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Unformatted text preview: hÝ × ic×ÙÑÑ eÖ8ÓÑ eÛ Ó Ök# Ó ÐÙ Ø iÓÒ ×bÝ iÑC a ÐÐahaÒ § 11 Problem 23: We have F A = Y Δ L L ⇒ A = FL Y Δ L = (400 N)(2 . 00 m) (20 × 10 10 N / m 2 )(0 . 0025 m) = 1 . 6 × 10 6 m 2 . This is a wire of circular crosssection, so A = πr 2 ⇒ r = radicalbig A/π = 0 . 71 mm , so the diameter is 2 r = 1 . 4 mm. § 11 Problem 24 ( a ): The two rods are subject to the same force F = 4000 N. They have the same crosssectional area A = πr 2 = π (0 . 75 cm) 2 = 1 . 8 × 10 4 m 2 . The strain is Δ L L = F AY = braceleftbigg 1 . 1 × 10 4 , steel 2 . × 10 4 , copper. ( b ): The elongation is Δ L = parenleftbigg Δ L L parenrightbigg L = braceleftbigg . 083 mm , steel . 15 mm , copper. § 11 Problem 32 ( a ): Water has a compressibility of k = 45 . 8 × 10 11 / Pa, so its bulk modulus is B = 1 k = 2 . 18 × 10 9 Pa . We then have P = B Δ V V ⇒ Δ V = V P B = (1 . 00 m 3 )(1 . 16 × 10 8 Pa) 2 . 18 × 10 9 Pa = . 053 m 3 . ( b ): The mass is m = ρV = (1 . 03 × 10 3 kg / m 3 )(1 . 00 m 3 ) = 1 . 03 × 10 3 kg . The new volume is V + Δ V = 0 . 947 m 3 . Thus the new density is ρ new = m V + Δ V = 1 . 03 × 10 3 kg . 946 m 3 = 1 . 09 × 10 3 kg / m 3 ....
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This note was uploaded on 07/12/2008 for the course PHY 121 taught by Professor Chamberlin during the Summer '08 term at ASU.
 Summer '08
 Chamberlin
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