midterm 2 practice

# midterm 2 practice - brown(twb493 extra credit 03...

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brown (twb493) – extra credit 03 – turner – (57340) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A ball of mass 2 kg hangs from a spring whose stiffness is 870 N / m. A string (not shown in the figure) is attached to the ball and an unknown force F is pulling the string to the right, so that the ball hangs motionless, as shown in figure. In this situation the spring is stretched, and its length is L = 13 cm, while the distance d is given to be 5 cm. What would be the relaxed length of the spring, if it were detached from the ball and laid on a table? Answer in units of cm. Take g to be 9 . 8 m / s 2 . Correct answer: 10 . 5594 cm. Explanation: Given : m = 2 kg , L = 13 cm , k = 870 N / m , d = 5 cm , and g = 9 . 8 m / s 2 . Using the two lengths given (and the Pythagorean theorem), we can first obtain the height of the ball below the point of sus- pension to be 12 cm. If we call the angle between the spring and the vertical to be θ and the force exerted by the spring on the ball to be F s , then balancing the y-components of the forces acting on the ball gives us- F s cos θ = mg Here, cos θ = 12 / 13 (from the figure) From this, we get F s = mg cos θ The extension s of the spring can now be obtained (in units of m). s = F s k = mg k cos θ From this, the relaxed length of the spring can be obtained as- L 0 = L - s = 13 cm - 100 * 2 kg * 9 . 8 m / s 2 870 N / m * (12 / 13) L 0 = 10 . 5594 cm 002 10.0points For a p - p pair, let the ratio of the magnitude of the electric force to gravitational force be: a = F e pp F g pp = 1 . 20 × 10 36 Determine the corresponding ratio for an α - α pair, here α stands for an α particle, i.e.the helium nucleus, which has 2 protons and 2 neutrons. You may use the approximation, proton mass is the same as the neutron mass and the same as the nucleon mass, i.e. m p = m n = m N .

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