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course Phy 202
submitted 01/31/2012
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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In the lab on 01/09 we discussed the potential energy is stored energy of position. Change in potential energy is equal and opposite to the work done by conservative force such as gravity. Using a bottle with a hose down the middle we can measure the percent atm. Ten cm up the tube equals 1% atm. When the bottle is not squeezed the energy is potential, but when it is squeezed it is then converted to kenetic energy. Once the water is collected and we stopped squeezing the potential energy is increased from the original PE because of less volume and more pressure. Through this experiment we learned that conservative forces restore energy.
In the lab on 01/11 we timed oscillation of a plastic stick balanced on dominoes on top of cups.
Trial 1 33.4 sec 44 oscillations
Trial 2 30.1 40
Trial 3 21.3 29
Trial 4 24.0 32
Conclusion is that the trials were not harmonic
#$&* Please note the following: Your response to the question should be inserted in the ‘middle line’ which is the line following the **** and before the #$&*. 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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PE = 10 g (9.8 m/s^2) (40 cm)
PE = 10 g (9.8 m/s^2) (0.4 m)
PE = 39.2 g*m^2/s^2
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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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The KE should be the same as the change in PE so the KE should be 0.01 Joules
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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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Velocity = SQRT [2(980 cm/s^2)(60cm)
v = 342.9 cm/s
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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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I really do not know what you are looking for here but here is what i got.
The halfway point between the two dominoes is 20 cm equalling 0.20 m.
We do not have the mass of the plastic strip nor do we know the constant k. so the equation would be.
acceleration = - (k/m) (0.20 m)
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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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one fourth of the way between the dominoes would be 10 cm equalling 0.10 m. The amplitude at 0.20 meters is 0.02 m so the amplitude at 0.10 m would be 0.01 m.
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The information given in this document includes the following, which is applicable here:
'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.'
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Based on your estimate what is the equation of motion of this point?
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The equation of motion would be
a = -(k/m) (0.10 m)
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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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The entire strip oscillates with a constant frequency. So whe you need to do here is estimate the amplitude and use the information in my preceding note.
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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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`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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mass*velocity + mass final*velocity final = 0, because the ball is bouncing off the wall elastically.
(10 g)(500 cm/s) + (10 g) v final = 0
-500 cm/s = Velocity final
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The ball's speed is the same after collision as before.
The direction of motion reverses.
So the velocity after the collision is the negative of the velocity before.
What therefore is the change in the ball's momentum?
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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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I am stuck on these next two questions and how they relate to the first one.
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How fast is it going when it is at rest?
How fast is it going after it leaves the wall?
What therefore is the momentum change?
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What is its momentum change between its first contact and its last contact with the wall?
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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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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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according to the notes given to us additional energy of 2/2 * 8 Joules is needed to be added to do the work. But I do not understand how to solve the question asked.
so PV = nRT, 2 L = n (8 J)(10 celcius)
n = 0.025
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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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volume = pi r^2 h
volume = pi (1.5 cm)^2 (10 cm)
volume = 70.6 cm^3
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What percent is this of the volume of a 500-milliliter bottle?
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500 mL = 500 cm^3 so,
70.6 / 500 = 14.1% of 500 mL bottle.
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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% of 500 mL bottle is 5mL so,
5 mL = pi (1.5)^2 h
h = 0.7 cm in lenght of the tube will match 1% of the 500 millileter.
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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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`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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it should be equally spaced between each domino so the first domino will be at point 0 cm, next 80 cm, 160 cm, 240 cm, 320 cm, 400 cm.
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So the dominoes could be spaced at 80 cm intervals.
That's one spacing.
You need four more.
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Good answers on some of these questions.
You need to give some of the others a little more thought.
See if you can make at least a little more progress on the remaining questions, especially the ones on which I inserted notes.
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&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
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