course Phy 201
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10:11:14 query Note that there are 10 questions in this assignment. At a certain instant the speedometer of a car reads 5 meters / second (of course cars in this country generally read speeds in miles per hour and km per hour, not meters / sec; but they could easily have their faces re-painted to read in meters/second, and we assume that this speedometer has been similarly altered). Exactly 4 seconds later the speedometer reads 25 meters/second (that, incidentally, indicates very good acceleration, as you will understand later). At what average rate is the speed of the car changing with respect to clock time?
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RESPONSE --> The average rate of the car is is changing in speed from 25 m/sec - 5 me/sec = 20m/sec aVe= 4sec/ (20 m/sec) aVe = 5 m/ sec
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10:11:23 The rate of change of the speed with respect clock time is equal to the change in the speed divided by the change in the clock time. So we must ask, what is the change in the speed, what is the change in the clock time and what therefore is the rate at which the speed is changing with respect to clock time? The change in speed from 5 meters/second to 25 meters/second is 20 meters/second. This occurs in a time interval lasting 4 seconds. The average rate of change of the speed is therefore (20 meters/second)/(4 seconds) = 5 meters / second / second. This means that on the average, per second, the speed changes by 5 meters/second.
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RESPONSE --> Correct
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10:12:47 `q002. Explain in commonsense terms of the significance for an automobile of the rate at which its velocity changes. Do you think that a car with a more powerful engine would be capable of a greater rate of velocity change?
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RESPONSE --> When the velocity changes the car goes either faster or slower over a certain distance with the same amount of time. A car with a more powerful engine would be capable of a greater rate of velocity change in a smaller time frame then an old car with little pickup.
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10:12:59 A car whose velocity changes more rapidly will attain a given speed in a shorter time, and will be able to 'pull away from' a car which is capable of only a lesser rate of change in velocity. A more powerful engine, all other factors (e.g., weight and gearing) being equal, would be capable of a greater change in velocity in a given time interval.
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RESPONSE --> Correct
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10:14:30 `q003. Explain how we obtain the units meters / second / second in our calculation of the rate of change of the car's speed.
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RESPONSE --> we obtain units meters/second/second as AVe = seconds (diffrence between initial and final time) / (meters/sec or the distance travelled).
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10:14:38 When we divide the change in velocity, expressed in meters/second, by the duration of the time interval in seconds, we get units of (meters / second) / second, often written meters / second / second.
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RESPONSE --> correct
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10:15:48 `q004. The unit (meters / second) / second is actually a complex fraction, having a numerator which is itself a fraction. Such a fraction can be simplified by multiplying the numerator by the reciprocal of the denominator. We thus get (meters / second) * (1/ second). What do we get when we multiply these two fractions?
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RESPONSE --> when we multiply these two fractions (meters / second) * (1/ second) our answer is meters/ sconds^2.
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10:17:09 Multiplying the numerators we get meters * 1; multiplying the denominators we get second * second, which is second^2. Our result is therefore meters * 1 / second^2, or just meters / second^2. If appropriate you may at this point comment on your understanding of the units of the rate of change of velocity.
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RESPONSE --> correct
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10:20:25 `q004. If the velocity of an object changes from 10 m/s to -5 m/s during a time interval of 5 seconds, then at what average rate is the velocity changing?
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RESPONSE --> If the velocity of an object changes from 10 m/s to -5 m/s during a time interval of 5 seconds, -5 m/s - 10 m/s = 15 m/s aVe = 5 sec/ 15 m/s aVe = 3 m
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10:20:31 We see that the velocity changes from 10 meters/second to -5 meters/second, a change of -15 meters / second, during a five-second time interval. A change of -15 m/s during a 5 s time interval implies an average rate of -15 m/s / (5 s) = -3 (m/s)/ s = -3 m/s^2. This is the same as (-3 m/s) / s, as we saw above. So the velocity is changing by -3 m/s every second.
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RESPONSE --> correct
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10:23:57 `q005. You should have noted that velocity can be positive or negative, as can the change in velocity or the rate at which velocity changes. The average rate at which a quantity changes with respect to time over a given time interval is equal to the change in the quantity divided by the duration of the time interval. In this case we are calculating the average rate at which the velocity changes. If v represents velocity then we we use `dv to represent the change in velocity and `dt to represent the duration of the time interval. What expression do we therefore use to express the average rate at which the velocity changes?
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RESPONSE --> Dv = vf - v0 is the chance in velocity
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10:25:49 The average rate would be expressed by [ave rate of velocity change with respect to clock time] = `dv / `dt. The expression [ave rate of velocity change with respect to clock time] is pretty cumbersome so we give it a name. The name we give it is 'average acceleration', abbreviated by the letter aAve. Using a to represent acceleration, write down the definition of average acceleration. The definition of average acceleration is aAve = `dv / `dt. Please make any comments you feel appropriate about your understanding of the process so far.
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RESPONSE --> The definition of average acceleration is aAve = `dv / `dt, what i had originalyl caluculated was the change in velocity regarding accelerated motion.
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10:29:21 `q006. If a runner is moving at 6 meters / sec at clock time t = 1.5 sec after starting a race, and at 9 meters / sec at clock time t = 3.5 sec after starting, then what is the average acceleration of the runner between these two clock times?
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RESPONSE --> average acceleration = average rate of velocity change = change in velocity / change in clock time Change in velocity is 'dv = 9 m - 6m = 3m Change in time is 'dt = 3.5 sec - 1.5 sec= 2 sec aVe = 'dv/'dt aVe = 3m/2sec aVe = 1.5 m/sec
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10:31:41 `q006a. What is the change `dv in the velocity of the runner during the time interval, and what is the change `dt in clock time during this interval?
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RESPONSE --> I do not understand what this question is asking.
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10:32:22 `q006b. What therefore is the average rate at which the velocity is changing during this time interval?
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RESPONSE --> The average rate at which velocity is changing during the time interval is the change in quantity/change in clock time
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10:32:30 We see that the runner's velocity changes from 6 meters/second to 9 meters/second, a change of `dv = 9 m/s - 6 m/s = 3 m/s, during a time interval their runs from t = 1.5 sec to t = 3.5 sec so that the duration of the interval is `dt = 3.5 s - 1.5 s = 2.0 s. The rate at which the velocity changes is therefore 3 m/s / (2.0 s) = 1.5 m/s^2. Please comment if you wish on your understanding of the problem at this point.
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RESPONSE --> Correct
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10:34:50 `q007. On a graph of velocity vs. clock time, we can represent the two events of this problem by two points on the graph. The first point will be (1.5 sec, 6 meters/second) and the second point will be (3.5 sec, 9 meters / sec). What is the run between these points and what does it represent? What is the rise between these points what does it represent? What is the slope between these points what does it represent?
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RESPONSE --> The run is the clock time of 2 seconds while the rise is the change in velocity of 3 meters/sec. The slope is rise/run = 2 sec/ 3m/sec = 1.5 m/sec^2
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10:34:58 The rise from the first point to the second is from 6 meters/second to 9 meters/second, or 3 m/s. This represents the change `dv in velocity. The run is from 1.5 seconds to 3.5 seconds, or 2 seconds, and represents the change `dt in clock time. The slope, being the rise divided by the run, is 3 m/s / (2 sec) = 1.5 m/s^2. This slope represents `dv / `dt, which is the average acceleration during the time interval. You may if you wish comment on your understanding to this point.
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RESPONSE --> correct
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10:37:57 `q008. In what sense does the slope of any graph of velocity vs. clock time represent the acceleration of the object? For example, why does a greater slope imply greater acceleration?
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RESPONSE --> the slope of any graph of velocity vs. clock time represents the acceleration of the object because it determins the change of rate by calucltaing change of velocity / change in clock time. A greater slope implies that the car accelerates and goes over a greater distance over a shorter period of time.
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10:38:04 Since the rise between two points on a graph of velocity vs. clock time represents the change in `dv velocity, and since the run represents the change `dt clock time, the slope represents rise / run, or change in velocity /change in clock time, or `dv / `dt. This is the definition of average acceleration.
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RESPONSE --> correct
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10:42:04 `q009. This is the same situation as in the preceding problem: An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of velocity vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph).
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RESPONSE --> In the graph, the x-axis would be clock time while the y-axis would be velocity. The run would be on the x-axis and the rise on the y-axis. The car would be increasing at a decreasing rate due to the resistance experienced. The slope would be decreasing.
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10:43:02 Your graph should have velocity as the vertical axis and clock time as the horizontal axis. The graph should be increasing since the velocity starts at zero and increases. At first the graph should be increasing at a constant rate, because the velocity is increasing at a constant rate. The graph should continue increasing by after a time it should begin increasing at a decreasing rate, since the rate at which the velocity changes begins decreasing due to air resistance. However the graph should never decrease, although as air resistance gets greater and greater the graph might come closer and closer to leveling off. Critique your solution by describing or insights you had or insights you had and by explaining how you now know how to avoid those errors.
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RESPONSE --> I was incorrect in that the graph should never decrease, although as air resistance gets greater and greater the graph might come closer and closer to leveling off. I will now be able to apply what air resistance would do in repsonse to changing vlocities.
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10:50:07 `q010. An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of acceleration vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph).
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RESPONSE --> This graph would be increasing as velocity starts at zero. Velcity increases at a constant rate and then increase at a decreasing rate as air resistance is experienced. The graph will not decrease however as the velocity continues to increase.
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10:50:14 Your graph should have acceleration as the vertical axis and clock time as the horizontal axis. At first the graph should be neither increasing nor decreasing, since it first the acceleration is constant. Then after a time the graph should begin decreasing, which indicates the decreasing rate at which velocity changes as air resistance begins having an effect. An accurate description of whether the graph decreases at a constant, increasing or decreasing rate is not required at this point, because the reasoning is somewhat complex and requires knowledge you are not expected to possess at this point. However it is noted that the graph will at first decrease more and more rapidly, and then less and less rapidly as it approaches the t axis. In answer to the following question posed at this point by a student: Can you clarify some more the differences in acceleration and velocity? ** Velocity is the rate at which position changes and the standard units are cm/sec or m/sec. Acceleration is the rate at which velocity changes and its standard units are cm/s^2 or m/s^2. Velocity is the slope of a position vs. clock time graph. Acceleration is the slope of a velocity vs. clock time graph. **
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RESPONSE --> Ok
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³î¹ñ€ù¿…±ÒÓòèþÖ˜‰]–í]‘帬ú Student Name: assignment #004 004. Units of volume measure
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10:56:32 `q001. There are 10 questions and 5 summary questions in this assignment. How many cubic centimeters of fluid would require to fill a cubic container 10 cm on a side?
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RESPONSE --> 10 cubic centimeters would need 1 cubicliter of fluid
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11:02:13 The volume of the container is 10 cm * 10 cm * 10 cm = 1000 cm^3. So it would take 1000 cubic centimeters of fluid to fill the container.
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RESPONSE --> Ok
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11:03:48 `q002. How many cubes each 10 cm on a side would it take to build a solid cube one meter on a side?
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RESPONSE --> To build a solid cube one meter on a side would need 10 cm * 10 cm *10 cm= 1000cm^3
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11:04:17 It takes ten 10 cm cubes laid side by side to make a row 1 meter long or a tower 1 meter high. It should therefore be clear that the large cube could be built using 10 layers, each consisting of 10 rows of 10 small cubes. This would require 10 * 10 * 10 = 1000 of the smaller cubes.
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RESPONSE --> correct- but would the units be ^3?
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11:06:17 `q003. How many square tiles each one meter on each side would it take to cover a square one km on the side?
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RESPONSE --> a kilometer is 1000 meters
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11:06:43 It takes 1000 meters to make a kilometer (km). To cover a square 1 km on a side would take 1000 rows each with 1000 such tiles to cover 1 square km. It therefore would take 1000 * 1000 = 1,000,000 squares each 1 m on a side to cover a square one km on a side. We can also calculate this formally. Since 1 km = 1000 meters, a square km is (1 km)^2 = (1000 m)^2 = 1,000,000 m^2.
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RESPONSE --> It therefore would take 1000 * 1000 = 1,000,000 squares each 1 m on a side to cover a square one km on a side.
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11:08:04 `q004. How many cubic centimeters are there in a liter?
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RESPONSE --> 1 inch = approx 2.54 centimeters. Therefore 1 cubic inch = approx. 16 cubic centimeters 1000 cubic centimeters/16 cubic centimeters = approx 62 cubic inches in one liter.
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11:08:46 A liter is the volume of a cube 10 cm on a side. Such a cube has volume 10 cm * 10 cm * 10 cm = 1000 cm^3. There are thus 1000 cubic centimeters in a liter.
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RESPONSE --> 1 inch = approx 2.54 centimeters. Therefore 1 cubic inch = approx. 16 cubic centimeters 1000 cubic centimeters/16 cubic centimeters = approx 62 cubic inches in one liter could have been solved in a much easier fashionh by combining 10 cm * 10 cm * 10 cm = 1000 cm^3
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11:09:34 `q005. How many liters are there in a cubic meter?
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RESPONSE --> A cubic meter holds 1,000 liters
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11:09:41 A liter is the volume of a cube 10 cm on a side. It would take 10 layers each of 10 rows each of 10 such cubes to fill a cube 1 meter on a side. There are thus 10 * 10 * 10 = 1000 liters in a cubic meter.
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RESPONSE --> correct
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11:11:49 `q006. How many cm^3 are there in a cubic meter?
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RESPONSE --> There are three centimeter measurements, one for each of the three sides of the cube. 1.00 m3 = 1.00 x 106 cm3
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11:12:34 There are 1000 cm^3 in a liter and 1000 liters in a m^3, so there are 1000 * 1000 = 1,000,000 cm^3 in a m^3. It's important to understand the 'chain' of units in the previous problem, from cm^3 to liters to m^3. However another way to get the desired result is also important: There are 100 cm in a meter, so 1 m^3 = (1 m)^3 = (100 cm)^3 = 1,000,000 cm^3.
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RESPONSE --> I am miscaluclating where i am not interpreting the chains of units well. this is something i need to go back and work on
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11:15:09 `q007. If a liter of water has a mass of 1 kg the what is the mass of a cubic meter of water?
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RESPONSE --> If one milliliter of water has a mass of one gram, then one liter of water should have a mass of 1000 grams or one kilogram. It is 1000 milliliters or one liter.
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11:15:16 Since there are 1000 liters in a cubic meter, the mass of a cubic meter of water will be 1000 kg. This is a little over a ton.
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RESPONSE --> Ok
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11:17:32 `q008. What is the mass of a cubic km of water?
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RESPONSE --> the density of water is 1000 kilograms per cubic meter and so the mass of a cubickm of water is 10 * 10* 10= 1000 km
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11:17:55 A cubic meter of water has a mass of 1000 kg. A cubic km is (1000 m)^3 = 1,000,000,000 m^3, so a cubic km will have a mass of 1,000,000,000 m^3 * 1000 kg / m^3 = 1,000,000,000,000 kg. In scientific notation we would say that 1 m^3 has a mass of 10^3 kg, a cubic km is (10^3 m)^3 = 10^9 m^3, so a cubic km has mass (10^9 m^3) * 1000 kg / m^3 = 10^12 kg.
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RESPONSE --> 1 m^3 has a mass of 10^3 kg, a cubic km is (10^3 m)^3 = 10^9 m^3, so a cubic km has mass (10^9 m^3) * 1000 kg / m^3 = 10^12 kg would be the porper caluclation
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13:38:44 `q009. If each of 5 billion people drink two liters of water per day then how long would it take these people to drink a cubic km of water?
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RESPONSE --> If each of 5 billion people drink two liters of water per day then how long would it take these people to drink a cubic km of water with 5,000,000,000 * 2 ml * 365 = 3.65 * 10^12
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13:39:44 5 billion people drinking 2 liters per day would consume 10 billion, or 10,000,000,000, or 10^10 liters per day. A cubic km is (10^3 m)^3 = 10^9 m^3 and each m^3 is 1000 liters, so a cubic km is 10^9 m^3 * 10^3 liters / m^3 = 10^12 liters, or 1,000,000,000,000 liters. At 10^10 liters per day the time required to consume a cubic km would be time to consume 1 km^3 = 10^12 liters / (10^10 liters / day) = 10^2 days, or 100 days. This calculation could also be written out: 1,000,000,000,000 liters / (10,000,000,000 liters / day) = 100 days.
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RESPONSE --> A cubic km is (10^3 m)^3 = 10^9 m^3 and each m^3 is 1000 liters, so a cubic km is 10^9 m^3 * 10^3 liters / m^3 = 10^12 liters, with the time consumed I shoudl not have multiplied the 365 days.
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13:50:56 `q010. The radius of the Earth is approximately 6400 kilometers. What is the surface area of the Earth? If the surface of the Earth was covered to a depth of 2 km with water that what would be the approximate volume of all this water?
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RESPONSE --> Sphere Volume = 4/3 * p * r^3 Sphere volume = 4/3 * 3.14 * 6400^3 sphere volume = 1.09 1.09 * 2 = 2.18 km^3
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13:51:21 The surface area would be A = 4 pi r^2 = 4 pi ( 6400 km)^2 = 510,000,000 km^2. A flat area of 510,000,000 km^2 covered to a depth of 2 km would indicate a volume of V = A * h = 510,000,000 km^2 * 2 km = 1,020,000,000 km^3. However the Earth's surface is curved, not flat. The outside of the 2 km covering of water would have a radius 2 km greater than that of the Earth, and therefore a greater surface area. However a difference of 2 km in 6400 km will change the area by only a fraction of one percent, so the rounded result 1,020,000,000,000 km^3 would still be accurate.
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RESPONSE --> V = A * h would have been a more appropirate formula for this question
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13:54:01 `q011. Summary Question 1: How can we visualize the number of cubic centimeters in a liter?
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RESPONSE --> we can visualize the number of centimeters in a liter by taking 1000 cubic centimeters of fluid to fill the container
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13:55:25 Since a liter is a cube 10 cm on a side, we visualize 10 layers each of 10 rows each of 10 one-centimeter cubes, for a total of 1000 1-cm cubes. There are 1000 cubic cm in a liter.
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RESPONSE --> Ok
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13:55:47 `q012. Summary Question 2: How can we visualize the number of liters in a cubic meter?
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RESPONSE --> It takes ten 10 cm cubes laid side by side to make a row 1 meter long or a tower 1 meter high
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13:55:50 Since a liter is a cube 10 cm on a side, we need 10 such cubes to span 1 meter. So we visualize 10 layers each of 10 rows each of 10 ten-centimeter cubes, for a total of 1000 10-cm cubes. Again each 10-cm cube is a liter, so there are 1000 liters in a cubic meter.
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RESPONSE --> Ok
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13:56:28 `q013. Summary Question 3: How can we calculate the number of cubic centimeters in a cubic meter?
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RESPONSE --> It takes ten 10 cm cubes laid side by side to make a row 1 meter long or a tower 1 meter high. Severazl smaller cubes are needed to make the larger whole
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13:56:39 One way is to know that there are 1000 liters in a cubic meters, and 1000 cubic centimeters in a cubic meter, giving us 1000 * 1000 = 1,000,000 cubic centimeters in a cubic meter. Another is to know that it takes 100 cm to make a meter, so that a cubic meter is (100 cm)^3 = 1,000,000 cm^3.
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RESPONSE --> Ok
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13:58:10 `q014. Summary Question 4: There are 1000 meters in a kilometer. So why aren't there 1000 cubic meters in a cubic kilometer? Or are there?
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RESPONSE --> To cover a square 1 km on a side would take 1000 rows each with 1000 such tiles to cover 1 square kilometer. If this is cubed than 1000 * 1000 * 1000 = 1,000,000,000 km not equalling out to a cubic kilometer
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13:58:20 A cubic kilometer is a cube 1000 meters on a side, which would require 1000 layers each of 1000 rows each of 1000 one-meter cubes to fill. So there are 1000 * 1000 * 1000 = 1,000,000,000 cubic meters in a cubic kilometer. Alternatively, (1 km)^3 = (10^3 m)^3 = 10^9 m^3, not 1000 m^3.
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RESPONSE --> Ok
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13:58:50 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I organized the knowledge of the principles of the assignment by taking note of the theories I did not understand and I will address them accordingly.
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13:58:55 This ends the fourth assignment.
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RESPONSE --> Ok
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