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m2 6 Fig awre .please explain your answer and draw graph. Also label the answers thoroughly. Answer both for thumbs up:))

2. The magnitude of the force of attraction between the positively charged proton and the negatively charged electron in a Hy10. A steel ball of mass 0.514 kg is fastened to a cord 67.8 cm long and is released when the cord is horizontal. At the bottri ure

m2 6 Fig awre .
2. The magnitude of the force of attraction between the positively charged proton and the negatively charged electron in a Hydrogen aton is given by f(r) = ke2/r2 where e is the charge on the proton, k is Coulomb's constant, and r is the distance between the proton and the electron. Assume that the protorn can't move. The electron is initially moving in a circular orbit around the proton with radius r and then "jumps" suddenly to another circular orbit of smaller radius, r2, as in Figure1. (a) Calculate the change in the kinetic energy of the system using Newton's second law. (b) Calculate the change in the potential energy of the atom. (c) By how much has the total energy of the atom changed? This energy is given off in the form of electromagnetic radiation (light)
10. A steel ball of mass 0.514 kg is fastened to a cord 67.8 cm long and is released when the cord is horizontal. At the bottom of its path, the ball strikes a 2.63 kg steel block initially at rest on a frictionless surface, as shown in Figure 6. The collision is elastic. Find (a) the speed of the ba and (b) the speed of the block, both just after the collision. (c) Suppose instead that one-half the kinetic energy is converted to internal energy as a result of the collision. Find the final speeds now.
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please explain your answer and draw graph. Also label the answers thoroughly. Answer both for thumbs up:)) m2 6 Fig awre . 2. The magnitude of the force of attraction between the positively charged p...
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