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course Phy 231
10/23/2011 @ 7:56p.m.
Physics I Class 111005You should use your text as a reference in solving the following, which are due next Wednesday:
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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30.48cm
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The number of feet in a meter.
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3.28ft
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The number of meters in a mile.
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1609m
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The number of nanometers in a mil (a mil is 1/1000 of an inch).
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2538nm
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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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1000ml
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How many milliliters are there in a cubic meter?
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1 x 10^6 ml
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How many liters are there in a cubic kilometer?
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1,000,000l
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How many cubic meters are there in a cubic mile?
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4 x 10^9 m^3
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@& You haven't shown how you used the given information to get these results.*@
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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64.35g
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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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1/8th
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4. Using common knowledge and the fact that 1 inch = 2.54 centimeters, express a mile/hour in centimeters / second.
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44.7cm/sec
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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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2:1
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The ratio of its width to its thickness.
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2.5:1
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The volume of a domino.
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12.5cm^3
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The percent uncertainty in your results, according to your estimates of the uncertainty in your measurements.
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10% on thickness, 4% on width, 2% on length
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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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18 steel balls, 540 BBs
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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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Mass = 1/4g, V = 2cm^3.
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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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500J (seems too large…)
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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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A million grains/liter, and maybe 1 x 10^22 grains/100m of beach
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@& That would imply 10^16 liters, which would be 10^13 m^3. If the depth of the sand was equal to the diameter of the Earth, and the beach was 100 m wide as well as long, you would still come up fa short of 10^13 m^3 of sand.*@
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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2x the number of grains as stars in the galaxy
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@& 10^22 is not double 100 billion.
10^22 is the square of 100 billion.
If you double 100 billion you get 200 billion.*@
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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.9x the number of grains as 100billion galaxies (seems wrong)
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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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Divide by 100,000 (This is dividing by 1g / 1000kg and 1cm/100m)
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@& Cm and m are linear units.
cm^3 and m^3 are not in the same ratio as cm and m.*@
10. A rubber ball of diameter 2.5 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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443cm/sec (used vf^2 = v0^2 + 2a’ds)
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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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Half a centimeter, -490cm/sec^2, .1sec
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@& In .1 sec at 443 cm/s, the ball would move 44 cm, way greater than its diameter.
Its average velocity during compression would be less than 443 cm/s, but .1 sec would still give the top of the ball plenty of time to pass through the bottom and several cm into the floor.*@
How much KE does it lose, per gram of its mass, during the compression?
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‘dKE = -98,000g*cm^2/sec^2
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How much KE does it gain, per gram of its mass, as it decompresses?
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It seems like it would compress and decelerate to 0cm/sec, which would mean all of the compression is stored as PE which would mean the ‘dKE is just a bit more than the PE since some is lost to various non-conservative forces.
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@& You can figure out how fast it's moving when it leaves the floor.*@
How much momentum does it have, per gram of its mass, just before it first reaches the floor?
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443 g*cm/sec
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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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370g*cm/sec
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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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20sec
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How far behind will I be at that instant?
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120m
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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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22sec, 21m/sec
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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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F = 850N (1700kg * .5m/sec^2) W = F*’ds = 850N * 441m = 374,850J
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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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You hit your breaks at T = 40sec, so you need a deceleration of -5m/sec^2, which is -8500N, which is 5x the mass of your truck.
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@& The mass of my truck is 1400 kg.
8500 N / (1400 kg) = 6 N / kg = 6 m/s^2, roughly.
Newtons don't measure the same thing as kg so your ratio isn't properly constructed.*@
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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1.4m = v0(.32sec) + ½(9.8m/sec^2)(.32sec)
V0 = 2.8m/sec, so vf = 5.95m/sec.
vAve from drop to top of window is 1.4m/sec, which means ‘dt = .29sec, which means ‘ds = .4m.
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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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B = breath.
1b/stroke
2m/stroke
1b/2m
1b/2sec
.5b/sec
400 watts/breath
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@& Pretty good on a number of questions. You missed the boat on a few, though, so be sure to check my notes.*@