Experiments 16-22

course Phy 202

Here is my experiments 16-22, 23 and 24 are just to be watched. I have looked back and found only 4 experiments that I have not done, that must be submitted. I will go back and do the Kin model experiment this week. I need to know if these and the others I have turned in give me the required 70 in this part of the class. Please email me and let me know so I can plan time to do more if need be.

Your work on these experiments is good. See my notes.

If you complete the kinetic model experiment successfully you will have a passing lab grade.

I plan to take the third test Thurs. Dec 1. This is two days later than I had originailly planned but the experiments took longer than I expected. I have a note that says I will be able to graduate if I have an official transcript from virginia highlands. Will there be any way possible to do this or will you and your comptroler's calls be sufficient. Another Problem I have encountered is with the hand held generator. Not the generator its self, but the receipt. My mother deciede she would clean my room one day when I was not here and she burnt the package that the generator came in, along with the receipt I had placed there. Will I still be able to return this?

Good job.

If anything is not clear let me know, and include as many specifics as possible.

Here is my experiments 16-22, 23 and 24 are just to be watched. I have looked back and found only 4 experiments that I have not done, that must be submitted. I will go back and do the Kin model experiment this week. I need to know if these and the others I have turned in give me the required 70 in this part of the class. Please email me and let me know so I can plan time to do more if need be.

I plan to take the third test Thurs. Dec 1. This is two days later than I had originailly planned but the experiments took longer than I expected. I have a note that says I will be able to graduate if I have an official transcript from virginia highlands. Will there be any way possible to do this or will you and your comptroler's calls be sufficient.

You will not be able to get an official transcript until grades are officially posted. I can write a letter or get our registrar to send one certifying the grade. You will need to find out specifically who needs to receive that information and from whom they need the information, from our registrar or from me or from both.

Be sure you provide this information immediately after taking the test, and be sure I acknowledge it.

Another Problem I have encountered is with the hand held generator. Not the generator its self, but the receipt. My mother deciede she would clean my room one day when I was not here and she burnt the package that the generator came in, along with the receipt I had placed there. Will I still be able to retur

I can't answer that question, since it's a decision that has to be made by the bookstore. Call Pat in the VHCC bookstore and ask her about this.

The graphs did not come through with the text, so if you need these I will have to email them to you as an attachment. Thank you for your help.

Experiment 16

The crank was easier to turn when clamped to the plastic. I did more work per second, considering that I was turning the generator at a constant equal rate both times, when the clamps were together.

I would say that more current flowed through the wires when clamped together, because the current would travel down one and right back up the other lead.

I would characterize the relationship between the current flowing and difficulty of cranking as the harder the crank is to turn, the more power being made and distributed.

Right.

The current travels more easily when clamped to the plastic.

This contradicts what you just said. When more current flows it's harder to crank, because it takes energy to make current flow. The ease of cranking with the plastic is because no charge is flowing.

I would say that the circuit would resist the electricity flow more when clamped to the plastic. I would say there is more electrical resistance when the generator is easy to crank. 0.25a bulb 16 times in 10 seconds 0.15a bulb 17 times in 10 seconds 0.20a bulb 17 times in 10 seconds The 0.20a bulb cranks the easiest. I made the connection between the two bulbs by connecting the bulb to the generator as I would initially have, but then connecting one tab to the tab of the second bulb by an alligator clip. The generator is easier to crank with both bulbs connected, but it takes a faster crank to get the bulb to burn as bright as before. The .2 and .15a bulbs are the ones I have connected, the .15 is burning brighter than the .2. The bulb that burns brighter takes less force in a faster turn to burn brighter than the .2a bulb. In the parallel circuit the generator takes more force to turn. In order to make the .2a bulb, which has been my first bulb throughout, burn as bright, it takes a slower rate of turn. The parallel requires more force to turn. The series combination takes more speed cranking to achieve the same brightness. The bulbs seem to have the same brightness in both circuts. I think the parallel was producing more work because of the force required to turn the crank. The series circuit would have the greatest voltage, because it has a higher crank rate. The parallel would have the highest current because of it’s high force of crank. The Force is greater in a parallel circuit, and the rate of crank is greater in the series combination, Because the force is so much greater in the parallel and the rate of turn per second doesn’t seem that much greater, I would say that the Parallel circuit would have the highest Power.

Good.

Experiment 17 As I cranked the handle the force necessary decreased with time. Because force is proportional to current the current must also have been decreasing. When I released the handle the handle kept turning, I explain this through the current being released by the capacitor was being dissipated through work in turning the handle. I was not turning the handle, it dosen’t turn without energy, there were not ghosts turning the handle, there fore the energy required to turn the handle had to come from outside the generator, the capacitor.

Good.

In the series circuit with the bulb, generator, and capacitor as I cranked the bulb’s burning got dimmer, and dimmer, while the force cranking remains nearly the same. When I stop, the generator keeps turning some, but not as long as the bulb burns. The voltage would have to remain the same because the number of cranks per second would not change. The current would also remain the same, because the force per crank did not change. I could tell by the force I was exerting on the crank that the current remains the same. By looking at the bulb, it looks as if the current is decreasing, because it burns dimmer and dimmer. As I continue Cranking the generator, the voltage across the bulb must decrease, because the bulb burns dimmer and dimmer. The current must also decrease in the light bulb because it burns dimmer and dimmer. This is consistent with my preceding answer.

Right. As this happens the cranking gets easier. Be sure you have observed this.

The voltage across the capacitor must increase because it has decreased at another point in the circuit.

The voltage across the generator remains constant but cranking gets easier, indicating that less current is flowing, meaning that the capacitor must be producing an voltage which acts opposite that of the generator.

Experiment 18 The ends of the loop are drawn toward the magnet, with the crank in one direction, they are pushed away with the crank in the opposite direction, the same is true with the magnet reversed.

The position of the wire changes, when the crank is turned in one direction, the wire is pushed away, if turned the other direction it is drawn to the magnet. If the magnet is turned upside down, the direct opposite reaction from above occurs.

The length from the table top to the wire is 10 cm. The displacement for 1 rev sec is about .75 cm. The displacement for 2 rev sec is about 1.0 cm. The displacement for 3 rev sec is about 1.5 cm

When I hold the generator in my left hand with the crank to the right and the strips and magnet are on my left, when I crank away from me the strips move away from me, when I crank toward me, the strips move toward me. If I reversed the magnet the reaction of the strips to my cranking would directly opposite. The ratio is displacement from equilibrium divided by the length of the pendulum is equal to the force divided by the weight of the pendulum. For 1 rev/sec we have .75 cm/ 10 cm = x/ 6 g. x = .45 g For 2 rev/sec we have 1 cm/ 10cm = x/6g = .6 g For 3 rev/sec we have 1.5cm/10cm = x/6g = .9 g

Grams don't measure force. The weight of the pendulum is the force exerted on it by gravity, which is 9.8 m/s^2 * .006 kg = .06 N, approx..

So where you have 6 g in your calculations, you should have .06 N. This would give you forces of about .0045 N, .006 N and .009 N.

Predictions 1. In orientation 1 depending on the way I hold the magnet, I believe that on side of the loop will move toward the magnet and one will move away. 2. In orientation 2 again depending on the way I hold the magnet I believe that the loop will move toward the end of the magnet, which ever it is drawn to by the current. 3. In orientation 3 I believe that the loop will be drawn toward one end of the magnet depending on which is drawn by the current.

What is actually happening is that the magnetic force is perpendicular to the magnetic field and to the current. The magnetic field is perpendicular to the surface of the magnet, so the force is parallel to the surface of the magnet. This causes it to appear to be drawn toward one end or another of the magnet.

The segments of the loop will work in only the plain that is not held by the straw. In orientation 1 the loop moves back away from me telling me the force is drawn toward the back side of the magnet In orientation 2 the movement is the best in this orientation. The loop moves best away from me and seems to move about 2-3 cm at the end of the dowel. In orientation 3 the movement is small and hard to see but the right side of the loop is drawn toward the magnet and the other side pushed away. This lets me know the force running through the wire is drawn toward the right side of the magnet. I do not have a pendulum, or remember seeing one in the video’s I watched. But I will answer the questions the best of my knowledge. 1. The force on the individual parts of the loop reacts to the different ends of the magnet. If this did not happen in this manner the loop would not move because being held by a straw lets the loop only rotate. If the magnet ends were the same charge, the loop would not move. With the magnet charged differently on each end the current running through the strips pull toward one end of the magnet and push away from the other. 2. The position of the pointer would represent the direction of the current based on the lay of the magnet. If this were known, then the direction the pointer moves would let us know which way the current was moving. 3. I would design a meter with a pointer with a aluminum strip with a magnet laid under it with two clips connected to each end of the aluminum strip. Experiment 19 The meter I bought from walmart does not have the exact settings as the lab says but I will use the closest possible. 10 V 1 cycle per second = 2.2 v 2 cycle per second = 4 v 3 cycle per second = 6 v 4 cycle per second = 8 v my meter is at 250 ma I got around 5 cranks in 10 seconds Cranking rate is .5cranks per second The closest approximation to the actual volts for .5 cranks per second is 120 ma From my graph above the voltage for .5 cranks per second would be 1 v and from the beep experiment we would have the approximate amperage of 120 ma A smaller current means a greater resistance so we would dived the v by the ma to get a larger resistance. So 1v/120ma = 0.00833

this would be .00833 v / ma, but a ma is a milliamp or .001 amp so you have .00833 v / (.001 amp) or 8.33 V / amp.

This is the same as 8.33 ohms.

My estimated cranking rate per second was 4 cranks per second My voltage across the bulbs was 4 volts Across the .25 bulb my voltage was about 1.8 Across the .15 bulb my voltage was about 4.6 The total of the two bulbs separate is much higher than both bulbs together. With both there would have been 6.4 volts the difference between the two may be a good answer for this question or 2.4 volts. The resistance across the .25 bulb was 25.2 Ohms The resistance across the .15 a bulb was 42 ohms For the .25 a bulb the measured voltage was 1.8 and the resistance is 25.2 Ohms. So the current would be 1.8/25.2 = .0714 a For the .15 bulb the measured voltage was 4.6 and the resistance is 42 Ohms so the current would be 4.6/42 = .110 a The answers I calculated are not equal and should not be equal because the bulbs allow different amperages through. My meter is at the 250 ma setting this is the closest I have. The current flowing through the circuit is about 110 mA This current is fairly close to the current calculated through the .15 bulb and pretty far away from the .25 bulb. Voltage would be resistance * current For the .25 bulb the voltage would be 25.2 * .110 = 2.77 A For the .15 bulb the voltage would be 42 * .110 = 4.62 A For the generator we would have .00833 * .110 = .000916 A This seems small if most of the resistance is in the generator The measured voltage for 4 cranks per second was 8 V the total around the whole circuit was 7.39 so they are fairly close. I still think there is something wrong with my generator calculation.

It's that factor of 1000.

This circuit takes more force to light the bulbs My meter is at 10 V, my cranking rate is 2 per second. The voltage across the .15 a bulb in parallel was 2.4 V The voltage across the .25 a bulb in parallel was 2.6 V Current is voltage divided by resistance so the current is 2.6/25.2 = .103 for .25 2.4/42 = .057 for .15 total current should be .168 for the two bulbs and the generator. The measured cranking rate for the generator was 4 V The measured rate through the bulbs was 5 V so they are not equal. The .25 bulb is brighter in parallel, this bulb also carries the most current. The .15 bulb has the greatest resistance. The brighter the bulb the more current that passes the dimmer the bulb the greater resistance. The parallel would expend less energy, because the cranking rate is 2 less per second and still produces similar amperage. For the single bulb that replaces the two bulbs the resistance would have to be 5v/.168 = 29.76 For 1 cycle per second we could connect the bulbs in parallel so that each bulb is getting equal power and not limited by the prior bulb, and measure the current through each bulb and then measure the current through the meter to see if they are equal. In my procedure The current for the .15a bulb is .150 mA The current for the .25 a bulb is .165 mA The current for the generator is .225 So the totals across the bulbs do not match the current across the generator. My procedure I connected the bulbs in parallel with the .25a bulb first and the .15 bulb behind. The meter was connected to the .15a bulb first and set on 250 mA and cranked at 1 cycle per second. This was also done for the .25a bulb and the results for each were recorded. I then connected the generator and cranked slowly at first to be sure the meter could handle the cranking rate, then cranked at 1 cycle second to get the measurement of the current that passes through the generator. My results were not as anticipated. The voltage across the lights combined were greater than the generator it’s self.

Experiment 20 Coulomb’s Law When I bring the non stick side of the tape toward the piece stuck to the table the piece on the table is attracted to the piece in my hands. The third piece I tore off the role attracts the piece hanging from the table. Since the new piece attracts the old piece on the table, the piece on the block would seem to repel the new piece of tape. My prediction was correct. I would predict that another piece of scotch tape would react the same as the third piece I tore off. I was wrong, the piece repelled both the pieces on the table top. My piece on the block has lost it’s charge now. I would think that two pieces of tape pulled off the roll would be alike because they are obtained in the same manner, by pealing the new piece off a non-sticky side of a like piece on the roll. The charges on the back of the two pealed peaces would be unalike because they attract. I would not think there could be three charges, but I also do not know how I would test for this other than continuing pealing tape to see if anything different happens, whether some odd pieces attract or repel each other and have no effect on the others. My lab kit did not have the pendulums included but I will watch the clips to see how the experiment is done. I have seen that when two pieces of tape with unalike charges get closer the draw between the two is greater.

Experiment 21

The pieces of tape are attractive and the piece closes to me repels the tape and the piece furthest attracts. I picked a piece of tape of the second two pieces, this piece repelled the pipe. I predict this piece will repel the closest piece of tape to me and attract the furthest from me. The other piece of tape will be drawn to my pipe, and attracted to the closest piece of tape to me, and repelled from the furthest from me. I was correct in both of my predictions. We can make these predictions from the knowledge that if one piece of tape has a positive charge it will be attracted to a negative charge and repelled from a positive charge. With the aluminum foil around one end of the pipe, the aluminum end does not have any affect on the closest piece of tape to me, and a very small attraction to the other piece. I still do not believe that there is a 3rd charge but this reaction could be evidence supporting the argument. The properties of this third type of charge would have no reaction to whatever charge the first piece of tape has and be somewhat opposite the second piece of tape. You could test the existence of a third charge by having a known charge of positive and negative and test different properties to find a 3rd charge. In the first charge, the first piece of tape did not have any reaction. In the second charge the second piece of tape had an opposite charge and thus an attraction. My finger attracts the first piece of tape, and thus I would believe that the other would be repelled. In all actuality the second piece of tape was attracted as well. With the aluminum foil between the pipe and the tape, there is no reaction between the tape and the Pvc pipe. Again with the cylinder there was no reaction I did not get any reaction from the aluminum foil piece of pipe that used to be on my pvc pipe. When I place the charged pipe next to the tape it repels the tape. The aluminum seems to keep the pipe from attracting the tape what so ever.

Experiment 22

My times between the volts are as listed below: 10 V 0 0 5 V 1.95 s 1.95 3 V 3.50 s 5.45 2 V 3.71 s 9.16 1 V 6.77 s 15.93 The largest value of V is 10 and has a time of 0 When t = RC we have e^-1, I am taking for granted that e is the constant 2.718….. So we end up with 10V * e^-1 = 3.68 V This takes somewhere in the neighborhood or 3 seconds. RC would have to equal the time or 3 The farad is 1 so we have 3 = RC or 3 = R*1 So R would be 3. This number is much higher for the resistance of the generator than was determined in the last experiment. I would predict that the capacitor will discharge more quickly because the work involved in turning the handle. The times past for the Voltage are as follows: 10v 0 5v 55.11 3v 102.18 2v 130.17 1v 156.78 at around 80 seconds to reach this voltage of 3.68 R would then be 80 through the formula t = RC This R is extremely high compared to holding the handle. The dial on the meter is pulled in the wrong direction when the capacitor is allowed to discharge through the meter. The time I am estimating is about .75 cranks per second For half the time cranking it took 5.25 seconds For half of this time it took 9 seconds For the discharge the initial mA is 100 The times are as follows: 100mA 0 50mA 8.06 25mA 13.5 12.5mA 19.89 the time for t=RC is about 12.62 seconds through interpellation so for c = 1 r= t or 12.62 The crank did not turn the entire time, but I did the best I could determing the times. Times are as follows 100ma,(did not actuailly show this time) 0 50 ma 4.77 25ma 18.19 12.5ma 27.45 the clock time for this is 16.03 seconds and thus so would be R