Homework 11 Solutions - 11 Problem 23 We have L F =Y A L A=...

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§ 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 cross-section, 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 cross-sectional 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 . 0 × 10 - 4 , copper. ( b ): The elongation is Δ L = parenleftbigg Δ L L parenrightbigg L = braceleftbigg 0 . 083 mm , steel 0 . 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 = - 0 . 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 0 . 946 m 3 = 1 . 09 × 10 3 kg / m 3 . § 11 Problem 47 ( a ): If the beam is 1.50 m long and the cable is 2.00 m long, then the cable must make an angle θ = Cos - 1 (1 . 50 / 2 . 00) = 41 . 4 from the horizontal. The two wires are symmetrically disposed about the center of the sign, so by symmetry each has a tension equal to half the sign’s weight, or 137 N. The weight
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