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Name___________________________________ Physics 120 Quiz #3, February 3, 2012 Please show all work, thoughts and/or reasoning in order to receive partial credit. The quiz is worth 10 points total. 1. Consider a gold wire that is suspended from the ceiling with dimensions 1cm by 1cm by 1m. A mass of 10kg is suspended from the wire stretching it by 12.6µm. a. What is the interatomic spacing (dIAB) between the gold atoms in the wire? (Hints: Assume that you have a 1cm3 volume to do the calculation, the density of gold is 19300 kg/m3 and the molar mass of gold is 0.197 kg/mol. € Assuming a 1m3 volume the total mass of gold (ρAl = 19300 kg/m3) is 19300kg. The number of atoms in this mass is given by 1mole # atoms atoms , where the × 6.02 ×10 23 = 5.9 ×10 28 volume volume =M T ×m × N A = 19300kg × 0.197kg molar mass of gold is m = 197g/mol. Then the number of atoms on a side of length L = 1m, is the cubed root of the number of atoms in the volume. Thus the 3 5.9 ×10 28 = 3.89 ×10 9 atoms number of atoms on a side is N atoms side = side . Lastly, in the side of length L = 1m, there are N(atoms/side) each spaced by dIAB. Thus the length of an interatomic bond is L 1m L = N × dIAB → dIAB€= = = 2.57 ×10−10 m. 9 N 3.89 ×10 b. What is the mass of a single atom of gold? € € matom = 0.197kg 1mol × = 3.27 ×10−25 kg 23 1mol 6.02 ×10 atoms c. What is Young’s modulus for gold? Stress = Y × Strain → Y = Y= Stress FL mgL = = Strain AwireΔL AwireΔL 10kg × 9.8 sm2 ×1m (0.01m) 2 −6 ×12.6 ×10 m = 7.8 ×1010 mN2 d. What is the speed of sound in gold? € vs = € 3 k IAB YdIAB dIAB = = matom matom 7.8 ×1010 mN2 × (2.6 ×10−10 m) −25 3.27 ×10 m 3 = 2048 ms Useful formulas: p = γmv k eff , parallel = n parallel kindividual 1 γ= keff ,series = 2 v 1− 2 c vi + v f v avg = 2 Fg = mg kindividual n series stress = Ystrain → vs = GM1 M 2 Fgravity = rˆ12 r122 Fspring = −ks; s = ( L − Lo ) sˆ W = ∫ F ⋅ dr =ΔKE = −ΔU F ΔL =Y A L kIAB d matom k IAB = Yd GM1 M 2 r U g = mgy Ug = − 1 U s = ks2 2 1 KE = mv 2 2 KE = (γ −1) mc 2 Useful Constants € Momentum Principle: Position-update: € p f = pi + Fnet Δt; Δt = large Δt p f = pi + Fnet dt; dt = = small n rf = ri + v avg Δt = ri + p p2 m 1+ 2 2 mc Δt rf = ri + v f dt; dt = = small n g = 9.8 sm2 2 G = 6.67 "10#11 Nm kg 2 Δt; Δt = large %o = 8.85 "10#12 NmC 2 µo = 4$ "10#7 TmA c = 3"10 8 ms h = 6.63"10#34 Js €€ Energy principle: Geometry /Algebra Circles Triangles A = πr 2 A = 12 bh 0.511MeV c2 937.1MeV m p =1.67 "10#27 kg = c2 948.3MeV mn =1.69 "10#27 kg = c2 931.5MeV 1amu =1.66 "10#27 kg = c2 23 N A = 6.02 "10 me = 9.11"10#31 kg = Spheres A = 4 πr 2 V = 43 πr 3 Quadratic equation : ax 2 + bx + c = 0, whose solutions are given by : x = −b ± b 2 − 4ac 2a Vectors € € 2 1eV =1.6 "10#19 J ΔE = W = ΔU g + ΔU s + ΔKE C = 2πr 1e =1.6 "10#19 C 1 C2 k= = 9 "10 9 Nm 2 4$%o magnitude of a vector : a = ax2 + ay2 + az2 writing a vector : a = ax ,ay ,az = a aˆ = ax iˆ + ay ˆj + az kˆ Ax 2 + Bx + C = 0 & x = ! #B ± B 2 # 4AC 2A