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20:26:18 `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 --> A = 10 CM3
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20:38:46 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 MULTIPLY THE AMOUNT BY ITSELF 3 TIMES TO GET THE VOLUME AREA OF A CUBE.
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20:43:58 `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 --> FIRST YOU HAVE TO KNOW HOW MANY CM WOULD EQUAL A METER. I DO N OT KNOW THIS. THEN YOU WOULD DIVIDE THAT BY 10 CM.
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20:44:27 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 --> OK
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20:45:26 `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 --> SAME METHOD DIFFERENT NUMBERS
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20:45:50 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 --> OK
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20:46:34 `q004. How many cubic centimeters are there in a liter?
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RESPONSE --> YES.... IT MUST BE A LOT YET I DON'T KNOW
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20:46:59 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 --> OK
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20:48:03 `q005. How many liters are there in a cubic meter?
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RESPONSE --> 1 I AM GUESSING
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20:58:13 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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20:58:43 `q006. How many cm^3 are there in a cubic meter?
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RESPONSE --> I DON'T KNOW
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20:59:05 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 --> OK
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20:59:56 `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 --> METER MINUS KG
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21:00:21 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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21:01:19 `q008. What is the mass of a cubic km of water?
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RESPONSE --> 1000 LITERS
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21:01:36 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 --> OK
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21:03:48 `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 --> CONVERT CUBIC KILOMETERS INTO LITERS AND DIVIDE
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21:03:59 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 --> OK
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21:04:26 `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 --> I DO NOT KNOW
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21:04:39 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 --> OK
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21:04:48 `q011. Summary Question 1: How can we visualize the number of cubic centimeters in a liter?
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RESPONSE --> HOW
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21:04:58 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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21:05:35 `q012. Summary Question 2: How can we visualize the number of liters in a cubic meter?
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RESPONSE --> OK
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21:05:59 06-24-2005 21:05:59 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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NOTES -------> OK
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21:06:03 06-24-2005 21:06:03 `q013. Summary Question 3: How can we calculate the number of cubic centimeters in a cubic meter?
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NOTES ------->
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21:06:06 06-24-2005 21:06:06 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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NOTES ------->
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21:06:09 06-24-2005 21:06:09 `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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NOTES ------->
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21:06:13 06-24-2005 21:06:13 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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NOTES ------->
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21:06:16 06-24-2005 21:06:16 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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NOTES ------->
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21:06:29 This ends the fourth assignment.
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RESPONSE --> OK
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__________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com wHXԱ˧Riz assignment #008 핆{lݘϱ˵xPF Physics I Class Notes 07-09-2005
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21:41:44 07-09-2005 21:41:44 What is the meaning of the x intercept of the graph of force vs. slope?
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NOTES ------->
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21:42:03 07-09-2005 21:42:03 ** On this graph the x axis represents slope, the y axis represents acceleration. The x axis point occurs when y = 0, so this point represents a slope for which acceleration is zero. Assuming that 'down the incline' is the positive direction, everywhere to the right of this point the acceleration is positive and everywhere to the left is negative. The x intercept therefore represents the maximum slope for which friction is great enough to prevent the ball from accelerating down the slope. Since friction is resisting the weight component down the incline, the x intercept represents the slope when friction and the weight component tending to accelerate the cart down the slope are exactly in balance, equal and opposite. **
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NOTES ------->
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21:42:08 07-09-2005 21:42:08 When we obtain a linear relationship between force and acceleration, is it plausible that the constant term in the equation is, within experimental uncertainty, zero?
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NOTES ------->
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21:42:11 07-09-2005 21:42:11 ** The constant term in the equation will be zero only if the graph passes through the origin. This would imply that any nonzero force will produce a nonzero acceleration, and that for example there is either no friction present or friction has been compensated for in some way. **
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NOTES ------->
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21:42:16 07-09-2005 21:42:16 For the force vs. acceleration relationship for the car, why should we expect that the acceleration corresponding to a net force equal to the car's weight is the acceleration of gravity?
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NOTES ------->
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21:42:19 07-09-2005 21:42:19 ** When the car is dropped and allowed to fall freely the net force acting on it is its weight, which is the force exerted on its mass by gravity. So the acceleration will be that of gravity. **
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NOTES ------->
.................................................bpwȲƪaJzIf
assignment #008 핆{lݘϱ˵xPF Physics I Class Notes 07-09-2005 ֎ְkDg assignment #009 핆{lݘϱ˵xPF Physics I Class Notes 07-09-2005
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21:47:37 07-09-2005 21:47:37 Do the data seem to indicate, within the limits of the errors inherent in the experiment, that acceleration is the same on a constant incline regardless of where the cart is or how fast it is going?
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NOTES ------->
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21:47:40 07-09-2005 21:47:40 ** The usual results do show that the acceleration is the same, within about 10%, for any section of the ramp. Since different sections can have different lengths, which are associated with different average and final velocities, this indicates that within these limits the acceleration does appear to be constant. **
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NOTES ------->
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21:47:43 07-09-2005 21:47:43 Where does the force that accelerates the cart come from?
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NOTES ------->
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21:47:47 07-09-2005 21:47:47 ** The force is the result of the gravitational force exerted on the cart by the Earth. On an incline a component of this force acts in the direction down the incline. The greater the incline the greater the proportion of the total gravitational force that acts down the incline. **
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NOTES ------->
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21:47:49 07-09-2005 21:47:49 List all the forces that act on the cart, and discuss how they affect its motion.
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NOTES ------->
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21:47:52 07-09-2005 21:47:52 ** gravity pulls straight down, but the ramp isn't able to push straight up. That leaves a component of the gravitational force parallel to the ramp; the greater the slope the greater this component. Friction opposes motion. If there is a force holding the car back, then it needs to be included in the list, as it is in your list. **
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NOTES ------->
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21:47:55 07-09-2005 21:47:55 Explain why a `ds is equal to the change in 1/2 v^2.
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NOTES ------->
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21:47:57 07-09-2005 21:47:57 ** By the fourth equation of motion vf^2 = v0^2 + 2 a `ds. Thus a `ds = (vf^2 - v0^2) / 2. Since vf^2 - v0^2 is the change in v^2, it follows that a `ds is half the change in v^2. So a `ds is proportional to v^2. **
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NOTES ------->
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21:48:00 07-09-2005 21:48:00 Explain in detail why we expect Fnet `ds to be proportional to the change in v^2.
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NOTES ------->
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21:48:03 07-09-2005 21:48:03 ** a `ds is proportional to the change in v^2 (see the preceding question), and Fnet * m = a. So the change in Fnet * `ds is the change in m a * `ds, which for a given object is proportional to the change in v^2. **
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NOTES ------->
.................................................|ZCBaψ
assignment #008 핆{lݘϱ˵xPF Physics I Class Notes 07-09-2005 ڼٔ assignment #009 핆{lݘϱ˵xPF Physics I Class Notes 07-09-2005 VzwËMy assignment #009 핆{lݘϱ˵xPF Physics I Class Notes 07-09-2005
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21:55:58 07-09-2005 21:55:58 Do the data seem to indicate, within the limits of the errors inherent in the experiment, that acceleration is the same on a constant incline regardless of where the cart is or how fast it is going?
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NOTES ------->
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21:56:01 07-09-2005 21:56:01 ** The usual results do show that the acceleration is the same, within about 10%, for any section of the ramp. Since different sections can have different lengths, which are associated with different average and final velocities, this indicates that within these limits the acceleration does appear to be constant. **
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NOTES ------->
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21:56:04 07-09-2005 21:56:04 Where does the force that accelerates the cart come from?
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NOTES ------->
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21:56:06 07-09-2005 21:56:06 ** The force is the result of the gravitational force exerted on the cart by the Earth. On an incline a component of this force acts in the direction down the incline. The greater the incline the greater the proportion of the total gravitational force that acts down the incline. **
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NOTES ------->
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21:56:09 07-09-2005 21:56:09 List all the forces that act on the cart, and discuss how they affect its motion.
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NOTES ------->
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21:56:12 07-09-2005 21:56:12 ** gravity pulls straight down, but the ramp isn't able to push straight up. That leaves a component of the gravitational force parallel to the ramp; the greater the slope the greater this component. Friction opposes motion. If there is a force holding the car back, then it needs to be included in the list, as it is in your list. **
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NOTES ------->
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21:56:15 07-09-2005 21:56:15 Explain why a `ds is equal to the change in 1/2 v^2.
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NOTES ------->
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21:56:17 07-09-2005 21:56:17 ** By the fourth equation of motion vf^2 = v0^2 + 2 a `ds. Thus a `ds = (vf^2 - v0^2) / 2. Since vf^2 - v0^2 is the change in v^2, it follows that a `ds is half the change in v^2. So a `ds is proportional to v^2. **
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NOTES ------->
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21:56:20 07-09-2005 21:56:20 Explain in detail why we expect Fnet `ds to be proportional to the change in v^2.
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NOTES ------->
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21:56:22 07-09-2005 21:56:22 ** a `ds is proportional to the change in v^2 (see the preceding question), and Fnet * m = a. So the change in Fnet * `ds is the change in m a * `ds, which for a given object is proportional to the change in v^2. **
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NOTES ------->
.................................................ȋcB ߤTÂ
assignment #009 핆{lݘϱ˵xPF Physics I Class Notes 07-09-2005 rJ~ assignment #009 핆{lݘϱ˵xPF Physics I Class Notes 07-09-2005
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21:58:49 07-09-2005 21:58:49 Do the data seem to indicate, within the limits of the errors inherent in the experiment, that acceleration is the same on a constant incline regardless of where the cart is or how fast it is going?
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NOTES ------->
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