The kinetic energy reaches its minimum amount. C. The kinetic energy is converted to potential energy. D. b and c ____ 8. A ball is thrown in the air. When does it have its maximum amount of potential energy?
The first part of the procedure is to investigate a = F/m by holding F constant This investigation involving the Atwood machine will investigate the different aspects of a = F/m. The first part of the procedure holds m constant while studying the relationship of F to a by altering the F of the system. Looking at the equation a = F/m, acceleration and force should be directly proportional to one another. F=ka and a=F/k. The second part of the procedure holds F constant and investigates the relationship of a to m by altering the total mass of the system.
(1) 2. The temperature of a 2.0 g sample of aluminium increases from 25°C to 30°C. How many joules of heat energy were added? (Specific heat of Al = 0.90 J g−1K−1) A. 0.36 B.
Planck's constant: the constant relating the change in energy for a system to the frequency of the electromagnatic radiation absorbed or emitted, equal to 6.626 X 10^-34 J 5. Quantization: the concept that energy can occur only in discrete units called quanta 6. Photon: a quantum of electromagnetic radiation 7. Photoelectric effect: ejection of electrons from a substance by incident electromagnetic radiation, especially by visible light 8. E=mc^2: Einstein's equation proposing that energy has mass; E is energy, m is mass, and c is the speed of light 9.
_________________________ ____ 21. The amount of energy a photon carries is determined by its mass.
Accelerated Motion Due to Gravity | Lab 6 | Objective: In this experiment we will determine that the displacement of a free falling body from the rest is directly related to the square of the falling time and that the acceleration due to gravity does not depend on the mass of the object. We will also determine the experimental value of the acceleration due to gravity. Theoretical Background: We can determine the position and velocity of a freely falling object by using the kinematic equations for motion with constant acceleration, since the acceleration due to gravity is constant near the surface of the Earth. As the motion is on the vertical axis, we will designate the displacement by y in the kinematic equations: v=v0+gt, y=v0t+gt22, v2=v02+2gy, where v0 is initial velocity. Since we are considering an object that is dropped, the initial velocity v0 would equal to zero (v0=0) in the kinematic equations.
638. The e/m value of cathode rays(rays of electrons) in a discharge tube is: A. 1. 6022x 1019 Coulombs Kg-1 B. 1.
AP 10006 Test (Total 100 Marks) Question 1(26.28) Consider the following figure. (a) Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in the figure above. Take C1 = 6.00 µF, C2 = 3.0 µF, and C3 = 2.00 µF. (15 marks) (b) What charge is stored on C3 if the potential difference between points a and b is 100.0 V? (15 marks) Answers (a) C1 and C2 in series = 2uF Cupper = 2uF + 2uF + 2uF = 6uF Clower = 3uF + 3uF = 6uF Cequ = 3uF (b) Qupper = Qequ = Cequ V = 3uF100V = 300uC Vupper = Qupper/Cupper = 300uC/6uF = 50V Charge stored in C3 = 2uF50V = 100uC (15 marks) (15 marks) Question 2 (27.53) A charge Q is placed on a capacitor of capacitance C. The capacitor is connected into the circuit as shown in the figure below, with an open switch, a resistor, and an initially uncharged capacitor of capacitance 4C.
2.Force is equal to mass times acceleration. 3.For every action there is an equal and opposite reaction. The athletes let go of the hammer when it is at the side rather than when it faces the direction they want it to land, because it travels on a tangent and id the release it where they want it to land it’ll travel to the side and not get counted as a throw. Inward net force causes circular motion. And no matter how long or short the circular motion is there is an inward acceleration.
The following formula was used: cm=mwcwT3-T1+mcccT3-T1 mmT2-T3 Where cw is the specific heat capacity of water and cc is the specific heat capacity of calorimeter. Substituting the values, we get an answer of 998.81 J/kg·C0 with a percent error of 10.98% determined by the formula: % error=theoretical-experimantalexperimental ×100 Table 1. Specific Heat of Metal Mass of Sample (mm) | 0.01655 kg | Mass of Inner Vessel of Calorimeter (mc) | 0.04723 kg | Mass of Inner Vessel of Calorimeter with Water | 0.24242 kg | Mass of Water inside the Inner Vessel of Calorimeter (mw) | 0.19519 kg | Initial Temperature of Water and Inner Vessel of Calorimeter (T1) | 330C | Initial Temperature of Sample (T2) | 860C | Equilibrium Temperature of Sample, Water, and Inner Vessel of Calorimeter (T3) | 290C | Calculated Specific Heat of Sample (cm) | 998.81 J/kg·C0 | Accepted Value of Specific Heat | 900 J/kg·C0 | % Error | 10.98% | Possible sources of error for this experiment may be due to changes in the temperature due to heat transferring to the environment and human error (wrong readings for temperature and mass). In the second activity, the latent heat of fusion of ice was computed (Table 2). Latent heat is associated with a phase change.