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course PHY 241
10/17/20113:43 am
By rights this should have been finished last wednesday, I just was focused so much on trying to finish my late work that it got lost in the flood. It's here now, though I had one particular question I could not answer." "Text-related problems:
1. An inch is 2.54 centimeters. How can you use this information along with common knowledge to find the following?
The number of centimeters in a foot.
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A foot is 12 inches. As such, 2.54 cm/inch * 12 inch = 30.48 cm. The inches cancel out.
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The number of feet in a meter.
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One meter = 100 cm. One foot = 30.48 cm thus, 1 foot/30.48 cm * 100cm = 3.281 feet
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The number of meters in a mile.
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One mile = 2675 feet one meter = 3.281 feet thus, 1 meter/3.281 feet * 2675 feet = 815.3 meters
@& A mile isn't 2675 feet.*@
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The number of nanometers in a mil (a mil is 1/1000 of an inch).
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10 nm/cm, 30.48 cm/inch, 1000 mil/inch, cm/mil = 0.00305mil/cm, nm/mil = 3278.69 nm
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2. A cube 10 centimeters on a side would hold 1 liter of water. A cube 1 centimeter on a side would hold 1 milliliter of water. Show how this information along with common knowledge, allows you to answer the following questions:
How many milliliters are in a liter?
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ten times as much volume in a liter as there is in a milliliter. 10^3 = 1000. 1000 ml/liter
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How many milliliters are there in a cubic meter?
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1 ml = 1 cm^3. 1 meter^3 = 100^3 cm = 1,000,000cm^3 thus, 1,000,000 ml/m^3
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How many liters are there in a cubic kilometer?
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1000 ml/liter, 1,000 liters/m^3, 1 km^3 = 1000^3 m = 1,000,000,000m^3 thus, 1,000,000,000,000 liters/km^3
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How many cubic meters are there in a cubic mile?
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1 mile = 1609.344 meters. 1 mile^3 = 1609.344^3 = 4,168,180,000 m^3.
4,168,180,000 meters^3/mile^3
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3. Steel has a density between 7 grams / cm^3 and 8 grams / cm^3. The larger steel balls we use in the lab have diameter 1 inch. Some of the smaller balls have diameter 1/2 inch.
What therefore is the mass of one of the larger balls?
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2.54 cm/inch. Assume 7 grams/cm^3. 7 grams/cm^3 * 2.54 cm = 17.78 grams/cm^2
@& You need the volume of the ball in cm^3 to multiply by the 7 grams / cm^3.
17.78 g / cm^2 is the correct result of your calculation, but your calcuation doesn't give you the mass (as is also indicated by the units of your result).*@
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What would the mass of the smaller ball be as a fraction of the mass of the larger ball?
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½ the mass of the larger ball. Assuming they share the same density.
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@& Half the diameter does not imply half the volume.*@
4. Using common knowledge and the fact that 1 inch = 2.54 centimeters, express a mile/hour in centimeters / second.
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one mile = 185.3 meters. 100cm/meter = 18,530 cm. 60 min/hour. 60 sec/min. 60*60 = 3600. 18,530 cm/3600 seconds = 1853 cm/360 seconds
@& Good, except that a mile is not 185. meters.
Done correctly this comes out around 44 cm/s.*@
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5. Using your measurements of a domino, find the following:
The ratio of its length to its width.
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The ratio of its width to its thickness.
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The volume of a domino.
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The percent uncertainty in your results, according to your estimates of the uncertainty in your measurements.
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6. Estimate how many of the large steel balls would fit into a drinking cup. Then based on your estimate and the fact that the small green BB's in the lab have diameters of 6 millimeters, estimate how many of those BB's would fit into a cup.
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mass of steel ball = 17.78 grams/cm^2 estimated size of drinking cup = . . . ?
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7. Estimate the volume and mass of a single Cheerio. As a point of reference, an average almond has a mass of about a gram.
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Guesstimated dimensions: length = 1.5 cm. width = 1.5 cm. Height = 0.5 cm. Volume = 1.125 cm^3. Guesstimate mass= 0.5 grams.
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If half the mass of the Cheerio consists of carbohydrates, and if a gram of carbohydrate has a food energy of about 4 000 Joules, then what is your estimate of the food energy of a single Cheerio?
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½ 0.5 = 0.25 0.25 * 4000 = 1000. 1000 Joules
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8. Estimate the number of grains of typical desert sand in a liter. Then estimate the number of liters of sand on a 100-meter stretch of your favorite beach.
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Guesstimate number of grains in cm^3 = 100, 1,000,000 cm^3 /m^3. 1 cm/ml. 1000ml/liter. 100 *1000 = 100,000 grains per liter. 1,000,000 cm^3/m^3 * 100 grains/cm^3 = 100,000,000 grains. Liter/100,000 grains * 100,000,000 grains =
1000 liters
@& That estimate is reasonable.
You haven't estimated the number of m^3 of sand on that stretch of beach.
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Compare the number of grains of sand with the number of stars in our galaxy, that number estimated to be about 100 billion.
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100,000 grains/1,000,000,000 stars = 1 grain/10,000 stars
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Compare the number of grains with the number of stars in the universe, which contains over 100 billion galaxies whose average size is about the same as ours.
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1 grain/10,000 stars, 100,000,000,000 galaxies, 1,000,000,000 stars/galaxy, 100,000,000,000,000,000,000 stars * 1 grain/10,000 stars =
1 grain /1,000,000,000,000,000 stars
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9. Water has a density of 1 gram / cm^3.
Using this information how would you reason out the density of water in kilograms / meter^3?
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1000 grams/kg, 100cm/meter, 1000^3 = 1,000,000,000. 100^3 = 1,000,000
1,000,000,000 kg^3/1,000,000 meters^3 = 1000kg^3/meters^3
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10. A rubber ball of diameter 25 cm is dropped on the floor from a height of 1 meter, and bounces back up to a height of 70 cm.
What is the ball's speed when it first contacts the floor, and what is its speed when it first loses contact with the floor on its rebound?
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a = 9.8 m/s^2, distance fallen = one meter, v0 = 0 m/s
@& From that information, with the equations of motion, you can get the velocity of the ball at this instant.*@
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Make a reasonable estimate of how far the center of the ball moves as it compresses before starting its rebound.
What do you think is its average acceleration during its compression?
How long do you think it takes to compress?
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during the split-second when its velocity is zero
@& That's when the compression ends. The compression starts when the ball first contacts the floor.*@
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How much KE does it lose, per gram of its mass, during the compression?
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take the acceleration of gravity, find the force. Then, take the new acceleration on the way up (find by taking time and distance) and find the force of that. The difference is the force lost.
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@& Kinetic energy is not force.
You're thinking along the right lines but need to become more specific, and incorporate the definitions of work and kinetic energy.*@
How much KE does it gain, per gram of its mass, as it decompresses?
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How much momentum does it have, per gram of its mass, just before it first reaches the floor?
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take the mass times the velocity
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How much momentum does it have, per gram of its mass, just after it first leaves the floor on its way up?
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take the mass times the velocity on the way up.
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11. I'm pulling out a parking place on the side of the street, in a pickup truck with mass 1700 kg (including the contents of the truck, which among other things includes me).
I wait for a car to pass before pulling out, then pull out while accelerating at .5 m/s^2. At the instant I pull out, the other car is 20 meters past me and moving at 10 meters / second. If that car's speed and my acceleration both remain constant, then
How long will it take me to match its speed?
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20 seconds
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How far behind will I be at that instant?
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an additional 10 meters, 30 meters total
@& In 20 seconds that car would have traveled 200 meters. My truck would have traveled much less than that, and the difference will be much more than 30 meters.*@
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How much longer will it take me to catch up, and how fast will I be going when I do?
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roughly 10.5 seconds. Roughly 25 meters/second
@& You're going to need to show how you reason this out.*@
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How much work will the net force on my truck have done by the time I catch the other car?
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1700 kg * 0.5 m/second^2 = 850 newtons. 850 * 10 seconds = 8500 Joules
@& 850 N * 10 s is not 8500 Joules. It's 8500 N * s.
This calculation gives you the impulse of the force, not the work done by the force.*@
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If I hit my brakes when I'm 20 meters behind that car, then how much force will be required to slow me down sufficiently that I don't catch up with the car? How does this force compare with the weight of my truck?
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850 newtons. Opposite of acceleration. Roughly half the weight of the truck.
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12. A ball is dropped from rest from a window, and passes another lower window in .32 seconds. That window is 1.4 meters high. From what height was the ball dropped?
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a = 9.8 meters/second^2. 9.8 * .32 = 3.136 meters. 1.4 + 3.136 = 4.536 meters
@& 9.8 m/s^2 * .32 s is 3.1 m/s. This is not a displacement in meters.*@
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13. To maintain a speed of 1 meter / second a swimmer must generate 200 watts of power. The swimmer breathes once every stroke and covers a distance of 2 meters per stroke. To sustain this pace the swimmer must inhale enough air with every stroke to support the production of the necessary energy. How much energy must be produced in for each breath?
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one breath per stroke, two meters per stroke, 200 watts per meter. 400 watts per breath.
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"
@& Good work on many questions, but there are fundamental errors on others.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
*@