Conclusion: Category 1: Momentum was found that after the collision was less than before the collision by 10%. This was not what has been expected‚ so the difference was fairly significant. This happened because of friction‚ when the two pucks collided‚ they have lost a bit of their momentum‚ so the momentum after the collision differed. Kinetic energy differed more than what was expected‚ it was significantly less after the collision‚ the difference before and after the collision was 63.7%‚
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DESIDOC Journal of Library & Information Technology‚ Vol. 30‚ No. 2‚ March 2010‚ pp. 3-14 © 2010‚ DESIDOC REVIEW PAPER Application of Bradford’s Law of Scattering to the Physics Literature: A Study of Doctoral Theses Citations at the Indian Institute of Science K.G. Sudhier Department of Library& Information Science University of Kerala‚ University Library Building Thiruvananthapuram–695 034‚ Kerala E-mail: kgsudhier@gmail.com ABSTRACT One of the areas in bibliometric research concerns
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- - Name: - Teacher: - Scientists: Ernest Orlando Lawrence Andre-Marie Ampere Subject: Physics Class: 12A Due Date: 20th August 2012 Name: - Teacher: - Scientists: Ernest Orlando Lawrence Andre-Marie Ampere Subject: Physics Class: 12A Due Date: 20th August 2012 Physics ERT: Magnetism Physics ERT: Magnetism Table of Contents Table
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The specific latent heat of vaporization of a substance is the quantity of energy required to A. raise the temperature of a unit mass of a substance by one degree Celsius. B. convert a unit mass of liquid to vapour at constant temperature and pressure. C. convert a unit mass of solid to vapour at constant temperature and pressure. D. convert a unit mass of liquid to vapour at a temperature of 100°C and a pressure of one atmosphere.
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DP Physics IA Thermal physics: Specific Heat Capacity of Metals Introduction: In this experiment we are going to measure the specific heat capacity of a unknown metal. To measure the specific heat capacity we will heat up the metal to certain temperature and release the metal in beaker filled with water. By knowing the mass and temperature of water filled in beaker‚ we will be able to calculate the specific capacity of unknown metal by change in temperature of beaker willed
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The physics concepts that were illustrated in our egg drop project were momentum and energy. In the project‚ we saw how momentum increases as a normal egg without anything to slow it down would crack. That was because the longer that the egg fell‚ the more momentum the egg gains. We also saw that by adding resistance during the fall‚ such as a plastic bag “parachute”‚ would slow down the egg enough so that it would not crack when it reached impact with the ground. The other concept that was discussed
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Page 12 1 a) Between t = 30 and t = 45 mins b) 7.5 mins c) i) distance travelled = area under graph between t = 0 and t = 12½ mins ii) average speed = total distance travelled total time for journey = total area under graph 60 mins 2 a) ∆v = 32 m/s a =10 m/s² t = ∆v = 32 = 3.2 s a 10 b) 3 a) OP – constant acceleration PQ – constant acceleration (greater than OP) QR – constant speed RS – constant deceleration b) O and S c) 6 m/s d) 70 s e) Total distance travelled = area under
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Zach Hicks (Z1603274) Karen Richards Section F (0900 - 1150) 11/30/12 The Waiter Problem Solving Part A: Center of Mass in One Direction 1. Since the mass of the plate is a uniform‚ the center of mass is in the center of the plate. The plate has diameter d = 28 cm‚ thus the center of mass is 14 cm from the edge of the plate. In the lab‚ two scenarios will be analyzed; 1) when the glass is standing up‚ and 2) when the glass is laying on its side. When the glass is standing up‚ its mass
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An investigation into the specific heat capacity of a solid Introduction: In this experiment I investigated the specific heat capacity of brass and aluminium‚ in order to do this I arranged a series circuit with a heater‚ the heater would go inside the brass and aluminium and heat it for an amount of time in which I decided‚ I also used the voltmeter in parallel. This arrangement made me able to arrange the equation VIT= mcΔθ‚ since the electrical energy in would be equal to the mass times the
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Circular motion‚ gravitational field‚ and oscillation 1.Define gravitational field strength: Force per unit mass 2. Explain why there is a point between two point masses at which the gravitational field strength is zero. Because gravitational fields of two masses are in opposite directions‚ so there is a point where it is zero What is meant by angular velocity The rate of change of the angular displacement of an object as it moves along curved path. Generally when asked to test
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