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Unformatted text preview: Version PREVIEW – Homework 8 – Savrasov – (39825) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. Dust Grains rms 001 10.0 points Dust grains of diameter 0 . 7 μ m and density 1000 kg / m 3 are in equilibrium with air ( A = 28 . 9) at a temperature of 279 K . What is the rms speed of the dust particles? 1 m = 10 6 μ m. Correct answer: 8 . 02155 mm / s. Explanation: Let : r = 0 . 35 μ m = 3 . 5 × 10 − 7 m , ρ = 1000 kg / m 3 , T = 279 K , and k = 1 . 38065 × 10 − 23 J / K . m = ρV = ρ 4 3 π r 3 = ( 1000 kg / m 3 ) 4 π (0 . 35 μ m) 3 3 = 1 . 79594 × 10 − 16 kg , so v rms = radicalbigg 3 k T m = radicalBigg 3 (1 . 38065 × 10 − 23 J / K) (279 K) (1 . 79594 × 10 − 16 kg) = 0 . 00802155 m / s = 8 . 02155 mm / s . Internal Energy of Helium 002 10.0 points The universal gas constant is 8 . 31451 J / K · mol. Calculate the change in internal energy of 8 mol of helium gas when its temperature is increased by 6 . 2 K. Correct answer: 618 . 6 J. Explanation: Given : n = 8 , R = 8 . 31451 J / K · mol and Δ T = 6 . 2 K . The internal energy of the monoatomic ideal gas is U = 3 2 nRT So, Δ U = 3 2 nR Δ T = 3 2 (8) (8 . 31451 J / K · mol) (6 . 2 K) = 618 . 6 J . Molecule Escaping the Earth 003 (part 1 of 2) 10.0 points If it has enough kinetic energy, a molecule at the surface of the Earth can escape the Earth’s gravitation. The acceleration of gravity is 9 . 8 m / s 2 , and the Boltzmanns’ constant is 1 . 38066 × 10 − 23 J / K. Using energy conservation, determine the minimum kinetic energy needed to escape in terms of the mass of the molecule m , the freefall acceleration at the surface g , and the radius of the Earth R...
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This note was uploaded on 01/10/2010 for the course PHY na taught by Professor Richardscalettar during the Fall '09 term at UC Davis.
 Fall '09
 RichardScalettar
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