says m^3 and I believe it should be cm^3.
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22:12:12 `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters?
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RESPONSE --> V = Ah V = (48 m^2)(2 m) V = 96 m^3
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22:12:22 Using the idea that V = A * h we find that the volume of this solid is
V = A * h = 48 m^2 * 2 m = 96 m^3. Note that m * m^2 means m * (m * m) = m * m * m = m^2.......!!!!!!!!...................................
RESPONSE --> OK.
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22:15:18 `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters?
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RESPONSE --> V = Ah V = (20 m^2)(40 m) V = 800 m^3
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22:15:36 V = A * h applies to uniform cylinders as well as to rectangular solids. We are given the altitude h and the base area A so we conclude that
V = A * h = 20 m^2 * 40 m = 800 m^3. The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This is the case for uniform cylinders and uniform prisms.......!!!!!!!!...................................
RESPONSE --> OK.
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22:17:03 `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?
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RESPONSE --> V = pi * r^2 * l V = (3.14)(5 cm^2)(30 cm) V = 471 cm^3
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22:19:47 The cylinder is uniform, which means that its cross-sectional area is constant. So the relationship V = A * h applies.
The cross-sectional area A is the area of a circle of radius 5 cm, so we see that A = pi r^2 = pi ( 5 cm)^2 = 25 pi cm^2. Since the altitude is 30 cm the volume is therefore V = A * h = 25 pi cm^2 * 30 cm = 750 pi cm^3. Note that the common formula for the volume of a uniform cylinder is V = pi r^2 h. However this is just an instance of the formula V = A * h, since the cross-sectional area A of the uniform cylinder is pi r^2. Rather than having to carry around the formula V = pi r^2 h, it's more efficient to remember V = A * h and to apply the well-known formula A = pi r^2 for the area of a circle.......!!!!!!!!...................................
RESPONSE --> V = pi * r^2 * l V = (3.14) * (5 cm)^2 * (30 cm) V = 2355 cm^3
I understand this formula, but did not raise 5 to the power of 2 the first time..................................................
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22:22:52 `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?
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RESPONSE --> V = pi * r^2 * l V = (3.14) * (5 cm)^2 * 15 cm V = 1177.5 cm^3
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22:24:58 People will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km). Different cans have different dimensions, and your estimate will depend a lot on what can you are using.
A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches. This can would have volume V = A * h, where A is the area of the cross-section. The diameter of the cross-section is 3 inches so its radius will be 3/2 in.. The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3. Approximating, this comes out to around 35 in^3. Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^2.......!!!!!!!!...................................
RESPONSE --> OK.
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22:27:45 `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?
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RESPONSE --> V = 1/3 * Ah) V = 1/3 * (50 cm^2) * 60 cm V = 1000 cm^3
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22:28:27 We can't use the V = A * h idea for a pyramid because the thing doesn't have a constant cross-sectional area--from base to apex the cross-sections get smaller and smaller. It turns out that there is a way to cut up and reassemble a pyramid to show that its volume is exactly 1/3 that of a rectangular solid with base area A and altitude h. Think of putting the pyramid in a box having the same altitude as the pyramid, with the base of the pyramid just covering the bottom of the box. The apex (the point) of the pyramid will just touch the top of the box. The pyramid occupies exactly 1/3 the volume of that box.
So the volume of the pyramid is V = 1/3 * A * h. The base area A is 30 cm^2 and the altitude is 60 cm so we have V = 1/3 * 50 cm^2 * 60 cm = 1000 cm^3.......!!!!!!!!...................................
RESPONSE --> OK.
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22:30:15 `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters?
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RESPONSE --> V = 1/3 * Ah V = 1/3 * (20 m^2) * 9 m V = 60 m^3
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22:30:35 Just as the volume of a pyramid is 1/3 the volume of the 'box' that contains it, the volume of a cone is 1/3 the volume of the cylinder that contains it. Specifically, the cylinder that contains the cone has the base of the cone as its base and matches the altitude of the cone. So the volume of the cone is 1/3 A * h, where A is the area of the base and h is the altitude of the cone.
In this case the base area and altitude are given, so the volume of the cone is V = 1/3 A * h = 1/3 * 20 m^2 * 9 m = 60 m^3.......!!!!!!!!...................................
RESPONSE --> OK.
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22:32:41 `q008. What is a volume of a sphere whose radius is 4 meters?
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RESPONSE --> V = 4/3 * pi * r^3 V = 4/3 * 3.14 * (4 m)^3 V = 267.95 m^3
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22:33:04 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. In this case r = 4 m so
V = 4/3 pi * (4 m)^3 = 4/3 pi * 4^3 m^3 = 256/3 pi m^3.......!!!!!!!!...................................
RESPONSE --> OK.
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22:35:13 `q009. What is the volume of a planet whose diameter is 14,000 km?
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RESPONSE --> V = 4/3 * pi * r^3 V = 4/3 * 3.14 * (14,000 km)^3 V = 1.15 * 10^13 km^3
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22:37:45 The planet is presumably a sphere, so to the extent that this is so the volume of this planet is V = 4/3 pi r^3, where r is the radius of the planet. The diameter of the planet is 14,000 km so the radius is half this, or 7,000 km. It follows that the volume of the planet is
V = 4/3 pi r^3 = 4/3 pi * (7,000 km)^3 = 4/3 pi * 343,000,000,000 km^3 = 1,372,000,000,000 / 3 * pi km^3. This result can be approximated to an appropriate number of significant figures.......!!!!!!!!...................................
RESPONSE --> V = 4/3 * pi * r^3 V = 4/3 * 3.14 * (7,000 km)^3 V = 1.44 * 10^12 km^3
I mistakenly used diameter instead of radius. I need to be more careful when reading the problem..................................................
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22:38:31 `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?
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RESPONSE --> We multiply the area of the base by the length of the cylinder.
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22:38:49 The principle is that when the cross-section of an object is constant, its volume is V = A * h, where A is the cross-sectional area and h the altitude. Altitude is measure perpendicular to the cross-section.
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RESPONSE --> OK.
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22:39:35 `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?
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RESPONSE --> We take 1/3 of the area of the base times height.
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22:39:49 The volumes of these solids are each 1/3 the volume of the enclosing figure. Each volume can be expressed as V = 1/3 A * h, where A is the area of the base and h the altitude as measured perpendicular to the base.
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RESPONSE --> OK.
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22:40:23 `q012. Summary Question 3: What is the formula for the volume of a sphere?
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RESPONSE --> V = 4/3 * pi * r^3
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22:40:32 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere.
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RESPONSE --> OK.
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22:42:19 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I have tried to make a connection between the different formulas for volume (for example: the formula for the volume of a cone is 1/3 the formula for the volume of a cylinder).
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22:42:28 This ends the second assignment.
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RESPONSE --> OK.
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Randomized Problems: Assignment 2 Problem 1 If the velocity of the object changes from 4 cm / sec to 16 cm / sec in 8 seconds, then at what average rate is the velocity changing? Average rate of change = ?s / ?t Average rate of change = (16 cm/s 4 cm/s) / 8 s Average rate of change = (12 cm/s) / 8 s Average rate of change = 1.5 cm/s/s A ball rolling from rest down a constant incline requires 8.2 seconds to roll the 97 centimeter length of the incline. What is its average velocity? vAve = ?s / ?t vAve = 97 cm / 8.2 s vAve ? 11.83 cm/s An object which accelerates uniformly from rest will attain a final velocity which is double its average velocity. What therefore is the final velocity of this ball? final velocity = 2 (vAve) final velocity = 2 (11.83 cm/s) final velocity = 23.66 cm/s What average rate is the velocity of the ball therefore changing? Average rate of change = ?s / ?t