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course Phy 201
Oct 23 11:45I'm sorry this is so late. I've had it worked out in the packet that I printed for a while now. I've just had the hardest time remembering to submit it.
The next packet is coming tomorrow and then I should be back up to speed. Finally.
Thanks Mr. Smith!" "Physics I Class 111005
You 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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1 inch = 2.54 cm
1 ft = 12 inches = 30.48 cm
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The number of feet in a meter.
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1 m = 100 cm
100 cm = 39.37 in
39.37 in = 3.3 ft
divide 39.37 by 12
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The number of meters in a mile.
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1 mile = 5280 ft = 1600 m
divide 5280 by 3.3
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The number of nanometers in a mil (a mil is 1/1000 of an inch).
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1/1000 of an inch = 25 400 nanometers
I looked this one up
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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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1 L = 1 000 mL
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@& You haven't shown how you used the given information to get this.*@
How many milliliters are there in a cubic meter?
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1 m^3 = 100 cm^3 = 100 mL
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@& 100 one-cm cubes won't fill a 1-meter cube. Not even close. You might need the better part of an hour to shovel 1-cm cubes into a 1-meter cube.*@
@& How would you go about building a solie 1-meter cube out of 1-cm cubes?*@
How many liters are there in a cubic kilometer?
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1 cubic kilometer = 1 000 000 000 000 L
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@& How did you use the given information to get that?*@
How many cubic meters are there in a cubic mile?
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1 cubic mile = 5280 cubic ft = 1600 m^3
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@& It would take more than 1600 one-meter cubes to fill a container 1 mile wide, 1 mile long and 1 mile hight. Way more.
1600 of those cubes would just make a row 1 meter wide and 1 meter high, one mile long. Clearly wouldn't even put a dent in the volume of a 1-mile cube.*@
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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V = 4/3 pi r^3
= [4 pi (1^3)] /3
= 4 pi / 3
@& You never included units in your calculation.
The radius is not 1 cm so the volume is not 4 pi / 3 cm^3, as it would have to be to give 33.5 g..*@
m = dv
= 8 g/cm^3 * 4pi/3
= 33.5 g/cm^3
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@& You need to use units throughout.*@
What would the mass of the smaller ball be as a fraction of the mass of the larger ball?
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m = 8g/cm^3 * [4 pi ((1/2)^3)] / 3
= 4g/cm^3
4/33.5 = 0.12 so about 1/10
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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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1 mile / hour
5280 ft / 60 min
160934.4 cm / 3600 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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2.5 cm : 0.9 cm
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The ratio of its width to its thickness.
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isn't the width and the thickness considered to be the same thing
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@& You have three dimensions, which I refer to here as length, width and thickness. You could call them length, width and altitude if you prefer. Which is which depends on how you visualize the domino. It will balance in any edge, or lie flat, so any of the three measurements could be the altitude. The other two would be length and width.*@
The volume of a domino.
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V = l * w * h
= 2.25cm * 0.9cm * 5.1cm
= 11.475 cm^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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Not sure how to calculate this.
I would estimate it to be less than 5% though...
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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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Steel balls : 12
Volume of 1 steel ball : (4/3) pi (r^2) = (4/3) pi ((.5 in)^2) = 85 mm^3
Volume of one BB : "" = (4/3) pi ((3mm)^2) = 4 mm^3
about 21 BBs would fit
What about the negative space in between the steel balls?
Wouldn't that allow more BBs to fit into the cup
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@& Several BB's could fit into the volume of one ball, so 21 BB's is a pretty low estimate.
What is the ratio of the ball's diameter to that of a BB?*@
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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.1 g
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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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1 g = 4 000 Joules
.1 g = 400 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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100 000 grains of sand in a L
how deep is the sand on the beach, and how wide?
to fill a volume with a volume wouldn't you need to know the volume of the thinkg being filled?
If I had to guess, I would guess 1 000 000 L
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@& Those are the right questions. You're asked for an estimate, so you should estimate some reasonable width and depth.*@
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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my estimate was lower
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@& It's going to be lower or higher. The question is how much lower, or how much higher.*@
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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in our galaxy : 100 billion starts
100 billion galaxies : (100 billion)^2 stars
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@& Good so far.
What is 100 billion squared?*@
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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1 gram / cm^3
0.001 kilograms / 0.01 meters^3
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@& Your numerator is good, but a cm^3 is very, very much less than .01 meters^3.*@
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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Initial V = 0 cm/sec
a = 980 cm / sec^2
Final velocity = ?
Delta S = 100 cm
Delta t = ?
Final V^2 = Initial V^2 a delta s
= 443 cm/sec
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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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o.25 cm
980 cm/sec^2
1/20 of a second
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@& At 400 cm/s, in 1/20 sec the ball would travel 20 cm. That's way more than its thickness, so even at half that speed the top of the ball would end up on the other side of the bottom.*@
How much KE does it lose, per gram of its mass, during the compression?
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KE = 1/2 m v^2
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1/5 of its KE <--- just a guess
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@& During the compression the ball slows from 400+ cm/s to 0. What is the KE of a 1 gram mass at each of those speeds?*@
How much KE does it gain, per gram of its mass, as it decompresses?
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1/7 of its KE <--- again just a guess
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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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p = m v
= m (443cm/sec)
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@& Good, but the question says 'per gram of its mass'. So you could use a mass of 1 gram.*@
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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V_f is the velocity when it hits the floor
What do I use to represent the instant before and after it hits the floor?
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@& You know how high it bounces so you can figure out how fast is has to be going when it leaves the floor.*@
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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.5 m/sec^2 * ___ = 10 m/s
20 seconds
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How far behind will I be at that instant?
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it has traveled 200 m
...
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@& It will. But I will have also moved.
How far will I have traveled?
So how far behind will I be?*@
How much longer will it take me to catch up, and how fast will I be going when I do?
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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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'dW = F * 'ds
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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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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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delta t = 0.32 sec
V_o = 0 m/s
V_f = ?
delta S = ?
a = 9.8 m/s^2
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@& You have the right setup. Use the equations of motion to solve for v0 and vf.*@
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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200 watts ---> 1m / sec
2m / stroke
400 watts?
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@& Right. Good.*@
"
@& Check my notes. You'll probably want to do more work on some of these problems, and/or correct some errors.*@