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INTRODUCTION This piece of physics coursework is an investigation, which will try to find out the factors affecting the resistance of a wire and reasons behind these. Mechanism of conduction: - In metals there is conduction most metals follow ohmic-law but in plastics there is no conduction. If a metal is good conductor of heat then it should also be a good conductor of electricity and visa-versa. In an atom of a metal there are bound electrons, nucleus and free electrons; a plastics atom has no free electrons, while metals have free electrons. The bound electrons are electrons with negative charge in an orbit around the nucleus; the bound electrons cannot leave their orbit. The free electrons are not in orbit but free to go where ever in the metal. The free electrons are the charge carriers, the charge is a negative charge in electricity and in conduction of heat the free electrons carry heat energy, which mean they are responsible for electrical conduction and heat conduction. As the plastics have no free electrons there can be no conduction, but as metals have free electrons this allows conduction to happen in the metals. An atom of a metal: - - + - - - An atom of a plastic: - + - There are no free electrons in the plastic so there will not be any conduction, because the free electrons are the charge carriers, without charge carriers no conduction. The free electrons are sub-atomic particles and have a negative charge so when there is a potential difference voltage the free electrons will move to the positive in one direction this is because opposite charges attract. The free electrons are moving in one direction but randomly, the free electrons don't all travel in straight lines but they are going in the same direction. This is a crystal of a metal or a space lattice: - The atoms are fixed meaning that they have a regular arrangement and that they are rigid and orderly. There will be collisions between the free electrons and the atoms. The atoms may not move and are rigid but they do vibrate. The reason why the free electrons are called charge carriers is because current is the flow of charge and free electrons carry this flow of charge. Free electrons move at high speeds as they travel large distances between collisions with the metal atoms, they can transfer energy at very quickly. Electrical conduction happens best in metals like silver, copper, aluminium, iron. Poor conductors and good resistors of electrical energy are glass, plastics, wood. Good conductors have low resistance. Bad conductors have high resistance. Resistance: - Resistance is the opposing of the flow of current. Electrical resistance is the property where electric flow or electric energy is transformed into heat energy this heat energy will then be opposing the electrical energy. Resistance involves the collisions between the free electrons and the vibrating atoms, to be more precise the collisions between free electrons and the bound electrons which makes up the structure of the conductor. The collisions with the free electrons and the vibrating atoms will slow down the flow of free electrons therefore slow down the flow of electricity. The resistance of a wire is directly proportional to its length and inversely proportional to its cross sectional area. The temperature and the number of collisions are related, if the temperature rises then the atoms will vibrate more. The more vibrations of the atoms, would decrease the space so there will be more collisions and the more collisions will slow down the charge carriers free electrons so there will be more resistance. From this diagram we can see that the charge carriers the free electrons are having more collisions because the atoms are vibrating more because of the higher temperature. P2a SAFETY Safety will always be a vital factor when carrying out any experiment. This experiment will not involve the use of electricity through the mains. This amount of electricity will be around 240volts coming through the mains, this is too much because it is dangerous enough too kill you. So the voltage should be lowered in this experiment so that is why cells are used and not the mains, this also is could because a low voltage would keep the temperature of the wire the same. It is much safer to experiment with circuits powered by low voltage source. In the experiment to make the level of voltage to be safe I would use battery cells which have low voltages, the lowest voltage the cells could be 2 volts and the highest would be 5volts so this source would be much safer. Another safety precaution that I will take is will I will not work near any desktops with water taps. The reason for this is that if water goes into the circuit or more importantly into the electrical socket it could results to hazards like fire hazards, so I will for safety do the experiment on a side desktop away from gas taps and water taps. P4a FAIR TEST For this experiment to be a fair and get reliable results I will have to undertake a fair text. The fair test will have to undertake these following things: - The non-variables or the things that I will keep the same through out the experiment will be the temperature; at room temperature cross sectional area of wire which is 0.5mm or 0.05cm, the material of the wire will be kept the same which will be Nichrome. I will also keep the same amount of cells which will be three cells throughout the experiment; keeping the same amount of cells the same will make the experiment a fair test because if the amount of cells were changed each time then this will affect the results because if the cells were changed the starting voltage of the power supply would be different so you the results would not be reliable. When the temperature of the wire is increased then the atoms of the wire will vibrate more and this will decrease the space in the space lattice so there will be more collisions and there will be more resistance. We know that if the temperature is not kept the same throughout the experiment it will not be a fair test, because of the changing temperature. If you keep the non-variables the same then you can get correct and fair results as you change the variable which in this case is the length of the wire. Each length will be repeated twice and the average taken to get fair results. P4b LIST OF EQUIPMENT AND APPROPRIATENESS Equipment/Instruments: - "¢ Ammeter "¢ Voltmeter "¢ Crocodile clips "¢ Three cells "¢ Nichrome wire "¢ Meter ruler "¢ Board to put cells in. For the measurement instruments they will need a least count for appropriateness and consistency. The least count for the meter ruler will be 1mm or 0.1cm. The voltmeter will be in volts and the least count will be 0.1volts. For the ammeter the measurement will be amperes and the least count will be 0.1amperes. P6a THEORY AND PREDICTION The factors that affect the resistance of a wire include the following: - "¢ Material "¢ Temperature "¢ Cross sectional area of wire "¢ Length of a wire The material of a wire has three things that affect the resistance: 1. The type of metal: - each different types of metal are structured differently; their space lattice or crystals are different to one another. The interatomic distance was big then there would be fewer collisions so there will be less resistance. If the interatomic distance in the structure of the metal was small there would be more collisions therefore more resistance because the free electrons flow has been slowed. 2. The amount of free electrons: - this also affects the resistance, if the amount of free electrons in one metal is more than in another metal then the metal with more free electrons will have less resistance. The more the number of free electrons then less resistance, the less the number of free electrons then the more the resistance. The reason for this is that if there are more free electrons means that more electrical charge is been carried so there is less resistance. 3. Arrangement of atoms in the metal: - if the structure of the atom or arrangement is different then this can affect the resistance. If the arrangement of the atoms has an atom in the middle there would be more collisions so there will be more resistance. If the structure had no atom in the middle there would be fewer collisions therefore less resistance. This diagram shows two differently arranged atoms one is the body-centred cubic bcc arrangement the other is faced-centred cubic FCC arrangement. The bcc arranged atom has less atoms and only one atom in the middle so it would have less collisions than the FCC arrangement because the FCC arrangement has more atoms and more in the middle, this will increase the amount of collisions and then there will be more resistance. The temperature will affect the resistance of a wire. A falling temperature will increase the conductivity. Most of the resistance to the motion of the free electrons therefore resistance of the wire is the thermal vibrations of the atoms. If the temperature is reduced to absolute zero then thermal vibrations will stop, then this will stop the resistance due to temperature. The cross-sectional area of a wire will affect the resistance. A larger cross sectional area will mean a larger current I and that makes smaller resistance. A smaller cross-sectional area has a smaller current which is a bigger resistance. The resistance of a wire is inversely proportional to its cross-sectional area. The length of a wire affects the resistance. If I were to double the length of the wire this would double the collisions, so a longer wire will have a bigger resistance. Doubling the length would double the number of collisions and therefore will double the resistance. This means that the resistance is directly proportional to the length of the wire. When the length is more than before there are more number of free electrons and when there is more free electrons there will be more collisions, the more collisions will slow down the charge carriers free electrons and this would cause bigger resistance. I predict that if the length was doubled so will the resistance, I can also predict that if the length was tripled the resistance will also triple, this is because of the rule the resistance of a wire is directly proportional to the length of the wire. L no: of collisions/sec. Resistance=R 2L 2N Resistance= 2R p8a Ohms law: - these are laws that relate to current, potential difference, and resistance. The resistance is measured in ohms, the potential difference is measured in volts and the current is measured in amperes amps. Ohms law states the amount of steady current through materials is directly proportional to the potential difference Voltage. An example that proves this is that if the potential difference voltage between two ends of a wire is tripled, the current is tripled. Ohms law also says that I=V/R, or the current in the conductor equals the potential difference voltage across the conductor divided by the resistance of the conductor. It also says that the potential difference across a conductor equals the current in the conductor times its resistance, V=IxR. Ohms law also explain that resistance of a material is its potential difference voltage divided by the current, R=V/R. The larger the resistance the greater the voltage is needed to push each ampere through it: there is a resistance of one ohm and a voltage p.d of one volt will drive a current of one ampere through it. Conductors that obey ohms law are called ohmic conductors; conductors that don't obey ohms law like semiconductors are called non-ohmic conductors. Resistivity: - is the electrical resistance of a conductor of unit cross-sectional area and unit length. A property of each material resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents. The unit of resistivity is ohm-meter or ohm-centimetre. A high resistivity is in poor conductors, a poor conductor has high resistance, so high resistivity has high resistance and a conductor with low resistivity will have low resistance. Resistivity P is quantitatively equal to the resistance R of e.g. a wire times its cross-sectional area A and then divided by its length L. P =R A /L. Resistance = R= P L /A p8a Method: - 1. First I will connect the cells to the circuit board where the cells are placed. 2. Then I will then use the three pairs of connecting wire crocodile clips to connect the cells to the ammeter and the wire and back to the cells to make a complete circuit. I will also add a voltmeter opposite to the power pack/in parallel. 3. Thirdly the wire will be on a board and on that there will be a ruler so I can measure the different lengths. 4. Next, when I have measured one length I will break the circuit by taking off a connecting wire, this is so that the wire does not change temperature and affect results. 5. Then when I am ready to take the next length I will connect the circuit and move the crocodile clip to the new length. 6. I will do all the lengths three times to find reliable readings. I will also have a good range of lengths so I can find a full set of results to see how the length of a wire affects the resistance of the wire. Cells Voltmeter V Before the power is put on the ammeter is at zero. The point where the crocodile clip and the wire meet is at zero. The ammeter has two scales and the first one is 0.00amps to 1.00amps this is for a low voltage that I am using so this is the appropriate scale that I should use as I am using low voltage cells. The voltmeters least count is 0.01volts and before the experiment the voltmeter was at absolute zero. In my method I could have used digital ammeter but there was not one provided, a digital display ammeter would give me even more accurate readings. p6b NO: OF READINGS The cross-sectional area will remain at a constant of 0.05cm through out the experiment. READING NUMBER ONE READING NUMBER TWO READING NUMBER 3 AVERAGE RESISTANCE Average= 1.d.p Length of Nichrome wire L+-0.01cm cm Voltage Volts v+- 0.01v 2d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p 05.00cm 10.00cm 15.00cm 20.00cm 25.00cm 30.00cm 35.00cm 40.00cm 45.00cm 50.00cm 55.00cm 60.00cm 65.00cm 70.00cm 75.00cm 80.00cm 85.00cm 90.00cm 95.00cm 100.00cm This is my table where I am going to show all my results for my experiment. I will take the readings from the voltmeter to find the potential difference voltage, I will take the ampere readings to find the current I. The table has three sets of reading to put in because this means I am repeating the readings and trying to find consistent results and reliable results, any anonymous results will be marked with an asterisk and then repeated again. The average will be used to find average resistance for the three sets of results and see if they are reliable e.g. 3.1Ω, 3.2Ω, 3.1Ω average will be 3.1Ω+3.2Ω +3.1Ω=9.4Ω divided by 3= 3.1Ω. The average can let me set a graph showing how the average resistance against the length of a wire. The final column will let me work out the resistance once I have collected all the data. The table has a suitable range for the length of the wire, this is because this is my variable and I need to get a full range of results. The table has the least counts of scale units and the decimal places I am using. p8b PRELIMINARY WORK The dissipation of electric energy in the form of heat, even though small, affects the amount of electromotive force, or driving voltage, required to produce a given current through the circuit. In fact, the electromotive force V measured in volts across a circuit divided by the current I amperes through that circuit defines quantitatively the amount of electrical resistance R. Precisely, R = V/I. Thus, if a 12-volt battery steadily drives a 2-ampere current through a length of wire, the wire has a resistance of 6 volts per ampere, or 6 ohms. Ohm is the common unit of electrical resistance, equivalent to one volt per ampere and represented by the capital Greek letter omega Ω. The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. Resistance also depends on the material of the conductor. The Electrical resistance of a conductor is dependent on cross-sectional area and length. A characteristic property of each material, resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents. High resistivity designates poor conductors. Most resistance is down to the motion of free electrons comes from the thermal vibration of the atoms; if the temperature is reduced to almost absolute zero, where thermal motion essentially stops, conductivity can increase several thousand times over. The value of resistivity depends also on the temperature of the material; tabulations of Resistivities usually list values at 20 C. Resistivity of metallic conductors generally increases with a rise in temperature; but resistivity of semiconductors, such as carbon and silicon, generally decreases with temperature rise. The resistivity of an exceedingly good electrical conductor, such as hard-drawn copper, at 20 C 68 F is 1.77 10-8 ohm-metre, or 1.77 10-6 ohm-centimetre. At the other extreme, electrical insulators have Resistivities in the range 1012 to 1020 ohm-metres. The unit of resistance is the ohm. In the metre-kilogram-second mks system, the ratio of area in square metres to length in metres simplifies to just metres. Thus, in the metre-kilogram-second system, the unit of resistivity is ohm-metre. If lengths are measured in centimetres, resistivity may be expressed in units of ohm-centimetre. OBSERVATIONS o4b o4a o6b o6b table of results: - The cross-sectional area will remain at a constant of 0.05cm through out the experiment. READING NUMBER ONE READING NUMBER TWO READING NUMBER 3 AVERAGE RESISTANCE Average= 1.d.p Ω Length of Nichrome wire L+-0.01cm cm Voltage Volts v+- 0.01v 2d.p Current I Amperes amps Least count 0.01amps 2.d.p Resistance Ohms R=resistance V= voltage p.d I= current R=V/I 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p Resistance Ohms R=resistance V= voltage p.d I=Current R=V/I 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p Resistance Ohms R=resistance V= voltage p.d I= current R=V/I 2.d.p 05.00cm 1.18v 2.22 0.53 Ω 1.18v 2.22 0.53 Ω *2.90v 1.18v *2.97 2.22 0.53 Ω 0.5 Ω 10.00cm 1.64v 1.91 0.86 Ω 1.64v 1.91 0.86 Ω 1.63v 1.90 0.86 Ω 0.9 Ω 15.00cm 1.99v 1.67 1.19 Ω 1.99v 1.67 1.19 Ω 1.99v 1.67 1.19 Ω 1.2 Ω 20.00cm 2.27v 1.49 1.52 Ω 2.27v 1.49 1.52 Ω 2.27v 1.49 1.52 Ω 1.5 Ω 25.00cm 2.49v 1.34 1.86 Ω 2.49v 1.34 1.86 Ω 2.51v 1.33 1.89 Ω 1.9 Ω 30.00cm 2.67v 1.22 2.19 Ω 2.67v 1.22 2.19 Ω 2.67v 1.21 2.21 Ω 2.2 Ω 35.00cm 2.82v 1.12 2.52 Ω 2.83v 1.11 2.55 Ω 2.82v 1.12 2.52 Ω 2.5 Ω 40.00cm 2.95v 1.03 2.86 Ω 2.95v 1.03 2.86 Ω 2.95v 1.03 2.86 Ω 2.9 Ω 45.00cm 3.06v 0.96 3.19 Ω 3.06v 0.95 3.22 Ω 3.06v 0.96 3.19 Ω 3.2 Ω 50.00cm 3.15v 0.90 3.50 Ω 3.15v 0.90 3.50 Ω 3.15v 0.90 3.50 Ω 3.5 Ω 55.00cm 3.24v 0.84 3.86 Ω 3.22v 0.83 3.88 Ω 3.22v 0.83 3.88 Ω 3.9 Ω 60.00cm 3.31v 0.75 4.41 Ω 3.31v 0.75 4.41 Ω 3.31v 0.75 4.41 Ω 4.4 Ω 65.00cm 3.38v 0.73 4.63 Ω 3.38v 0.73 4.63 Ω 3.38v 0.73 4.63 Ω 4.6 Ω 70.00cm 3.43v 0.71 4.83 Ω 3.43v 0.71 4.83 Ω 3.43v 0.71 4.83 Ω 4.8 Ω 75.00cm 3.49v 0.68 5.13 Ω *4.00v 3.49v *0.98 0.67 *4.08 Ω 5.21 Ω 3.49v 0.68 5.13 Ω 5.2 Ω 80.00cm 3.54v 0.64 5.53 Ω 3.54v 0.64 5.53 Ω 3.54v 0.64 5.53 Ω 5.5 Ω 85.00cm 3.58v 0.61 5.87 Ω 3.58v 0.61 5.87 Ω 3.58v 0.61 5.87 Ω 5.9 Ω 90.00cm 3.62v 0.59 6.14 Ω 3.62v 0.59 6.14 Ω 3.62v 0.59 6.14 Ω 6.1 Ω 95.00cm 3.66v 0.56 6.54 Ω 3.66v 0.56 6.54 Ω 3.66v 0.56 6.54 Ω 6.5 Ω 100.00cm 3.69v 0.54 6.83 Ω 3.69v 0.54 6.83 Ω *4.29v 3.70 *0.32 0.54 6.85 Ω 6.8 Ω o8a: - *note: the circled and crossed out results are anonymous so I repeated the reading to get a reliable reading and result, I did this because I thought that the reading was wrong and I needed to do it again, it was common sense because the anomalous results obviously didn't follow certain patterns in the table. I did the reading again and I got the same as I got in the other two sets of readings for that length. o6b: - the table is an accurate record of observations because each column of reading is to the same decimal place and is accurate. * In the planning I planned to use an analogue ammeter and voltmeter but I was able to use a digital ammeter and voltmeter as the teacher could get one in the last lesson of the practical. * I rounded the final average resistance to one decimal place because this is common sense and this will make it easier to plot the graph. ANALYSIS A.2a After completing the experiment and getting a full range of reliable results from the experiment, I have found out that in general that as the length of the wire goes up the resistance of the wire also goes up. This states that my theory and prediction is correct. I have found out that this general finding says that the resistance of the wire is proportional to the length of the wire and in many occasions the length is directly proportional to the resistance of the wire. A4.b PATTERN IN READINGS: - After collecting my results from the observations I carried out I can now analyse the results and make sense of the readings in the table of results. The table of results has a general pattern that suggests that as the length of the wire is increased the current decreases and therefore the resistance increases as my theory states through different rules. This general finding means that the resistance of the wire is proportional to the length of the wire. The readings from the table of result shows as the length of the Nichrome wire increases the current in the circuit decreases and as the theory explains the resistance should increase and this occurs in the table of readings. The readings in my results table show that the length is proportional to the resistance of a wire and in some cases nearly directly proportional and other cases exactly the length is directly proportional to the resistance of the wire. E.g. the result from 5cm to 10cm, the average resistance for 5cm is 0.5Ω and the average resistance for 10cm is 0.9Ω, this shows that the length of the wire is proportional to the resistance, if it was directly proportional then the resistance for 10cm will be 1.0Ω, this is because when the length is doubled then the resistance is doubled this then would be directly proportional, this result was not accurate by 0.1 because I could of made minor human error like not putting the exact length. The results from 30cm has the resistance of 2.2Ω and the and the doubled length that is 60cm has the resistance of 4.4Ω, this shows that as the length is doubled the resistance is doubled which means that the length is directly proportional to the resistance of the wire, these result from 30cm to 60cm proves that my theory that the length of wire is directly proportional to the resistance of a wire. The mistakes I made called experimental like not breaking the circuit when changing the length could of have affected the temperature of the wire therefore create more vibrations for the atoms in the wire and cause a bigger resistance then it should have been without the experimental error. A4a A6a A4b Graph of results: - A4b PATTERN ON GRAPH: - The pattern that appears on the graph of results tells us that the length of the wire is directly proportional to the resistance of the wire. The line of best fit provides the graph with the results. The results that prove that the resistance of a wire is directly proportional to its length include the result of the lengths 15cm to 30cm, 15cm has the resistance of 1.1Ω and the resistance of 30cm is 2.2Ω this shows as the length is doubled the resistance also is doubled which means that the resistance of the wire is directly proportional to the length of the wire, this result shows the graph of results to state my theory and to be correct as the result agrees with my theory. Other results in my graph also state that the resistance of the wire is directly proportional to its length. They include: - 26cm that had the resistance of 1.9Ω and then when the length was doubled to 52cm the resistance was 3.8Ω, this result shows that as the length is doubled the resistance is also doubled which means that the length is directly proportional to the resistance of a wire, the result again proves my prediction correct because I predicted that the length of wire is directly proportional to resistance of the wire, the results from the graph state this with these results. The other results from the graph also tell us that the length is proportional to the resistance of the wire they include: - 5cm R=3.5Ω and 10cm R=7.5Ω, 20cm R=1.5 & 40cm R=2.9Ω. These results that are not directly proportional but it is only by a few 0.1 ohms this could be down to not drawing a 100% straight line and this cause these results to be a bit off. A6b A8a: - explanation using theory From my readings in the experiment the results from table which I've conducted, I have found results that show the length of the wire is proportional to the resistance of the wire. On occasions in the readings I have seen that the length of the wire was directly proportional to the resistance of the wire. The results 30cm R=2.2Ω, 60cm R=4.4Ω shows that as the length is doubled the resistance is doubled as well which means that the length is directly proportional to the length as I predicted. This result proves my theory that if the length is doubled the number of collisions will also double then so will the resistance be doubled; this shows that collisions increase the resistance will also increase. This result proves my theory which states that if I were to double the length of the wire this would double the collisions, so a longer wire will have a bigger resistance. Doubling the length would double the number of collisions and therefore will double the resistance. In detail my theory explains this result, when the length is more than before there are more number of free electrons and when there is more free electrons there will be more collisions, the more collisions will slow down the charge carriers free electrons and this would cause bigger resistance. If a length was doubled this would double the amount of free electrons and double the amount of collisions between the free electrons and the vibrating atoms, this would slow down the free electrons the charge carriers x 2 times by two and therefore double the resistance, this is the case in the result 30cm R=2.2Ω, 60cm R=4.4Ω, so the results from the table of readings state my theory and prediction, this also proven by the diagram theory I did: - From the graph I have found out patterns that generally tell me that the length of the wire is directly proportional to the resistance of the wire. This was the case in the graph results: - 15cm to 30cm as my theory explains when the length is doubled the resistance should also doubled this was the case it went from 1.1Ω ohms to 2.2Ω, so this means that the length of the wire was directly proportional to the resistance of the wire, this states my theory which says 'I predict that if the length was doubled so will the resistance, I can also predict that if the length was tripled the resistance will also triple, this is because of the rule the resistance of a wire is directly proportional to the length of the wire.' The result 15cm=1.1 ohms and 30cm=2.2 ohms suggests that when the length is doubled the number of collisions will also double then so will the resistance be doubled, this shows that collisions increase the resistance will also increase. This result proves my theory which states that if I were to double the length of the wire this would double the collisions, so a longer wire will have a bigger resistance. Doubling the length would double the number of collisions and therefore will double the resistance. In detail my theory explains this result, when the length is more than before there are more number of free electrons and when there is more free electrons there will be more collisions, the more collisions will slow down the charge carriers free electrons and this would cause bigger resistance. If a length was doubled this would double the amount of free electrons and double the amount of collisions between the free electrons and the vibrating atoms, this would slow down the free electrons the charge carriers x 2 times by two and therefore double the resistance, this was proven by the results; 15cm R=1.1Ω and 30cm R= 2.2Ω, this is exactly what I predicted through my theory. The resistivity is quantitatively equal to the resistance, from the results this would be the resistance R of the wire times the cross-sectional area A divided by the length L of the wire. From this you can find out if the resistance is directly proportional to the length and inversely proportional to the cross-sectional area and see if your results are correct. P =R A /L. E.g. A=0.05cm × R=6.83Ω ÷ 100.00cm= 0.3415 ohmic-centimetre = resistivity. To prove the result is correct I will know find out the resistance R using this equation: - R= P L /A. P resistivity =0.3415 × 100.00cm ÷ A 0.05cm = 683 ohmic-centimetre, this is quantitatively equal to 6.83Ω. So, these results show my theory of resistivity is correct and that my results are correct and follow the theory. Results are correct because the resistivity and the resistance are quantitatively equal. This result also shows that my main prediction that is, the length of the wire is directly proportional to the resistance of the wire; this prediction was proven by this result as the theory of resistivity and the equations P =R A /L and R= P L /A were worked out using my result that showed that the resistance and resistivity were quantitatively equal this is proof that my results are reliable and follow the theory of rules in resistance and resistivity and states my prediction. A8b: - conclusion of results and analysis. Throughout the analysis of the results I have found many results from my readings in the table of results and the results from the graph, they show the relationship of the length of the wire and to the resistance of the wire. The general conclusion I have made after analysing these results is that the length of a wire is directly proportional to the resistance of a wire. The proof of this conclusion lies within the results in the readings and the graph which shows the relationship between the length of the wire and its resistance. The results from the table of readings and the graph do agree with my theory this shows that my results are proof of the theory. This is shown e.g. in the experiment the results from table which I've conducted, I have found results that show the length of the wire is proportional to the resistance of the wire. On occasions in the readings I have seen that the length of the wire was directly proportional to the resistance of the wire. The results 30cm R=2.2Ω, 60cm R=4.4Ω shows that as the length is doubled the resistance is doubled as well which means that the length is directly proportional to the length as I predicted. This result proves my theory that if the length is doubled the number of collisions will also double then so will the resistance be doubled; this shows that collisions increase the resistance will also increase. This result proves my theory which states that if I were to double the length of the wire this would double the collisions, so a longer wire will have a bigger resistance. Doubling the length would double the number of collisions and therefore will double the resistance. In detail my theory explains this result, when the length is more than before there are more number of free electrons and when there is more free electrons there will be more collisions, the more collisions will slow down the charge carriers free electrons and this would cause bigger resistance. If a length was doubled this would double the amount of free electrons and double the amount of collisions between the free electrons and the vibrating atoms, this would slow down the free electrons the charge carriers x 2 times by two and therefore double the resistance, this is the case in the result 30cm R=2.2Ω, 60cm R=4.4Ω. Also shown in from the graph I have found out patterns that generally tell me that the length of the wire is directly proportional to the resistance of the wire. This was the case in the graph results: - 15cm to 30cm as my theory explains when the length is doubled the resistance should also doubled this was the case it went from 1.1Ω ohms to 2.2Ω, so this means that the length of the wire was directly proportional to the resistance of the wire, this states my theory which says 'I predict that if the length was doubled so will the resistance, I can also predict that if the length was tripled the resistance will also triple, this is because of the rule the resistance of a wire is directly proportional to the length of the wire.' The result 15cm=1.1 ohms and 30cm=2.2 ohms suggests that when the length is doubled the number of collisions will also double then so will the resistance be doubled, this shows that collisions increase the resistance will also increase. This result proves my theory which states that if I were to double the length of the wire this would double the collisions, so a longer wire will have a bigger resistance. Doubling the length would double the number of collisions and therefore will double the resistance. In detail my theory explains this result, when the length is more than before there are more number of free electrons and when there is more free electrons there will be more collisions, the more collisions will slow down the charge carriers free electrons and this would cause bigger resistance. If a length was doubled this would double the amount of free electrons and double the amount of collisions between the free electrons and the vibrating atoms, this would slow down the free electrons the charge carriers x 2 times by two and therefore double the resistance, this was proven by the results; 15cm R=1.1Ω and 30cm R= 2.2Ω, this is exactly what I predicted through my theory. This result I worked out about resistivity also can prove that the results are proof of the theory. P =R A /L. E.g. A=0.05cm × R=6.83Ω ÷ 100.00cm= 0.3415 ohmic-centimetre = resistivity. To prove the result is correct I will know find out the resistance R using this equation: - R= P L /A. P resistivity =0.3415 × 100.00cm ÷ A 0.05cm = 683 ohmic-centimetre, this is quantitatively equal to 6.83Ω. So, these results show my theory of resistivity is correct and that my results are correct and follow the theory of resistivity and show results are correct so the results are the proof of the theory. EVALUATION E.2a: - This coursework that I undertook on the factors that affect the resistance of a wire was quite challenging but at the same time enjoyable to carry out. The reason for which I found it enjoyable was that the physics topic the coursework came under is one of my favourite areas as I have an interest within physics. The coursework to me was easy in parts but other parts of the coursework were challenging as I had to implicate the theory to real experiments and I also enjoyed the independent research that was needed. The coursework made me understand the factors affecting the resistance of a wire, and it also allowed me to understand certain laws in physics like the ohms law, and the theory of conduction. E.4a: - The coursework I took on was accurate this could be proven by the results and the graph. The points that lie on my graph is quite adequate out of twenty points of results ten where on the graph line of best fit the other eight were only a few decimal places on the graph below or above, they were slightly off the line of best fit, while only two were anomalous but by only small margins. The anomalous results were for 60cm and 10cm readings as the points were the furthest away from the graph or line of best fit. The accurateness was also proven by the graph passing through the origin which is the co-ordinates 0, 0, this level of accuracy suggest that my results and graph is reliable and accurate. E.4b: - evaluation of method For this coursework the method was very important because if the method was faulty and had a lot of gaps in it the results would be affected therefore most of the coursework. The method that I used was generally a good one to use as with anything it could have improved e.g. I could have used a power pack connected to the mains as the power supply, the power pack could easily change the voltage you want and keep it the same, this would also subsequently avoid complications of finding the voltage with a voltmeter and make the method quicker. A power pack would also allow me to switch of the circuit and not affect the temperature and then the results. In the method instead of taking the readings at every 5cm I could take it at every 2cm or even 1cm to give a better range and reliable results. Other improvement I could make to the method would be to make more accurate by putting the crocodile clips together without any part of the clip not connected. These changes and improvement would make my method much better and get rid of the faults that affected the results I got. E.6a: - evidence. A table to show the reliability of the results and the anomalous results present. LENGTH cm RESISTANCE From GRAPH Ω ohms RESISTANCE from EXPERIMENT Ω ohms DIFFERENCE 2.d.p % DIFFERENCE =diff÷ R graph × 100 1d.p * 5.00 0.35 0.50 0.15 *42 % 10.00 0.80 0.90 0.10 1 % 15.00 1.10 1.20 0.10 *9 % 20.00 1.50 1.50 0.00 0 % 25.00 1.82 1.90 0.08 4 % 30.00 2.20 2.20 0.00 0 % 35.00 2.52 2.50 0.02 1% 40.00 2.90 2.90 0.00 0 % 45.00 3.25 3.20 0.05 1 % 50.00 3.55 3.50 0.05 1 % 55.00 3.90 3.90 0.00 0 % 60.00 4.25 4.40 0.15 3 % 65.00 4.60 4.60 0.00 0 % 70.00 4.91 4.80 0.11 2 % 75.00 5.25 5.20 0.05 1% 80.00 5.60 5.50 0.10 1 % 85.00 5.90 5.90 0.00 0 % 90.00 6.20 6.10 0.10 1 % 95.00 6.50 6.50 0.00 0 % 100.00 6.80 6.80 0.00 0 % AVERAGE ERROR = 3.35 % Here you can see the average error in % is acceptable within experimental error, this is evidence to justify that my conclusion, because it is within the scope of experimental error so the results I got should be considered reliable because it is a low % of error, so this would prove that the length of the wire is directly proportional to the resistance of the Nichrome wire and also prove my analysis and conclusion to be reliable. In the table which shows the reliability of results also showed that my experiment had only two anomalous results, this is acceptable because within the experiment I could have made only one significant error to cause this anomalous result. The rest of the results were reliable as the table shows. Length = 35.00cm, Resistance = 2.50 Ω Doubled = 70.00cm, Resistance = 4.80 Ω 0.20 Ω off 5.0 Ω 0.20 ÷ 5.0 × 100% = 4 %. This is within experimental error. Length =30.00 cm, Resistance = 2.20 Ω Doubled = 60.00cm, Resistance = 4.40 Ω 0.00 Ω off 4.40 Ω 0.00 ÷ 4.40 × 100% = 0%. This is within experimental error, because there is no experimental error in this result. Length = 15.00cm, Resistance = 1.10 Ω Doubled = 30.00cm, Resistance = 2.20 Ω 0.00 Ω off 2.20 Ω 0.00 ÷ 2.20 × 100% = 0%. There is no experimental error here. Length = 35.00cm, Resistance = 2.50 Ω Doubled = 70.00cm, Resistance = 4.80 Ω 0.20 Ω off 5.00 Ω 0.20 ÷ 5.00 × 100% = 4%, again this is within experimental error. All these results I have showed show that the experimental error was low or there was no experimental error, so the results can be expected to be reliable and therefore the results are reliable enough to justify my conclusion. With the table showing the reliability of results with the difference and the individual evidence of results I have picked out it can be said my results are reliable. E6b: - further work. For the coursework if I were to do it again I would do further work to get more evidence. The further that I would carry out to get more evidence would consist of doing a wider range of lengths of wire and also to try other materials, like copper, constantan in the experiment, all this so I could get more evidence. Further work could also consists of choosing another factor and investigate how that factor affects the resistance of the wire. The factor could be the S.W.G or the thickness of the wire. The principle of the S.W.G gauge is that the electrical resistance of a wire changes when either stretched or compressed or dependent on the thickness. If I were to the experiment for the factor of the thickness I would need to have a new circuit, equipment and a new method for the experiment. The equipment that will be need is:- 1. Crocodile clips or connecting wire. 2. Power pack. 3. Ammeter. 4. Voltmeter. 5. Constantan or Nichrome wire depending on what you want to use. Method: - Power pack Voltmeter 1. First I would connect the power pack to the main power but not turning on the switch for safety purposes. 2. I then would set the power pack to A.C and the voltage that could be suitable a voltage between 4 volts to 10 volts could be used. 3. Then I would have to connect the circuit, using the connecting wire and crocodile clips. I will connect the voltmeter opposite to the power pack and the ammeter anywhere in the circuit. 4. then I would have to connect the circuit to the wire using the crocodile clips I will have to make sure that al of the crocodile clip is in contact with the wire at both sides. 5. I will take the readings for that specific thickness of the wire then remove the wire putting another thickness then taking the readings, I will take the readings to the appropriate range of thickness that would have to be chosen. 6. When changing the wires I will have to shut the power pack so the circuit doesn't over heat and also the wire doesn't heat up because this can affect the resistance. Table to show how I would collect the readings for the experiment: - READING NUMBER ONE READING NUMBER TWO READING NUMBER THREE AVERAGE RESISTANCE Average= 1.d.p Cross sectional area/thickness of wire A A+-0.1cm cm Voltage Volts v+- 0.01v 2d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p 05.00cm 10.00cm 15.00cm 20.00cm 25.00cm 30.00cm 35.00cm 40.00cm 45.00cm 50.00cm [7,657] WORDS
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INTRODUCTION This piece of physics coursework is an investigation, which will try to find out the factors affecting the resistance of a wire and reasons behind these. Mechanism of conduction: - In metals there is conduction most metals follow ohmic-law but in plastics there is no conduction. If a metal is good conductor of heat then it should also be a good conductor of electricity and visa-versa. In an atom of a metal there are bound electrons, nucleus and free electrons; a plastics atom has no free electrons, while metals have free electrons. The bound...
count 0.01amps

2.d.p R=V/I

Resistance

Ohms

R=resistance

V= voltage p.d

I= current

2.d.p Voltage

Volts

v+- 0.01v

2.d.p Current I

Amperes amps Least count 0.01amps

2.d.p R=V/I

Resistance

Ohms

R=resistance

V= voltage p.d

I= current

2.d.p

05.00cm

10.00cm

15.00cm

20.00cm

25.00cm

30.00cm

35.00cm

40.00cm

45.00cm

50.00cm

[7,657] WORDS

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Aim: To investigate the heat... Aim: To investigate the heat loss of the insulating materials, bubble wrap, cotton wool, blanket and foil Introduction: I will carry out this investigation by measuring the temperature of hot water in beakers insulated with these materials. Naturally the insulated beaker with the warmest water will have prevented the most heat loss. To understand how materials prevent heat loss I need to look at the three ways in which heat can be transferred or lost to the surroundings. One of the ways is conduction. This is when heat energy is transported along an object from the hotter region to the cooler region. The object itself does not move. So when an object is heated, its atoms start vibrating with the heat energy. The free electrons begin to diffuse through the object and collide into other electrons transferring kinetic energy. Substances that have particles close together, like metals are good conductors. Meanwhile gases have particles that further apart so they are poor conductors. Therefore things that are made of particles that are not close together, are poor conductors but good insulators such as gases. Another way is convection, which only happens in liquids and gases. This is when heated particles start moving faster so they move further apart. The heat also makes the particle expand so they become less dense than the unheated particles. As a result of this, the heated particles rise upwards taking the extra energy with them and are replaced by colder and denser regions. A form of convection is in evaporation where the particles in a liquid keep bumping into each other. Sometimes they collide with each other with kinetic energy. During these collisions some particles receive so much energy that they can break away from the liquid. The final way in which heat can be lost is through radiation. This is the transfer of heat energy by waves. Every object sends out infra red radiation but hotter objects send out more. This is completely different to conduction and convection as it does not require particles, so the so infra-red radiation can pass through a vacuum. Also dull and black surfaces emit more heat while shiny and white surfaces emit less. However this makes black and dull surfaces better absorbers Prediction: With the kinetic theory on mind, I predict that out of the materials of foil, bubble wrap, wool and blanket the bubble wrap will be the better insulator. This is because the bubble wrap has a lot of air trapped in its pockets. Air is a very good insulator as its particles are widely spread out therefore there is a less chance of them colliding and passing energy on. This means that there will be less energy transferred to the surroundings, as the bubble wrap would prevent heat loss mainly through conduction. Furthermore the material is made from plastic which in itself is a good insulator. As the aluminium foil is shiny it will reduce heats loss through radiation by reflecting heat back in, like a flask. However since it is a metal, some heat will be lost through conduction. Naturally the blanket material would be a good insulator as its purpose is to keep people warm. This because it has lots of layers and like the wool has air trapped between it fibres. So by reducing heat loss through conduction it should be a good insulator but not as good as the bubble wrap. This is because as it is a dull and black colour some heat will be lost through radiation. The cotton wool is also material that has a lot of trapped air in between its many fibres. Even though it is a good insulator, I feel that because of the thinness of the material the bubble wrap will still be the better insulator. All these materials except the foil are lagging which are insulating materials that have tiny air pockets that trap air to prevent heat from being conducted away. Safety: Since I am dealing with very hot water there are precautions I will need to take. Firstly I will abide to the general laboratory rules i.e. tidy workspace, loose clothing tucked away, and safety goggles. When I need the water for my experiment I will my teacher to pour the hot water. Preliminary investigation: For my investigation I need to know at what intervals I should record a temperature reading and for what period of time. Here are the results of my preliminary investigation: Time Secs Temperature °C 0 84.5 30 80.5 60 77.0 90 75.5 Time Mins Temperature °C 0 84 1 80 2 74 3 72 These results show that taking the temperature readings at 30 second intervals would make the results more accurate and to do this over a 5 minute period. I will also use 80 ml of boiling water and make sure all starting temperatures are the same. Range of results: My results will fit into a table like this, TIME SECS TEMPERATURE OF WATER IN BEAKER ºC CONTROL WOOL BUBBLE FOIL BLANKET Apparatus: In my experiment I will use the following equipments: Electric Kettle "“ To boil the water 5 Beakers "“ To hold the hot water for the experiments Aluminium Foil Bubble Wrap Cotton Wool Blanket Thermometer "“ To measure the temperature Stopwatch "“ To time the experiments Elastic Bands "“ To hold the materials in place Variables: In my experiment the volume of the water for each beaker will be 80 ml. Temperature readings will be taken at 30 second intervals for 5 minutes and I will make sure all beakers have the same starting temperature. Therefore there will 10 readings after the starting temperature. This will ensure my test is fair. The only thing that will be different is the material around the beakers. Method: First of all I will wrap four of the beakers in the materials with the elastic bands. The other beaker will stay uncovered as it is the controlled experiment. After placing the thermometer into the beaker, the 80 ml of water will be poured in and the stopwatch will be started straight away. Also the starting temperature recorded. After that the temperature reading will be recorded at 30 second intervals shown by the stopwatch. This will be done for 5 minutes and repeated for all the beakers. Results: TIME SECS TEMPERATURE OF WATER IN BEAKER ºC CONTROL WOOL BUBBLE FOIL BLANKET 0 83.0 83.5 83.0 83.5 83.5 30 80.5 80.0 81.5 79.5 80.5 60 77.0 78.0 80.0 77.5 78.5 90 75.0 76.0 78.5 76.0 76.5 120 73.5 74.5 76.5 74.0 75.5 150 72.0 73.5 75.5 73.5 74.5 180 70.0 72.5 74.5 72.5 73.5 210 69.0 71.5 74.0 71.5 72.5 240 68.0 70.5 73.0 70.5 71.5 270 66.5 70.0 72.5 69.5 71.0 300 65.5 69.5 72.0 68.5 70.5 Observations: For all the materials there is a sharp drop in temperature for the first minute and after that it gradually slows down. The starting temperatures for the insulated beakers are very close but at 5 minutes they the end readings are more spread out. As you can see on the graph the bubble wrap has the most gradual curve and consequently the slowest rate of heat loss. From the starting temperature of 83.0 °C, it dropped by 12.0 °C after 5 minutes. The blanket is the second best insulator with the temperature drop of 13 °C from a starting temperature of 83.5 °C. The cotton wool lost 14.0 °C from a starting temperature of 83.5°C The aluminium foil is the worst insulator from the materials as from a starting temperature of 83.5 °C it lost 15 °C The controlled experiment the beaker with no insulation has the steepest curve and therefore the fastest heat loss. From the starting temperature of 83.0 °C, it lost 17.5 °C after 5 minutes. Conclusion My prediction was correct as the bubble wrap material was the best insulator because it prevented the most heat loss being the best insulator. The bubble wrap was the best insulator due to its ability to reduce heat loss through conduction. This can be understood by looking behind theory of conduction. Since heat is energy associated with the motions of the particles making up a substance, it is transferred by these motions, shifting from regions of temperature, where particles are more energetic to regions of lower temperature. The rate of heat flow between the two regions is relative to the temperature difference and the heat conductivity of the substance. In solids the molecules are bound and add to the conduction of heat mainly by vibrating against neighboring molecules. However a more important mechanism is the migration of free electrons. Materials such as metals have a high free-electron density therefore they are good conductors of heat while non-metals do not conduct as well. Since gases and liquids have their molecules even further apart they are generally, very poor conductors which makes them good insulators. Therefore the bubble wrap which is made up of lots of air pockets has not only less free electrons to transfer heat but it has molecules the trapped air that are quite far apart. This makes harder for heat from the boiling water in the beaker to be conducted away through the bubble wrap and to the surroundings The blanket and cotton wool were also quite good insulators as they too have a lot of air trapped between the fibres. The foil was not as good because being a metal it had more free electrons with particles being closer together, for heat to be conducted away to other regions. However, since it was quite shiny it reflected some heat back in, to the beakers, such as the effect of a flask so it prevented heat loss better than the controlled beaker. The controlled beaker naturally let out the most heat since it had no form of insulation except the fact that it the beaker was made out of glass which in itself is a fairly good insulator. Evaluation The method that I used to carry out my investigation was fairy simple to carry out. It required me to pour out 80 mls of hot water into beakers insulated with different materials. The temperature of the water was then recorded over a 5 minute period at 30 second intervals. The results of my investigation were accurate as I took my readings at eye-level to the nearest half a degree centigrade. I had no anomalous results and the accuracy can be seen on my results graph where the readings are close to the best fit curve. However I could have been even more accurate as if the time h d been available tome, I would have repeated my experiments to get an average of the results. Another way in which I could have improved my experiment was if was available to me, I would have used a data logger with temperature probe. This would have given me very accurate readings every second. I could have also used polystyrene lids to reduce heat loss by convection in the beakers. To extend by investigation I could have used different beakers like copper cans to see how the material of the beaker affects the rate of heat loss. In addition I could look at the effect of the amount of layers of material insulating the beaker as well as looking at different volumes of boiling water.   

Aim: To investigate the heat loss of the insulating materials, bubble wrap, cotton wool, blanket and foil Introduction: I will carry out this investigation by measuring the temperature of hot water in beakers insulated with these materials. Naturally the insulated beaker with the warmest water will have prevented...

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A trolley is positioned the... A trolley is positioned the top of a ramp, the summit being Xcm from the ground. It is then released, rolling all the way down the ramp which is a set length, then comes in to contact with the ground and travels a distance of 2 meters before it passes a line. Once the trolley cleared the ramp we began timing, then once it reached the 2 meter waypoint we stopped the timer, by doing this we are able to determine the speed of the trolley. Based on existing scientific knowledge, I know that this experiment depends on a certain type of energy being converted into another type. When the trolley is raised to the top of the ramp, it gains a certain amount of gravitational potential energy. This is then converted into kinetic energy as the trolley moves down the ramp and reaches the ground. To see what factors may affect the way the experiment turns out, it may be useful to look at the formula for potential energy. P.E = MxHxG where m=mass, h=height and g=gravity Obviously, the more potential energy the trolley has got, the faster it will move down the ramp. So, theoretically, the only factors that can affect this experiment are the height and the mass. The gravity could be altered although we would have to travel to other planets which is far out of the schools budget, therefore the gravity will always stay constant: 1G. I will be investigating, by varying the height the summit of the ramp is raised off the ground. I will measure if the average speed increases or decreases. There will always be smaller forces that could slightly affect the result, such as friction between the ramp and the trolley's wheels, and air resistance. There is no way I can control any of these factors, but they shouldn't affect the results so much as to give completely anomalous readings for each experiment seeing as they remain a constant throughout the experiment. Planning When planning my experiment, I will need to take into consideration the following points: "¢ Safety "¢ Fair testing "¢ How many results I will take "¢ What range of variables I will experiment with "¢ Equipment Safety With this straightforward experiment there is not much that needs to be taken into consideration. No harmful substances are being used, neither are flames, solvents, or atomic-reactors. Although at the end of the 2 meter run way some sort of barrier will need to be placed to prevent the trolley from continuing its course and therefore thwart potential harm to any unsuspecting pedestrian. Fair Testing As with all scientific experiments, only one variable must be altered at one time. All the rest must remain constant to ensure good stable results. By using present knowledge, I know that the following factors can affect the outcome and must be controlled: Height of ramp. The height is the factor which I will be changing. This is included in the formula for potential energy. the height of the ramp should affect the speed of the trolley in some way. I will be changing this variable to do the test, it is my only variable. Mass of trolley. The mass is also included in the formula for potential energy and so could affect the speed of the trolley one way or the other. The mass will be kept the same throughout the experiment because we will be using the same trolley for all of them. Gravity. The gravity is the last portion of the formula for potential energy, which will affect the outcome if it is increased or decreased. The way to maintain this factor is to simply stay on the same planet and not to move the experiment area by to great a range gravity at equator less due to centrifugal force created by earths spin. Friction. As I mentioned that the only factors that should affect the outcome of the experiment would be mass, height and gravity. They make up the formula for the potential energy. But other factors may use some of this energy when it is being converted into kinetic energy as the trolley moves down the ramp. The friction between the wheels of the trolley and the surface of the ramp can consume some of the energy, also the friction between the axels and fixings, and the point at which the trolley makes contact with the ground will also cause friction. This can slow down the trolley, but only very slightly. To maintain the same friction for all the results we should use the same ramp throughout the experiment, and will not apply anything to the ramp or the wheels of the trolley. Air resistance. This is also a form of friction. There is very little we can do to control this factor, and its effects would be so insignificant it may not matter. Basically, we just need to make sure we have the same trolley and ensure that the surface area is constant. With these points in mind it is essential that we must keep the same trolley, use the same ramp and keep the mass constant. We will also have to keep the length of the runway the same. Ranges and amounts To make this investigation successful, we must choose a suitable range and amount of readings to record in order to come up with a useful set of results that are easy to plot a graph with and that make sense. For example, it would be pointless to experiment with heights ranging from 1cm to10cm because the trolley would not even reach the 2 meter waypoint. Instead a more sensible range, let's say from 10cm to 80cm, would be appropriate and should yield better results. We could take readings every 5cm, and take a minimum of three readings on each height to work out an average this makes sure that the end result more accurate. Therefore we will use the following variations in height the only variable we are experimenting with is height and it is the only value to be varied, everything else remains constant. 10cm 15cm 20cm 25cm 30cm 35cm 40cm 45cm 50cm 55cm 60cm 65cm 70cm 75cm 80cm Equipment To perform the experiment we will need the following equipment. Trolley "“ To roll down the ramp Ramp "“ For the trolley to roll down Meter Stick "“ To measure out 2 meter run way Chalk or some sort of marker pen "“ To mark the start and finish lines Stop Watch "“ To time the trolley Barrier "“ To stop the once it has reached the end of the runway Big metal thing with clamps on "“To hold one end of the ramp up and keep it stable From this experiment I expect to find out what factors affect the speed of a body when no manual force is applied to them i.e. pushing them. This experiment is being conducted to prove the potential and kinetic energy formulae which, once completed, can be used to calculate exactly the results of any situation using these theories. In fact, before recording the results I could use the formula to estimate what the outcome would be. Method The following is the method to the practical. it gives a step by step description of how we preformed the experiment. 1. Set up equipment ramp held up by metal clamp thing and measure 2 meters from the ramp marking off the 2 meter point and positioning the barrier. 2. Ensure the height of the ramp is the determined test height using the meter stick. 3. Position trolley at the top of the ramp 4. Release the trolley 5. Start the timer as the trolley makes contact with the floor 6. Stop the clock when the trolley reaches the finish line 7. Record the time taken for the trolley to reach the finish, next to the relevant height, in a table We will then record the results and proceed to raise the ramp another 10cm and continue the experiment, once we have attained a full set of results we will proceed to repeat the test another 2 times to give us the 3 tests needed to give a fair test. Prediction As I mentioned in the Introduction, the experiment is based on the potential energy at the top of the ramp being converted into kinetic energy at the bottom. I then deduced the following formulas by using my current scientific knowledge and attaining information from the internet. Potential Energy at the top = Kinetic Energy at the bottom Potential energy = work done Potential energy = weight x height lifted Weight in N = mass x 10 therefore: Gravitational P.E = Mass g height joules kg N/kg m I also found out some formulas to determine Kinetic Energy: K.E = ½ x mass x velocity squared K.E = ½mv2 Knowing this we can write: P.E = K.E Therefore mgh = ½mv2 The formula can be simplified by cancelling out the 2 m's and moving the ½ to the left side of the formula changing the x10 to x20, therefore we now have: 20h = v2 SQRT 20h = v SQRT = square root This formula will give us the average velocity for the trolley going down a ramp of h metres high. Once we have found this we can actually use the equation for average speed to find out how long it will take the trolley to reach the finish line and actually produce a theoretical result prior to conducting the experiment. Therefore this shows that the higher the ramp is raised, the higher the velocity of the trolley will be resulting in a quicker time to reach the finish line. I can also predict from this formula, that the shape of the graph v against h. As h increases, by 5cm each time seeing as this is how much we will increase our actually height by v will increase too, but not in proportion. This is due to the square root in the formula that we have to use to find v. The higher the height goes, the less gap there will be between the velocity of the present reading and the previous heights velocity reading. The graph should produce a curve. Therefore I can predict that the Increase in height of ramp = Increase in velocity of trolley. Results Height cm Time seconds 10 2.00 15 1.40 20 1.00 25 0.93 30 0.90 35 0.81 40 0.79 45 0.78 50 0.77 55 0.75 60 0.79 65 0.81 70 0.82 75 0.85 80 0.87 Height cm Time seconds 10 2.00 15 1.38 20 1.01 25 0.94 30 0.89 35 0.80 40 0.78 45 0.76 50 0.75 55 0.75 60 0.76 65 0.79 70 0.80 75 0.83 80 0.85 Height cm Time seconds 10 2.04 15 1.30 20 1.05 25 0.95 30 0.90 35 0.78 40 0.77 45 0.76 50 0.72 55 0.75 60 0.80 65 0.83 70 0.85 75 0.85 80 0.88 Average results i have also added a velocity column to the table I did not do this for the other ones because it would have taken too much time because we would be averaging the velocity values as well as the time values which would be easier to work out by getting the velocity from the averaged time values, and would not be needed because either way we get the average velocity, but this way is less time consuming. Height cm Time seconds Velocity M/S 10 2.01333 0.99338 15 1.36 1.47059 20 1.02 1.96078 25 0.94 2.12766 30 0.89666 2.23051 35 0.79666 2.51048 40 0.78 2.56410 45 0.76666 2.60872 50 0.74666 2.67861 55 0.75 2.66666 60 0.78333 2.55320 65 0.81 2.46914 70 0.82333 2.42916 75 0.84333 2.37155 80 0.86666 2.30787 From these results I have made a graph to show the curve of h against v Conclusion The graph has a few fluctuations but there were problems when the trolley hit the ground because it would lose some of its speed there, and the amount lost there was out of our control and not usually regular, so we tried to get the results as close as we could without better equipment. The curve in the graph is due to the increase in friction at where the trolley is hitting the ground, and along the ramp and ground which at some speeds slows the object down to a lower speed than objects dropped than lower heights; this is mainly due to the friction caused by the impact with the ground. I originally said that I could predict the outcome using the formula, but I never took in to account the friction caused when the trolley hit the ground, the formula would most probably predict the first few results seeing as they are not as greatly affected by friction as the rest. My prediction was proved correct as the graphs clearly show that the speed does indeed increase when the ramp is raised higher. This is due to the fact that more potential energy is given to the trolley as it is raised higher and height is part of the formula that makes up P.E: P.E = mgh P.E = mass x gravity x height So the higher an object is the more gravitational potential energy it gains. When it falls, its potential energy is converted into kinetic energy and since energy can neither be created or destroyed, only converted; it will move at a faster speed. So, to sum up, as you lift an object to a height the object gains gravitational potential energy. The higher you lift the object, the more energy you are using which is chemical energy to power your muscles and therefore the more potential energy the object is gaining. Potential energy is converted into kinetic energy completely so the object when released will move at a faster rate depending on how high it is lifted. The height does affect the speed at which a trolley travels down a ramp. The experiments went very well and ran efficiently, thanks to the plan we had drawn out beforehand. Although we had very limited time and were not able to produce any other results using different methods. Although the results did have quite a few inconstancies, this was due to many factors, firstly the impact point of the trolley hitting the ground, and secondly the timing was done by a human, and therefore is not completely accurate. I also tried some tests above 80cm to see what would happen, this eventually resulted in the destruction of the trolley, but the results did prove that the speed did decrease as it got higher due to the landing, and eventually at 100cm the trolley was unable to reach the end of the runway. These results I decided were not needed because they just prove that friction gets greater as the object increases in speed which I have already shown. If I were to do this experiment again, I would experiment with different surfaces of ramp, and better equipment, for example we could use 2 light gates for timing, one would be positioned at the start of the run and one at the end to completely automate the timing part f our experiment and help reduce error, also the computer would record the results for us and save time typing or writing them down. We could have also created some sort of curve at the bottom to prevent the extra friction caused as the trolley hits the ground, or possibly used a longer ramp and conduct the timing while the cart goes along the ramp therefore eliminating the landing stage completely also we could have made sure that the ramp was now leaning to a side which would encourage the cart to move to the left or right and lose time by travelling a longer distance. We could have also tried varying the mass of the cart, although the mass of an object does not affect the rate at which it falls as proven by the famous experiment done by Galileo which states that the mass of an object does not affect the rate at which it falls, and he proved this by dropping 2 balls of different weight off a tower and they both hit the ground at the same time. Although there may possibly be some variation in the speed due to friction because the greater weight would put more pressure on the wheels and put more pressure on the ramp therefore causing more friction, and also more friction at the point where the trolley hits the ground too. I will now test the formula and see how close our results were to the expected results. I will use the average table without the time because we will not need the time, only the speed seeing as the formula I am using does not calculate time. Height cm Velocity M/S 10 0.99338 15 1.47059 20 1.96078 25 2.12766 30 2.23051 35 2.51048 40 2.56410 45 2.60872 50 2.67861 55 2.66666 60 2.55320 65 2.46914 70 2.42916 75 2.37155 80 2.30787 Therefore the square root of 20x10 should give 0.99338 The actual answer is 14.14213 There is a significant difference, although I have not taken in to account the 2 meter distance and the friction it will cause and also the friction of the landing and the trip along the ramp, possibly if I do some more calculation we can work out how much energy the friction consumes. Predicted Velocity Height cm Velocity M/S 10 14.14214 15 17.32051 20 20 25 22.36068 30 24.49491 35 26.45751 40 28.28427 This would have been about the speed of the trolley as it reaches the bottom of the ramp, there is a significant loss of speed at the impact point and as the car moves along the ground, which is not as smooth as the ramp and also causes greater friction. If I combine the tables I should be able to see how much energy is lost to friction. Height cm Velocity M/S Calc. Velocity M/S Velocity lost due to friction 10 0.99338 14.14214 13.14876 15 1.47059 17.32051 15.84992 20 1.96078 20 18.03922 25 2.12766 22.36068 20.23302 30 2.23051 24.49491 22.25981 35 2.51048 26.45751 23.94703 40 2.56410 28.28427 25.72017 This shows that a lot of energy is lost to friction, if we would have improved the experiment by using a light gate and performing the experiment on the actual ramp itself we could have eliminated a lot of this friction, but not all of it because that would be virtually impossible, this experiment also shows me that friction has a much stronger effect on material than I had originally expected. If the results had been more accurate we could have worked out a formula for the amount of energy lost through friction, but our results would need to be very accurate to calculate this and we have run out of time and do not have the sufficient materials to do this.   

A trolley is positioned the top of a ramp, the summit being Xcm from the ground. It is then released, rolling all the way down the ramp which is a set length, then comes in to contact with the ground and travels a distance of 2 meters before it...

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