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course 201
10/18 9:20
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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2.54cm*12=30.48cm
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The number of feet in a meter.
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100cm=1m
100/30.38=3.28ft
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The number of meters in a mile.
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1mile=5280ft=1609.76m
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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)/(1*10^-9m)=1000000nm in a mile
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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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1ml=1cm^3
1000000
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How many liters are there in a cubic kilometer?
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How many cubic meters are there in a cubic mile?
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1609.76^3=4,171,414,966.19m^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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60.06g to 68.64g
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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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1/8
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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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100cm=1m;1609.76m=1 mile;=160976cm
60s=1 min; 60 min=1 hr= 3600s
44.72cm/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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5cm/2.5cm=2/1
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The ratio of its width to its thickness.
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2.5cm/.85cm=2.94/1
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The volume of a domino.
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.85*2.5*5=10.625cm^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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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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12 steel balls
1in=2.54cm=25.4mm
25.46/6mm=4.23
4.23^3=75.69*12=908+ 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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.25g
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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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.125/x
1/4000
1/.125=8
4000/8=500 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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1ml=1cm^3
1ml=100grains
1000ml=1l*100=100000grains per l
1000ml=1000cm^3=1000*10^-2=10m
100/10= 10L
if you spead it 1cm thick and 1 cm wide you would need 10l to strech the 100m beach
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@& There aren't many beaches 1 cm wide and 1 cm deep, but if there is such a beach you have a reasonable estimate.*@
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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1000000000 stars/100000*10 grains of sand= 1000 stars for every grain of sand
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@& 100 billion is bigger than that.
Otherwise your procedure is good.*@
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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1^20 stars/1^6 grains of sand= 1^14 stars for every grain of sand
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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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1000g=1kg
1000000cm^3=1 m
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@& Good, but you still don't have a density.*@
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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vf^2=0^2+2(10)(100)=44.72cm/s
o^2=vo^2+2(10)70=37.42cm/s
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@& You didn't use units, and therefore god an incorrect result. Right procedure, but without rigorous use of units you'll be prone to mistakes of this type.*@
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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1/10""
-44.72cm/s
.2s
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@& The 1/10 inch is a good estimate. However in .2 sec that ball would move nearly 10 cm, which is much more that 1/10 inch.
*@
How much KE does it lose, per gram of its mass, during the compression?
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ke1=1/2m(44.72)^2
ke2=1/2m(0cm/s)^2
ke2-ke1=-1/2m(44.72cm/s)^2
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How much KE does it gain, per gram of its mass, as it decompresses?
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ke1=1/2m(0)^2
ke2=1/2m(37.42cm/s)^2
ke2-ke1=1/2m(37.42cm/s)^2
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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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44.72(m)
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@& You need to use units more carefully.
*@
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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37.42(m)
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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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10m/s=.5dt=20s
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How far behind will I be at that instant?
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ds=10m/s(20s)=200m+20m=220m
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@& That's a good step in the right direciton, but the result is the distance of that car from my starting point. I'm no longer at my starting point.*@
How much longer will it take me to catch up, and how fast will I be going when I do?
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42s-20s= 22 more sec.
21m/s
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@& After 42 s I would be moving at 21 m/s.
How far will I be at that instant from my starting point?
How far will the car be from the starting point?
Those questions aren't hard to answer, but the overall question is still challenging. Answer those questions and you'll get better insight into what's necessary.*@
How much work will the net force on my truck have done by the time I catch the other car?
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1/2(1700)(21m/s)^2=374850kg m^2/s^2
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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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vo 21 m/s
vf 10 m/s
vave 15.5 m/s
dv-11 m/s
dt 1.29 s
ds 20 m
a -8.53 m/s^2
1700kg* -8.53m/s^2= -14496.12kg/(m/s^2)
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@& This is how long it takes me to get to where the car was when I hit the brakes. The car has since moved, so I will not have caught up with it.*@
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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window
vo 3.725 m/s
vf 5.025 m/s
vave 4.375 m/s
dv 1.3 m/s
dt .32 s
ds 1.4 m
a 4.063 m/s^2
@& The acceleration of the ball is 9.8 m/s^2 downward.
It's not clear here what quantities you are assuming and what quantities you are calculating, bit it does look as though the 4.06 m/s^2 was calculated rather than assumed.*@
total
vo 0 m/s
vf 5.025 m/s
vave 2.513 m/s
dv 5.025 m/s
dt 1.24 s
ds 3.11 m
a 4.063 m/s^2
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@& Your solutions are not consistent with the known acceleration of gravity. It's not clear what you regarded as given quantities or how you proceded from those quantities to your results.*@
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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1 m/s=200 watts
1 breath=1stroke= 2m
200watts*2m= 400 watts per breath
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&#Good responses. See my notes and let me know if you have questions. &#