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course Phy 242
1/24 11
Physics II Class 120111
(This document contains class notes of 1/09/12 and 1/11/12. It is duplicated under the date 120111).
Your best attempt on the questions posed here will be due by the end of the day on Friday, 1/20/12.
To raise a vertical water column 10 cm requires about an additional 1.0% of atmospheric pressure.
To achieve a 1% increase in the pressure of a confined gas at constant volume requires that its absolute temperature be raised 1%.
If the absolute temperature of a confined gas increases by 1% without changing its pressure, its volume increases by 1%.
For our purposes the freezing and boiling points of water at atmospheric pressure are defined, respectively, to be 0 Celsius and 100 Celsius. This is not the standard SI definition. You are not yet expected to be prepared to understand the standard SI unit, so for the moment we are using the ‘old’ definition. The two definitions agree to about 5 significant figures.
A mole of ideal gas at atmospheric pressure at 0 Celsius occupies 22.4 liters; for ballpark calculations you could use either 20 or 25 liters as a basis for estimation.
To raise the temperature of a gas we have to add energy to the gas, in order to speed up the particles. The approximate energy required raise the temperature of one mole of gas by 1 degree Kelvin depends on the nature of the gas and the situation, according to the following guidelines:
· If the gas is monatomic and kept at constant volume, the energy required is roughly 3/2 * 8 Joules, all of which goes into the translational kinetic energy of the atoms.
· If the gas is diatomic and kept at constant volume, the energy required is roughly 5/2 * 8 Joules. Of this 3/2 * 8 Joules goes into the translational kinetic energy of the molecules and 2/2 * 8 Joules goes into the rotational kinetic energy of the molecules.
· If a gas is allowed to expand at constant pressure, additional energy must be added to do the work of expansion. The additional energy is roughly 2/2 * 8 Joules. This applies to any gas, whether monatomic, diatomic or polyatomic.
· The 8 Joule figure is a rough figure. A more accurate figure is 8.31 Joules. This comes from the gas constant R, which is R = 8.31 Joules / (mole Kelvin).
The change in the potential energy of an object or system (designated `dPE) between two events is defined to be equal and opposite to the work done by conservative forces on the system between those events. Examples of conservative forces are gravitational, electrostatic and magnetic forces, as well as ideal elastic forces.
The weight of an object is the force exerted on it by gravity. The force exerted on an object by gravity in the vicinity of the Earth’s surface will, in the absence of other forces, accelerate that object toward the center of the Earth at 9.8 meters / second^2, which is also equal to 980 centimeters / second^2 and close to 32 feet / second^2. We use g to stand for this acceleration. Since the unopposed gravitational force gives the object this acceleration, the gravitational force on the object must be F_grav = m g, where m is its mass. Thus the weight of an object of mass m, near the surface of the Earth, is weight = m g.
If we raise the object its displacement is in the direction opposite its displacement, so that the gravitational force does negative work on it.
The equation of motion of an object undergoing simple harmonic motion is x(t) = A cos(omega * t + theta_0), where omega is the angular frequency of the motion, A the amplitude and theta_0 the initial angular position of the circular-model reference point. In the absence of other information theta_0 may be taken to be zero. In SI units the angular frequency is equal to the frequency of the motion in cycles per second, multiplied by the 2 pi radians in a cycle.
In an elastic collision, kinetic energy is conserved. An object which collides elastically with a much more massive object loses negligible kinetic energy.
`q000. Report the data you obtained in lab on 1/09 and 1/11. Include a brief but clear description of what you did, and report the data in a concise, organized, self-explanatory table.
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We had a 20cm tall water cylender with a hole approximately 1/4 inch in diameter. the system was held up 120cm high from floor, and we removed the finger from over the hole to release the water, and observed how far the stream went.
#$&* Please note the following: Your response to the question should be inserted in the ‘middle line’ which is the line following the **** and before the #$&*. Follow this practice throughout the course.
Never add anything to or delete anything from any line that begins with #$&* or ****.
`q001. What would be the potential energy change of a 10 gram mass of water whose vertical position changes by 40 centimeters?
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knowing that water = 1g/cm^2, and that potential energy is equal and opposite to the non conservative forces. so change in PE is .01kg*(9.81m/s^2)* (.4m) = 0.03924 Nm or .04 Joules or 392,400 g*cm^2/s^2
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You would have observed some water depths and distances, which should be reported.
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`q002. If the potential energy of 10 grams of water, which we will consider to be initially at rest, changes by 100 000 g cm^2 / s^2 (which is equal to .01 kg m^2 / s^2 or .01 Joules), and if this PE loss is converted to KE,
then how fast is the water moving?
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change in PE is equal and opposite to KE. so KE = -100,000 g cm^2/s^2 = 1/2(10g) Vf^2. solve for Vf to get 141.42 cm/s
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`q003. A stream of water spurts out of the side of a vertical cylinder, falling 60 centimeters while traveling 40 centimeters in the horizontal direction.
Assuming that the horizontal velocity of the water remains constant, what was the speed of the water as it exited the cylinder?
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using the 4th equation of motion, V = sqrt( 2(981cm/s^2)(40cm) ) = 280.41 cm/s
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That would be the velocity attained by an object which fell freely from rest through a vertical drop of 40 cm.
40 cm isn't in the vertical direction, so 40 cm isn't relevant to analysis of the vertical direction. 980 cm/s^2 is in the vertical direction, so it wouldn't appear int the same equation of motion with 40 cm.
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`q004. If the frequency of the oscillation of a strip of plastic between two dominoes separated by 40 cm is 5 cycles per second, with amplitude 2 cm, then assuming that each point on the strip undergoes SHM:
What is the equation of motion of the point halfway between the dominoes?
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if amplitude is 2cm, then total distance is 4cm. If it travels 5 cyc/sec with distance 4cm. It has velocity or I guess frequency of 20cm/s. equation of motion would be v = v_0 + a*t where a can be found by diving that frequency again by 5 seconds
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v = v_0 + a t applies to uniformly accelerated motion. It does not apply to SHM, where the net force and therefore the acceleration are constantly changing.
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and because were assuming SHM, we can assume constant acceleration, which is why again we can use that equation of motion. v_0 is 0 and t is given. Well now that i'm looking at it, I think the s=s_0 + v_0*t + 1/2a*t^2 equation could be used.
But our answer would still be the same.
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Calculate or estimate the amplitude of motion for a point on the strip which is ¼ of the way between the dominoes.
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if the half way point was 20cycl / sec. then the 1/4 point could be estimated at 5 cycl / sec. which would give a total distance of 1.25 cm. divide that by two to get amplitude and we have 0.75cm. which seems reasonable.
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The entire strip is oscillating at a single frequency. The amplitude of the oscillation varies from point to point, but the frequency does not.
The amplitude at the center is 2 cm. The amplitude at every other point is less than 2 cm. The endpoints are fixed; they aren't oscillating at all.
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Based on your estimate what is the equation of motion of this point?
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The 1st one. (delta s) equation. because were given all the same values as before, and I think the v = v_0 + a*t equation might work as well.
so it would be 10cm = 0 + 5cm/s * t + 1/2a*t^2
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Acceleration is not uniform.
The context here is SHM, with F_net = - k x and x(t) = A cos(omega t).
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What is the equation of motion of a point on the strip which is 1/6 of the way between the dominoes?
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40cm/6 = 6.67 cm and 20 cylc/sec / 6 =3.33 cm/s
6.67cm = 0cm + 3.33 cm/s * t + 1/2a*t^2
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If x is the coordinate of a point on the strip, as measured from the domino on the left, what is the equation of motion at this point?
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x=(40cm-x) + v_0 * t + 1/2a*t^2
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You are using the equations of uniformly accelerated motion for a situation where it 'ain't so'.
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`q005. If a ball of mass 10 grams, moving at 500 cm / s in a direction perpendicular to a wall, strikes the wall and bounces off elastically, then what is its momentum change from the moment it contacts the wall until it comes to rest?
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the momentum change is equal but opposite to the momentum it had before it hit the wall because it came to rest. so the momentum is 10g*(500cm/s) = -5,000g*cm/s
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What is its momentum change between the time it comes to rest and the time it loses contact with the wall?
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if your asking about the time it takes after it has came to rest, and the elastic force it has ""rebounding"" off the wall, then assuming it lost no KE, ( the conservative force was 100%) or coefficient of restitution was 1. then it is again equal and
opposite to its initial momentum of 5,000 g* cm/s
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What is its momentum change between its first contact and its last contact with the wall?
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it is 0. because it changed right back to its initial momentum. and lost no energy.
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When the ball came to rest was its momentum change in the direction of the wall or away from the wall?
When it rebounded from rest was the momentum change toward or away from the wall?
What to total change in momentum toward the wall, away from the wall, or was it zero?
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If the ball bounces elastically off of another wall and returns to the original wall every .1 second, what average force does it exert on that wall?
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the average force is the change in potential energy with respect to the change in clock time. so. 5,000g* cm/s / 0.1s = 50,000 g * cm
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Your units calculation isn't right.
Be sure you also have the right momentum change.
If both of these things are right then your result will be right.
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In the above, be sure you have treated velocity as a vector quantity. Mainly this means that you need to declare a positive direction and abide by your declaration.
`q006. If the temperature of the air in a 2-liter bottle increases by 10 degrees Celsius, how much energy is required? Air consists of a mix of gases, most of which are diatomic.
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assuming atmospheric pressure of 1 to begin with.
I know avogadro's number to be 6.022x10^23. and to find the number of mols, multiply the atoms diatomic mass by that, so two oxygen has diatomic mass of 36 (2x18).
this gives you the mass of a mole, not the number of moles
36g/mol * avogadro = 2.168*10^25.
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Avogadro's number is the number of particles in a mole. You wouldn't multiply the mass of a mole by the number of particles in a mole.
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Now I honestly dont know where to go with this.
?????????? guidance?
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In the absence of information to the contrary, you can assume that the air in the bottle is at atmospheric pressure at 0 Celsius.
The information given at the beginning of the document should then allow you to determine the number of moles.
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`q007. A plastic tube has inner diameter about 3 cm.
What is the volume inside a 10-cm length of this tube?
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using pi*r^2(h) r being 1.5cm, volume is about 70.69cm^3
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What percent is this of the volume of a 500-milliliter bottle?
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if the 500 milliiter bottle is full, then it is about 14%. ( 71 / 500 ) is about 0.14 or 14%
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How many 10-cm lengths of this tube would be required to match 1% of the volume of a 500-millileter bottle?
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1/14ths of a length. so 0.7 lengths. which is consistent with 71 cm^3
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If the air in a 500-milliliter bottle is heated until the water in the tube rises 10 cm, by what percent did the volume of the bottle increase, and by what percent do you think the temperature of the air had to increase?
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Again I think I would have to know more about the problem I had earlier with pressure and volume to get an answer.
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You have figured out the volume of a 10 cm length of the tube.
If that much water went into the tube, then that much water is no longer in the bottle. Which means that that much additional volume within the bottle is now occupied by the air.
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`q008. A plastic strip resting on a series of dominoes, including dominoes at its ends, is expected to resonate if the dominoes are equally spaced.
What are the first five domino spacings expected to produce harmonic resonance in a strip of plastic 4 meters long?
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4m / 5 spaces = 0.8m equal distances. I'm using the number of ""gaps"" inbetween the dominos.
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If there are 5 spaces between dominoes, then this is the domino spacing. So this is one possible spacing of the dominoes.
The question asked for five possible spacings. You need five spacings.
What are some other domino spacings that would answer the question?
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`q009. A certain aluminum rod is supported at the middle. When the end of the rod is struck sharply in a direction perpendicular to the rod, it produced a bell-like sound with a frequency of 300 cycles / second and amplitude .3 millimeters.
What is the maximum speed of the simple harmonic motion of a point on the end? (Note that you can determine this if you think about the SHM: what is the amplitude and frequency of the SHM of this point,
and how do you use the amplitude and frequency of the SHM to find the maximum speed?)
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amplitude is .3mm, so total distance is .6mm. 0.6mm * 300 cycl / s = 180 mm/ s which would indicate maximum speed.
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We're referring here to something that moves in SHM.
What is the radius of the reference circle, and what is the speed of the point moving around the reference circle?
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Good answers on some problems, but on some problems you aren't using the right model.
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