course Phy 201
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19:48:57 `q001. Note that there are 11 questions in this assignment. vAve = `ds / `dt, which is the definition of average velocity and which fits well with our intuition about this concept. If displacement `ds is measured in meters and the time interval `dt is measured in seconds, in what units will vAve be obtained?
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RESPONSE --> The vAve will be in meters per second.
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19:49:49 vAve = `ds / `dt. The units of `ds are cm and the units of `dt are sec, so the units of `ds / `dt must be cm / sec. Thus vAve is in cm/s.
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RESPONSE --> I read the 'ds as being in meters, but I used the correct form. sorry, it was a careless mistake
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19:51:10 `q002. If the definition equation vAve = `ds / `dt is to be solved for `ds we multiply both sides of the equation by `dt to obtain `ds = vAve * `dt. If vAve is measured in cm / sec and `dt in sec, then in what units must `ds be measured?
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RESPONSE --> 'ds is being measured in cm.
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19:51:19 Since vAve is in cm/sec and `dt in sec, `ds = vAve * `dt must be in units of cm / sec * sec = cm.
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19:53:16 `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.
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RESPONSE --> By multiplying the unit cm/sec by the united sec would give you the unit cm.
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19:54:46 When we multiply cm/sec by sec we are multiplying the fractions cm / sec and sec / 1. When we multiply fractions we will multiply numerators and denominators. We obtain cm * sec / ( sec * 1). This can be rearranged as (sec / sec) * (cm / 1), which is the same as 1 * cm / 1. Since multiplication or division by 1 doesn't change a quantity, this is just equal to cm.
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RESPONSE --> Yes I understand, my answer covers this but in a simpler form.
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19:55:51 `q004. If the definition vAve = `ds / `dt is to be solved for `dt we multiply both sides of the equation by `dt to obtain vAve * `dt = `ds, then divide both sides by vAve to get `dt = `ds / vAve. If vAve is measured in km / sec and `ds in km, then in what units must `dt be measured?
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RESPONSE --> The 'dt would be measured in seconds.
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19:55:55 Since `dt = `ds / vAve and `ds is in km and vAve in km/sec, `ds / vAve will be in km / (km / sec) = seconds.
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19:57:01 `q005. Explain the algebra of dividing the unit km / sec into the unit km.
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RESPONSE --> BY dividing the unit km/sec into the unit km, you would obtain 'dt or seconds.
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19:57:06 The division is km / (km / sec). Since division by a fraction is multiplication by the reciprocal of the fraction, we have km * (sec / km). This is equivalent to multiplication of fractions (km / 1) * (sec / km). Multiplying numerators and denominators we get (km * sec) / (1 * km), which can be rearranged to give us (km / km) * (sec / 1), or 1 * sec / 1, or just sec.
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20:02:35 `q006. If an object moves from position s = 4 meters to position s = 10 meters between clock times t = 2 seconds and t = 5 seconds, then at what average rate is the position of the object changing (i.e., what is the average velocity of the object) during this time interval? What is the change `ds in position, what is the change `dt in clock time, and how do we combine these quantities to obtain the average velocity?
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RESPONSE --> The vAve of the object in the time interval would be 10m - 4m = 6m, 5s - 2s = 3s. 6m/3s = 2m/s = vAve
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20:03:45 We see that the changes in position and clock time our `ds = 10 meters - 4 meters = 6 meters and `dt = 5 seconds - 2 seconds = 3 seconds. We see also that the average velocity is vAve = `ds / `dt = 6 meters / (3 seconds) = 2 meters / second. Comment on any discrepancy between this reasoning and your reasoning.
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RESPONSE --> My reasoning and yours correspond perfectly.
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20:05:15 `q007. Symbolize this process: If an object moves from position s = s1 to position s = s2 between clock times t = t1 and t = t2, when what expression represents the change `ds in position and what expression represents the change `dt in the clock time?
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RESPONSE --> The expression s2 - s1 would rep. the change in 'ds. The expression t2 - t1 would rep. the change in 'dt.
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20:05:23 We see that the change in position is `ds = s2 - s1, obtained as usual by subtracting the first position from the second. Similarly the change in clock time is `dt = t2 - t1. What expression therefore symbolizes the average velocity between the two clock times.
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20:09:27 `q008. On a graph of position s vs. clock time t we see that the first position s = 4 meters occurs at clock time t = 2 seconds, which corresponds to the point (2 sec, 4 meters) on the graph, while the second position s = 10 meters occurs at clock time t = 5 seconds and therefore corresponds to the point (5 sec, 10 meters). If a right triangle is drawn between these points on the graph, with the sides of the triangle parallel to the s and t axes, the rise of the triangle is the quantity represented by its vertical side and the run is the quantity represented by its horizontal side. This slope of the triangle is defined as the ratio rise / run. What is the rise of the triangle (i.e., the length of the vertical side) and what quantity does the rise represent? What is the run of the triangle and what does it represent?
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RESPONSE --> The rise of the triangle would be 6 meters. The run of the triangle would be 3 seconds.
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20:09:31 The rise of the triangle represents the change in the position coordinate, which from the first point to the second is 10 m - 4 m = 6 m. The run of the triangle represents the change in the clock time coordinate, which is 5 s - 2 s = 3 s.
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20:10:49 `q009. What is the slope of this triangle and what does it represent?
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RESPONSE --> The slope of the triangle would be 2m/s. This represents the vAve of the object moving on the graph.
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20:10:53 The slope of this graph is 6 meters / 3 seconds = 2 meters / second.
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20:12:37 `q010. In what sense does the slope of any graph of position vs. clock time represent the velocity of the object? For example, why does a greater slope imply greater velocity?
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RESPONSE --> Slope and velocity are both figured out by using the same formula because 'ds/'dt = vAve corresponds to rise/run = slope. So, if the velocity is high, the slope will be as well.
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20:13:02 Since the rise between two points on a graph of velocity vs. clock time represents the change in `ds position, and since the run represents the change `dt clock time, the slope represents rise / run, or change in position /change in clock time, or `ds / `dt. This is the definition of average velocity.
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20:24:15 `q011. As a car rolls from rest down a hill, its velocity increases. Describe a graph of the position of the car vs. clock time. If you have not already done so, tell whether the graph is increasing at an increasing rate, increasing at a decreasing rate, decreasing at an increasing rate, decreasing at a decreasing rate, increasing at a constant rate or decreasing at a constant rate. Is the slope of your graph increasing or decreasing? How does the behavior of the slope of your graph indicate the condition of the problem, namely that the velocity is increasing?
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RESPONSE --> The car starts out at 3cm at clock time 5 sec., then 12 cm at the clock time 10 sec. The car is increasing at an increasing rate, because the car is covering the same distance in a shorter amount of time. The slope is increasing
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20:24:35 The graph should have been increasing, since the position of the car increases with time (the car gets further and further from its starting point). The slope of the graph should have been increasing, since it is the slope of the graph that indicates velocity. An increasing graph within increasing slope is said to be increasing at an increasing rate.
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?}??i?f??????G|?Student Name: assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density
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22:06:46 `q001. There are 10 questions and 5 summary questions in this assignment. What is surface area of a rectangular solid whose dimensions are 3 meters by 4 meters by 6 meters?
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RESPONSE --> 2lw + 2lh + 2wh (2 * 3 * 4) + (2 * 3 * 6) + (2 * 4 * 6) SA = 108 m.
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22:06:50 A rectangular solid has six faces (top, bottom, front, back, left side, right side if you're facing it). The pairs top and bottom, right and left sides, and front-back have identical areas. This solid therefore has two faces with each of the following dimensions: 3 m by 4 m, 3 m by 6 m and 4 m by 6 m, areas 12 m^2, 18 m^2 and 24 m^2. Total area is 2 * 12 m^2 + 2 * 18 m^2 + 2 * 24 m^2 = 108 m^2.
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22:10:13 `q002. What is the surface area of the curved sides of a cylinder whose radius is five meters and whose altitude is 12 meters? If the cylinder is closed what is its total surface area?
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RESPONSE --> SA = 2Pir(h + r) SA = 2Pi * 5 (12 + 5) SA = 534.07 m^2
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22:19:12 The circumference of this cylinder is 2 pi r = 2 pi * 5 m = 10 pi m. If the cylinder was cut by a straight line running up its curved face then unrolled it would form a rectangle whose length and width would be the altitude and the circumference. The area of the curved side is therefore A = circumference * altitude = 10 pi m * 12 m = 120 pi m^2. If the cylinder is closed then it has a top and a bottom, each a circle of radius 5 m with resulting area A = pi r^2 = pi * (5 m)^2 = 25 pi m^2. The total area would then be total area = area of sides + 2 * area of base = 120 pi m^2 + 2 * 25 pi m^2 = 170 pi m^2.
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RESPONSE --> I didnt know the formual for a curved side. I now understand the open and closed approach after reviewing the solution
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22:21:59 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> The SA would equal 4 * Pi (r^2) 4 * Pi (1.5^2) = 28.274 cm^2
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22:23:16 The surface area of a sphere of radius r is A = 4 pi r^2. This sphere has radius 3 cm / 2, and therefore has surface area A = 4 pi r^2 = 4 pi * (3/2 cm)^2 = 9 pi cm^2.
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RESPONSE --> I thought radius would have been half the diameter, which was 3 so the radius would have been 1.5
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22:27:39 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> Using the Pythagorean Theorem a^2 + b^2 = c^2 5^2 + 9^2 = 106 = 10.296 ^2
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22:28:29 The Pythagorean Theorem says that the hypotenuse c of a right triangle with legs a and b satisfies the equation c^2 = a^2 + b^2. So, since all lengths are positive, we know that c = sqrt(a^2 + b^2) = sqrt( (5 m)^2 + (9 m)^2 ) = sqrt( 25 m^2 + 81 m^2) = sqrt( 106 m^2 ) = 10.3 m, approx.. Note that this is not what we would get if we made the common error of assuming that sqrt(a^2 + b^2) = a + b; this would tell us that the hypotenuse is 14 m, which is emphatically not so. There is no justification whatsoever for applying a distributive law (like x * ( y + z) = x * y + x * z ) to the square root operator.
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22:34:06 `q005. If the hypotenuse of a right triangle has length 6 meters and one of its legs has length 4 meters what is the length of the other leg?
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RESPONSE --> The other leg would equal 4.472 m.
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22:34:12 If c is the hypotenuse and a and b the legs, we know by the Pythagorean Theorem that c^2 = a^2 + b^2, so that a^2 = c^2 - b^2. Knowing the hypotenuse c = 6 m and the side b = 4 m we therefore find the unknown leg: a = sqrt( c^2 - b^2) = sqrt( (6 m)^2 - (4 m)^2 ) = sqrt(36 m^2 - 16 m^2) = sqrt(20 m^2) = sqrt(20) * sqrt(m^2) = 2 sqrt(5) m, or approximately 4.4 m.
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22:36:43 `q006. If a rectangular solid made of a uniform, homogeneous material has dimensions 4 cm by 7 cm by 12 cm and if its mass is 700 grams then what is its density in grams per cubic cm?
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RESPONSE --> The density would eqaul 2.083 g/cm^3
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22:36:50 The volume of this solid is 4 cm * 7 cm * 12 cm = 336 cm^3. Its density in grams per cm^3 is the number of grams in each cm^3. We find this quantity by dividing the number of grams by the number of cm^3. We find that density = 700 grams / (336 cm^3) = 2.06 grams / cm^3. Note that the solid was said to be uniform and homogeneous, meaning that it's all made of the same material, which is uniformly distributed. So each cm^3 does indeed have a mass of 2.06 grams. Had we not known that the material was uniform and homogeneous we could have said that the average density is 2.06 grams / cm^3, but not that the density is 2.06 grams / cm^3 (the object could be made of two separate substances, one with density less than 2.06 grams / cm^3 and the other with density greater than 2.06 g / cm^3, in appropriate proportions; neither substance would have density 2.06 g / cm^3, but the average density could be 2.06 g / cm^3).
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22:41:59 `q007. What is the mass of a sphere of radius 4 meters if its average density is 3,000 kg/cubic meter?
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RESPONSE --> The mass would be 12000 kg.
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22:42:44 A average density of 3000 kg / cubic meter implies that, at least on the average, every cubic meter has a mass of 3000 kg. So to find the mass of the sphere we multiply the number of cubic meters by 3000 kg. The volume of a sphere of radius 4 meters is 4/3 pi r^3 = 4/3 * pi (4m)^3 = 256/3 * pi m^3. So the mass of this sphere is mass = density * volume = 256 / 3 * pi m^3 * 3000 kg / m^3 = 256,000 * pi kg. This result can be approximated to an appropriate number of significant figures.
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RESPONSE --> Ah, I didnt apply the volume of a sphere I see the mistake.
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22:45:41 `q008. If we build a an object out of two pieces of material, one having a volume of 6 cm^3 at a density of 4 grams per cm^3 and another with a volume of 10 cm^3 at a density of 2 grams per cm^3 then what is the average density of this object?
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RESPONSE --> The average density would be 2 g/cm^3
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22:46:39 The first piece has a mass of 4 grams / cm^3 * 6 cm^3 = 24 grams. The second has a mass of 2 grams / cm^3 * 10 cm^3 = 20 grams. So the total mass is 24 grams + 20 grams = 44 grams. The average density of this object is average density = total mass / total volume = (24 grams + 20 grams) / (6 cm^3 + 10 cm^3) = 44 grams / (16 cm^3) = 2.75 grams / cm^3.
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RESPONSE --> I rounded down. sorry
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22:53:35 `q009. In a large box of dimension 2 meters by 3 meters by 5 meters we place 27 cubic meters of sand whose density is 2100 kg/cubic meter, surrounding a total of three cubic meters of cannon balls whose density is 8,000 kg per cubic meter. What is the average density of the material in the box?
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RESPONSE --> I dont get the cannon balls they are thrwoing me off.
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22:55:03 We find the average density from the total mass and the total volume. The mass of the sand is 27 m^3 * 2100 kg / m^3 = 56,700 kg. The mass of the cannonballs is 3 m^3 * 8,000 kg / m^3 = 24,000 kg. The average density is therefore average density = total mass / total volume = (56,700 kg + 24,000 kg) / (27 m^3 + 3 m^3) = 80,700 kg / (30 m^3) = 2,700 kg / m^3, approx..
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RESPONSE --> Ok this makes since now. I kept trying to figure up the 3 meters ina different way.
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23:02:52 `q010. How many cubic meters of oil are there in an oil slick which covers 1,700,000 square meters (between 1/2 and 1 square mile) to an average depth of .015 meters? If the density of the oil is 860 kg/cubic meter the what is the mass of the oil slick?
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RESPONSE --> There are 25500m^3 I dont know.
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23:04:08 The volume of the slick is V = A * h, where A is the area of the slick and h the thickness. This is the same principle used to find the volume of a cylinder or a rectangular solid. We see that the volume is V = A * h = 1,700,000 m^2 * .015 m = 25,500 m^3. The mass of the slick is therefore mass = density * volume = 860 kg / m^3 * 24,400 m^3 = 2,193,000 kg. This result should be rounded according to the number of significant figures in the given information.
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RESPONSE --> I get the first part bu where did the 24,400 m^3 come from
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23:04:49 `q011. Summary Question 1: How do we find the surface area of a cylinder?
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RESPONSE --> SA = 2Pi * r^2(h + r)
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23:05:36 The curved surface of the cylinder can be 'unrolled' to form a rectangle whose dimensions are equal to the circumference and the altitude of the cylinder, so the curved surface has volume Acurved = circumference * altitude = 2 pi r * h, where r is the radius and h the altitude. The top and bottom of the cylinder are both circles of radius r, each with resulting area pi r^2. {]The total surface area is therefore Acylinder = 2 pi r h + 2 pi r^2.
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23:06:08 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> 4 * Pi * r^2
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23:06:12 The surface area of a sphere is A = 4 pi r^2, where r is the radius of the sphere.
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23:07:14 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> Density is how much mass is contained in a given unit of volume.
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23:07:19 The average density of an object is its mass per unit of volume, calculated by dividing its total mass by its total volume. If the object is uniform and homogeneous then its density is constant and we can speak of its 'density' as opposed to its 'average density'
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23:08:27 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> We can multiply the mass * density to equal volume.
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23:10:00 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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RESPONSE --> I had these two backwarsd. I guess thats where my errors came from, I have made the necessary corrections.
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23:11:39 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I have had a major refresher when it comes to density, mass, and volume its been a while. I have began to be able to incprporate new ideas into these principles.
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