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course phy201
10-16 12:00 i have tried to submit this but couldn't get it to work. hopefully it will go through this time.
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/30.48cm=3.28ft
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The number of meters in a mile.
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5280ft=1mile 5280ft/3.28=1609.76meters
3.28ft=1meter
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The number of nanometers in a mil (a mil is 1/1000 of an inch).
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(3.937*10^-8)nanometers=1inch
(3.937*10^-8)*(1/1000)=3.978*10^-11
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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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1liter=1000milliliters
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How many milliliters are there in a cubic meter?
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100cm=1meter
so, 100^3=1000000ml=1meter^3
or, 10^3liters=1000liters=1meter^3
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How many liters are there in a cubic kilometer?
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1000^3 liters=1000000000000L=1 km^3
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How many cubic meters are there in a cubic mile?
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1609.76meters=1mile
1609.76^3 meters=4171414966 m^3=1 cu mile
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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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density= 7grams/cm^3 and 8grams/cm^3 diameter of large=2.54cm, radius=1.27cm
v=(4/3)pi*r^3 v=(4/3)3.14*(1.27cm)^3=8.580cm^3
mass=dv mass=7g/cm^3*8.580cm^3=60.06g or,
mass=8g/cm^3*8.580cm^3=68.64g therefore btwn 60.06g and 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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diameter of small=1.27cm, radius=0.635cm
v=(4/3)3.14*(0.635cm)^3=1.073cm^3
mass=7g/cm^3*1.073cm^3=7.511g or,
mass=8g/cm^3*1.073cm^3=8.584g therefore btwn 7.511g and 8.584g
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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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2.54cm*12inches=30.48cm in 1 ft. 30.48cm *5280ft= 160934.4cm in 1 mile
60 sec in 1 min. so, 60s*60min=3600s in 1 hr therefore we have 160934.4cm/3600s reduced to 44.704cm/1s
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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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1.875""/0.9375"" 2:1
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The ratio of its width to its thickness.
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(15/16)""/(1/4)"" 3.75:1
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The volume of a domino.
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v=1.875in*0.9375in*0.25in=
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The percent uncertainty in your results, according to your estimates of the uncertainty in your measurements.
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i am no more than 50% certain on my measurements because i didn't use the most percise measuring tools
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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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approx 10 large steel balls in an 8oz drinking cup
approx 1000 6 mm 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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i think without much certainty that is would be 1/2g, possibly less
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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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1/4g 4000joules*0.25 carbs=1000 joules of food energy
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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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billions
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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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more sand.
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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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i still think more sand because the sand on the beach is not just a thin layer, it would be a thick layer of sand on the 100-meters of beach
@& Right.
The beach would also have width.*@
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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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1g/cm^3 = 0.001kg/0.01m^3
@& I believe you calculated previously that 1 m^3 is 1 000 000 cm^3.
How would this affect your result?*@
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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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v0=0 vf^2=0^2+2(980cm/s^2)*100cm = 1960cm/s^2*100cm = 196000cm^2/s^2, then take square root=442.72cm/s
a=980cm/s^2 Vave=(442.72cm/s-0cm/s)/2 = 221.36cm/s 'dt=(442.72cm/s-0cm/s)/(980cm/s) = 0.452s
""ds=100cm 'dv=442.72cm/s-0cm/s = 442.72cm/s
when the balls 1st leaves the floor the in the upward direction it will be in the neg direction if we choose dwn as pos., so it would be moving at -442.72cm/s
@& The ball doesn't bounce as high as when it was dropped. The upward displacement is only 70 cm. So the velocity on rebound would be different.*@
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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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i don't know. milliseconds
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How much KE does it lose, per gram of its mass, during the compression?
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@& It starts the compression with a speed of about 440 cm/s. How much KE is that, assuming a gram of mass?
At the end of the compression the ball is at rest.
So how much KE was lost by a gram of the ball's mass?*@
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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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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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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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a=5m/s^2 vf^2=(0m/s)^2+2(5m/s^2)*20m = 10m/s^2*20m = 200m^2/s^2, take square root, =14.14m/s
mass=1700kg Vave=(14.14m/s-0m/s)/2 = 7.07m/s
v0=0m/s 'dv=14.14m/s-0m/s = 14.14m/s
'ds=20m 'dt=(14.14m/s-0m/s)/(5m/s^2) = 2.828s
@& Having traveled 20 meters at .5 m/s^2 I won't match his speed.
You calculated based on an acceleration of 5 m/s^2. That's OK, though my truck won't do that.
You found that the truck would be moving at 14.14 m/s. That's faster than the car is moving. So the truck will have matched the speed of the car before traveling 20 m.
The question is, how long does it take before the truck is traveling at 10 m/s. which is the speed of the car?*@
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How far behind will I be at that instant?
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10m/s*2.828s= 28.28m
@& The car ahead of me will have traveled 28.28 meters in that time. However I will have traveled 20 meters, according to your calculation. So how far ahead will the car be?*@
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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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i don't know how to do this one
@& Let's not worry about this question for the moment. You're doing pretty well with the preceding questions, and this one is a little harder. To avoid confusion, then, let's concentrate on the earlier questions.*@
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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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Fnet=(5m/s^2)*1700kg= 8500kg m/s^2 'dwork=8500kg m/s^2*20m=170000kg m^s/s^2
@& Good.*@
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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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'dt=0.32s Vave=1.4m/0.32s=4.375m/s
a=-9.8m/s^2 vf=2(4.375m/s)-0m/s=8.75m/s
'ds=1.4m 'dv=8.75m/s-0m/s=8.75m/s
v0=0
@& Good, but the initial velocity for this interval isn't zero. This interval just goes from the top of the window to the bottom.
You know the average velocity.
What can you calculate from the acceleration and the time interval?
How can you put this together with the average velocity to get the initial and final velocity of the ball as it passes the window?
Having found these results, you can now consider a different interval. I would suggest the interval between falling from the higher window (when its velocity was zero) and reaching either the top or the bottom of the lower window.*@
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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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don't know how figure this
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@& How long does a stroke last?
How many joules of energy would be required in this time?*@
@& Your problem solving continues to improve. You've done a credible job on most of these questions.
You should spend a little time reading my notes and making some revisions. Don't spend all day on it, though. I would recommend an upper limit of an hour.*@