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course PHY 202
2/15 2:46pm I left question 3 blank because I wanted to see if I was right on question 2. Then I will procede with question 3.
Ph2 Class 12130If you haven’t done so for another course, work through and submit the following documents, which review the calculation of areas and volumes. Answer and self-critique the questions you aren’t sure of; you can just OK those you are sure of:
areas
volumes
`q000. Report the data you obtained in today’s class, where we measured temperature vs. clock time for initially warm water in a cup. After some time elapsed we introduced some ice into the cup, and we later introduced a very warm steel washer, continuing to measure the temperature as a function of time. After providing your raw data, detail how you then used your data to calculate the specific heat steel. Include estimates of uncertainty.
Before the addition of anything:
Temperature (Celsius) Time (sec)
35 30
34.5 60
34.2 90
34 120
33.2 180
32.5 240
32 300
31 360
After the addition of ice:
Temperature (Celsius) Time (sec)
15 60
10 120
9 180
10 240
11 300
11 360
After the addition of the washer that was in hot water. The water temp was at 12.5 degrees Celsius when the water was added. The steel was in water that was about 68 degrees Celsius:
Temperature (Celsius) Time (sec)
14.5 60
15.5 90
16 120
16 150
16 180
Specific heat of steel in this experiment:
The end question is how many Joules did steel give up per gram per degrees Celsius? First we have to answer questions which precede this.
1. How many J did water pick up from steel? The specific heat of water is 4.19J/g*Celsius. The change in temperature when the steel was added after about 3 minutes was 3.5 degrees C. So, (4.19J/g*C)*30g (which was the mass of the washer I used)*3.5 degree C.
=360J that the steel gave up to the water
2. How many J did each gram give up to the water? 360J/30g=12J/g
3. How many degrees did steel temp change? 52 degrees C. It started at 68 degrees C and stopped at 16 degrees C.
4. How many J did steel give up per gram per degrees C?
12J/g/52degrees C=.45J/g*degrees C
q001: What is the weight of a column of water 10 cm high, in a vertical tube of diameter 3 mm?
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First, we need to find the volume of the tube which is a cylinder.
V=pi*r2*h=3.14*(.3cm)2*10cm
=2.83cm3
Now, using the density of water which is 1g/cm3, we can convert to grams: 2.83cm3*1g/cm3=2.83g
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Good, except that 3 mm is the diameter of the tube, not its radius.
The result is that your answer is 4 times as great as the correct answer, which is about .71 grams.
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How much force is required to support that column?
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Now to find the force we use ‘rho*g=.00283kg*9.8m/s2=.0277N
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If the column is supported by water at the base of the tube, what must be the pressure at the base?
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For this, we use P=rho*g*y where y is equal to the displacement of 10cm or .1m. P=.00283kg/m3*9.8m/s2*.1m=.00277N/m2.
I’m not sure if my units are correct on this. Earlier I solved for the weight of the water and got grams of water. When using rho, is that an expression per cubic meter?
I wanted to leave the previous question just to make sure, but as I was thinking more about it, I believe it’s correct because rho is dealing with density. That’s why we use that and not grams….I think.
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rho is the density of the fluid, in this case the density of water, which is 1 g / cm^3 or 1000 kg / m^3.
When you solved earlier you got a mass, not a weight.
If you figure out the weight, then divide by the cross-sectional area, you get the pressure change due to the water column.
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What pressure would be required to support a column 50 cm high?
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P=rho*g*y
=.00283kg/m3*9.8m/s2*.5m
=.013867N/m2
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rho is the density of the fluid, in this case the density of water, which is 1 g / cm^3 or 1000 kg / m^3.
When you solved earlier you got a mass, not a weight.
If you figure out the weight, then divide by the cross-sectional area, you get the pressure change due to the water column.
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Symbolic alternative:
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Let rho stand for the mass density of water and g for the acceleration of gravity.
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A column of water in a vertical cylindrical tube has height y. The tube has radius r.
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How much force is required to support the water in the tube?
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F=mg
I’m not clear on this question. Is it just a symbolic expression of the previous problems with numbers?
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m*g is the weight, but m needs to be expressed in terms of the cross-sectional area and the height of the column, and the density of the fluid.
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Using your preceding answer, what is the pressure at the base of the tube?
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P=rho*g*y
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`q002. Water exits a cylindrical container, whose diameter is 6 cm, through a hole whose diameter is 0.3 cm. The speed of the exiting water is 1.3 meters / second. How long would it take the outflowing water to fill a tube having cross-sectional diameter 0.3 cm and length 50 cm?
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For this, the first thing I did was to find the volume of the tube that we are filling. After that, I’m not sure what to do. I know that I need to have a time as my answer, but am not sure the calculations to get there.
Volume of tube=pi*r2*L=3.14*(.15cm)2*50cm=3.53cm3=.0353m3
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3.53 cm^3 is right.
A cm is .01 m so a cm^3 is (.01 m)^3 = .000001 m^3, not .03 m^3.
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Now, the volume of the tube has to be equal to the volume lost by the original cylinder. So, in order to find the height of the cylinder, we solve for h.
V=pi*r2*h
.0353m3=3.14*.03m2*h
h=12.49m
Now, the only way I know to solve for time (s) is to divide 12.49m/1.3m/s=9.6s
I know this is not correct but I’m stuck on this one.
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Some of the quantities you use here aren't right (see preceding notes) but the answer isn't in any case relevant to the volume of the cylinder.
The water comes out of a hole whose diameter matches that of the tube, moving at 1.3 m / s. So the water would fill a 1.3 meter length of the tube in a second.
The question is how long it takes to fill a 50 cm length of tube.
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What volume of water exits during this time?
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The volume has to be equal to the volume of the tube we are filling.
Volume of tube=pi*r2*L=3.14*(.15cm)2*50cm=3.53cm3=.0353m3
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By how much does the water level in the cylinder therefore change?
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Now you need to consider the cross-sectional area of the cylinder, whose diameter you recall is 6 cm.
How much would the water level change to vacate a cylinder with the cross-sectional of the cylinder, and the volume of the tube?
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What is the ratio of the exit speed of the water to the speed of descent of the water surface in the container?
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Symbolic alternative:
Water flows from a cylinder whose radius is R_cyl through a hole of radius r_hole, with water level y relative to the hole.
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A short tube of length `dL and diameter equal to that of the outflow hole extends horizontally from the hole.
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What mass of water is required to fill the tube?
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What is the PE change of the system during the interval required to fill the tube?
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Based on these results, assuming no dissipative forces to be present, what is the velocity of the water flowing from the cylinder?
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How long does it therefore take for the cylinder to fill?
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What is the expression for the volume rate at which water flows from the cylinder, with respect to time?
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What is the expression for the rate `dy / `dt at which the depth of water in the cylinder changes with respect to time?
`q003. Assume the water level in the container of the preceding question is 50 cm above the outflow hole.
What mass of water is required to fill the 50 cm tube?
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You previously figured out the volume of that tube. What mass of water would occupy that volume?
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As that mass of water exits the cylinder, what is the change in the gravitational PE of the system?
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What therefore is the speed of the exiting water?
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How long does it therefore take to fill the 50 cm tube?
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By what percent will the water level in the cylinder change during this time?
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At what rate is water flowing from the cylinder, in units of cm^3 / second?
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Symbolic alternative:
If you have completed the symbolic alternative to the preceding problem, you may use your results along with the given quantities to answer this question.
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`q004. In a cylinder containing a fixed large number of particles, all colliding elastically with one another and with the container, one end of the cylinder consists of a moveable piston.
If the piston is stationary, then does the speed of a particle increase, decrease or remain the same when it collides with the piston?
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It remains the same once it collides with the piston. The kinetic energy will go to zero when it hits the piston but will rebound and have the same kinetic energy that it had when it hit the piston.
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Good answer.
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If the piston is moved towards the other end of the cylinder:
Does the speed of a particle increase, decrease or remain the same when it collides with the piston? (and do you see a potential discrepancy between what you know happens and the previously given definition of an elastic collision?)
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The speed of the particle increases because it’s hitting the end of the piston and the end of the cylinder more frequently.
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The frequency of collisions is important for figureing out the average force, but it isn't relevant to the speed of the particle.
When the bb collides with a large mass which is moving toward it, what happens to its speed?
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Does this result in more pressure, no change in pressure or less pressure on the piston?
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As it moves toward the end of the cylinder, the pressure increases.
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Does the volume of the system increase or decrease in this process?
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The volume would decrease if the piston moves to the other end of the cylinder.
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`q005. The gas in a system is confined if no gas can enter or leave the system. So the number of particles in a confined gas does not change.
If the pressure in a confined gas is held constant while the temperature is increased what happens to its volume?
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The volume will increase because the molecules are hitting the confinement more.
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If the volume of a confined gas is held constant while the temperature is increased what happens to its pressure?
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The pressure will increase as the temperature increases.
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If the temperature in a confined gas is held constant while the volume is increased what happens to its pressure?
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If the volume increases, then the pressure will decrease because it takes longer for the molecules to hit the confined space that they are in.
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`q006. For a confined gas, a temperature change can result in pressure changes, volume changes, or both pressure and volume changes.
If the percent changes in the pressure, volume and temperature of a confined gas are small, then every 1% increase in pressure implies a 1% contribution to an increase in absolute temperature, while every 1% increase in volume implies a 1% contribution to a decrease in temperature. So for example a 1% increase in pressure accompanied by a 1% decrease in volume would imply a 2% increase in temperature, while a 1% decrease in pressure accompanied by a 1% decrease in volume would imply unchanged temperature.
If volume increases by 2% and pressure increases by 3%, what is the percent increase in temperature?
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It seems that volume and temperature are inversely related while pressure and temperature are directly related.
The temperature increases by 1%
V+1%=T-1%
P+1%=T+1%
So,
V+2%=T-2%
P+3%=T+3%
3%-2%=1%
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If volume decreases by 2% and pressure increases by 3%, what is the percent increase in temperature?
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The temperature will increase by 5%
V-2%=T+2%
P+3%=T+3%
+2%+3%=+5%
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Decreasing volume would be associated with decreasing temperature.
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What must happen to the volume when combined with a 3% increase in temperature, if the result is to be a 1% decrease in pressure?
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Here, because the temperature is directly related to the pressure, the net result has to be a -1% decrease in temperature. The net value of temperature combined with the 3% increase has to be at -4%. So with T being -4%, then V has to start at +4%. So, for the end result to be temperature equal to -1%, then volume has to be +4%.
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What must happen to the pressure when combined with a 3% increase in temperature, if the result is to be a 2% decrease in volume?
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Pressure must go down -1%. This would bring the 3% increase in temperature to 2% increase in temperature which results in a 2% decrease in volume
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Increasing temperature and decreasing volume are both associated with increasing pressure. The two effects would reinforce.
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`q007. On a set of x and y axes:
Sketch dots at (-5, 1), (0, 1) and (2, 0). Connect the dots from left to right with straight line segments, and when you get to (2, 0) keep going for a ways. Call this ‘path 1’.
Sketch a straight line from (-5, 1) to (0, 0) and keep going until the line intersects path 1, and keep going for a little ways. If you need to extend path 1 a bit in order for the lines to intersect, do so. Call your new path ‘path 2’.
At what coordinates do the paths intersect?
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They intersect at (4,-1)
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You can check your answer by finding the slopes of the various lines. For example the slope from (0, 1) to (2, 0) should be the same as from (2, 0) to (4, -1). This does check out.
The two slopes of the other line nearly check out, but not quite.
So you've got a good estimate, which means that you made a good sketch, which is all that was asked for here.
To be completely accurate you would have to do some algebra.
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Sketch a dot at (-2, 0). Sketch a line from (-5, 1) through this dot and continue until the line hits the y axis. Make a dot at this point. Then sketch a horizontal line starting at this last dot, directed to the right and continue the line until it intersects each of your other two paths.
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What are the coordinates of your points of intersection?
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The coordinate where it hits the y-axis is (0,-1) and it intersects the other paths at (14.5,-6)
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If it intersects both of the other paths at the same point, it has to do so at the point where they intersect.
Otherwise it will intersect the two lines in different points.
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Symbolic alternative:
Path 1 is defined as y = a if x < 0, y = a - b x if x > 0, where a and b are positive constants.
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Path 2 is the line y = m x, where m is a negative constant.
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For x < 0 path 3 is the line through (-a/b, 0) having slope m, where -b < m < 0. The path is continuous for all x, and for x > 0 the path is parallel to the x axis. At what points does this path cross each of the first two paths?
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I’ve thought and thought about this and do not know how to solve.
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This is an alternative. You did pretty well with the sketches.
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`q008. Two identically-shaped cylinders, each of diameter 2 centimeters, hang straight down into the water, extending 5 centimeters beneath the surface. Water pressure results in forces acting perpendicular to the surfaces of each cylinder. When all these forces on a cylinder are added up, the result is a buoyant force on that cylinder.
One cylinder is made of steel, the other of glass.
Does the pressure exerted at a point on the lower end of the cylinder depend on whether the material is steel or glass?
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No, because the cylinders are identically the same.
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Does the buoyant force on the cylinder depend on whether the cylinder is steel or glass?
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No. Archimedes’ principle says that the buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by that object. If the cylinders are identically shaped, then it doesn’t matter which material they are made from.
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If the glass or steel cylinder is removed, the surrounding water would quickly occupy the vacated cylindrical region. So in that sense we would now have a cylinder of water in place of the glass or steel cylinder. The ‘water cylinder’ experiences a buoyant force from the pressure of the water surrounding it, just as the glass or steel cylinder experienced a buoyant force. Is the buoyant force on the ‘water cylinder’ greater or less than that on the glass and steel cylinders?
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It is equal to the others
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The glass and steel cylinders would have sunk of the buoyant force was the only force acting on them. Something in addition to the buoyant was supporting them either from above or from below. However no additional force is required to support the ‘water cylinder’. How therefore does the buoyant force of the ‘water cylinder’ compare to the weight of the water in that cylinder?
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It is equal to the weight of the water
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How does the buoyant force on the glass or steel cylinder compare to the weight of the cylinder?
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The buoyant force would be rho*g of the steel or the glass
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You would multiply rho by the volume of the cylinder to get the mass of water displaced. Then multiply the mass by g to get the weight.
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How does the buoyant force on the glass or steel cylinder compare to the weight of the water displaced by the cylinder?
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The buoyant force on the glass or steel is equal to the weight of the water displaced.
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end document
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Good answers on many of the questions.
I do recommend a reasonable of time on some revisions, provided some of my notes ring a bell.
&#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.
Spend a reasonable amount of time on your revision, but don't let yourself get too bogged down. After a reasonable amount of time, if you don't have at least a reasonable attempt at a solution, insert the best questions you can showing me what you do and do not understand, and I'll attempt to clarify further. &#
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