course Phy 121 HٶYvStudent Name:
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20:36:08 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 --> average rate=change in position/change in clock time 80/4=20 m/s
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20:36:42 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 --> ok
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20:39:13 `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 --> Yes because it would have a higher rate of acceleration.
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20:40:32 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 --> ok that makes since when you account factors such as weight and gearing.
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20:43:31 `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 --> it is calculated by change in speed/change in clock time. change in speed will be m/s and change in clock time will be seconds. So therefore you get meter/second/second
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20:44:20 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 --> ok! I think I was just trying to explain to many things when the problem was a lot less complicated then i was making it.
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20:45:28 `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 --> m/s2
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20:46:04 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 --> I understood that one!
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20:48:26 `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 --> 5(m/s)/5s=1m
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20:49:13 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 --> I subtracted rather then doing the distance right, so therefore it threw off the whole problem.
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20:51:31 `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 --> vAve=dv/dt
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20:52:36 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 --> Yay!! haha except I should have put aAve instead of ave
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20:57:46 `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 --> (3/2)m/s
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20:58:31 `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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. 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. ok (3/2)
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20:58:46 `q006b. What therefore is the average rate at which the velocity is changing during this time interval?
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. 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. (3/2)
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20:59:40 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 --> ok. I think i just left out m/s^2
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21:04:57 `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 seconds and it represents clock tome The rise is meters/sec and it represents the velocity The slope is rise/run = 3/2
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21:05:33 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 --> I keep forgetting to use the correct units.
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21:07:24 `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 is rise/run which would be velocity divided by clock time. A greater slope represents greater acceleration.
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21:07:46 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 --> ok
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21:10:05 `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 --> At the beginning the graphy would be increasing at a constant rate. After it attains a certain velocity it will decrease after the air is factored in.
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21:10:28 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 --> OK
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21:11:28 `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 is the same problem as the previous problem.
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21:11:59 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 --> This question is repetitive
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jγȘ Student Name: assignment #004 004. Units of volume measure
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23:05:31 `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 --> i dont know.
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23:06:33 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 I understand now that you just have to multiply it out.
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23:07:37 `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 --> 10 cubes
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23:08:26 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 --> I only did part of the question
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23:10:28 `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 --> 1,000*1,000=1,000,000
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23:10:53 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 --> yay!
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23:11:34 `q004. How many cubic centimeters are there in a liter?
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RESPONSE --> 1000
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23:12:14 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 --> yay!
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23:13:10 `q005. How many liters are there in a cubic meter?
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RESPONSE --> 1,000
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23:13:29 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 --> ok
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23:14:41 `q006. How many cm^3 are there in a cubic meter?
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RESPONSE --> 1,000*1,000=1,000,000
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23:15:03 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 --> yay!
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23:16:45 `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 --> 1,000 kg
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23:17:07 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. I understand now.
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23:17:57 `q008. What is the mass of a cubic km of water?
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RESPONSE --> I don't know.
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23:18:43 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 --> I wasn't for sure how to do this one so therefore I just put in I do not know. Now i understand what you did.
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23:19:35 `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 --> 5,000
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23:20:15 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 --> I dont really understand this one.
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23:22:37 `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 --> 6400/2,000=3.2
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23:23:21 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 --> I was using the wrong formula.
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23:24:21 `q011. Summary Question 1: How can we visualize the number of cubic centimeters in a liter?
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RESPONSE --> one liter is equal to 1000 cubic centimeters.
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23:24:43 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 --> yay!
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23:26:12 `q012. Summary Question 2: How can we visualize the number of liters in a cubic meter?
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RESPONSE --> one liter is equal to 1000 cubic centimeters in volume and 0.01 liter is equal to 1 centiliter
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23:26:37 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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23:27:25 `q013. Summary Question 3: How can we calculate the number of cubic centimeters in a cubic meter?
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RESPONSE --> multiply the cubic centimeters by 1000
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23:28:06 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 --> its 1000*1000. Not just a 1000
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23:28:32 `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 --> There are
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23:28:59 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 --> tricky question
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23:29:30 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> By writing down the notes in my notebook and printing the memorizing idea page and synopsis page
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23:29:36 This ends the fourth assignment.
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RESPONSE --> ok!
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