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Entirely Multiple-Choice (on bubble sheet):
unit conversions:
1. D 6.6x10^-4 GW
2. C 0.72 cm^3
3. E 3.0x10^18 mm^3
4. B 5.5x10^4 um/ns
5. A "a" only (graph of constant v = horizontal graph, slope = 0)
6. E "a, b, & f" (constant accel -> graph with constant slope = straight line)
7. C "e & f" (negative accel -> graph with negative slope at all times)
8. D "e & f" (slowing down -> graph getting closer to zero at all times)
9. C 17.1 m/s (v_0 = sqrt(2*g*h_peak) )
10. A 1.7 s (t = v_0 / g)
11. D momentum of cannonballer (p = m*v; since v decreases over time, p also decreases)
12. C 700 N (F = (coeff of friction)*m*g)
13. B 0.35 (coeff of friction = F / (m*g) )
14. B 1.9 kJ (Work = F * d)
15. D 803 N (F_drag = weight = m*g)
16. D constant momentum (p = m*v, and v = constant)
17. C 2.0 m (want center-of-mass position at point where suspension cable attaches = 1.5 m)
18. D point D (E_tot is conserved, so K is largest when U=m*g*y is smallest)
19. B equal to v_A (E_tot is conserved; since U_C=U_A, we see that K_C=K_A, so v_C=v_A)
20. A 20. m (E_B = E_D, so (K + U)_B = (K + U)_D, and use U_B=m*g*y to find y_B)
21. B equal to (E_tot is conserved)
22. E cannot determine using only the information given (conserv of E gives us v^2_A - v^2_E, but NOT v_A - v_E!)
23. C greater than (friction causes (K + U) to decrease over time)
24. E 2.8 m/s (impulse = delta(p) = p_f - p_i; use p_f = m*v_f to find v_f)
25. B 75 N (F = impulse / t)
26. B equal in strength (Newton's 3rd Law: force of paddle on puck = equal & opposite to force of puck on paddle)
27. B 1/3 (T = 2*pi / omega)
28. B 1/3 (L is conserved, so I_init * omega_init = I_final * omega_final )
29. C (L is conserved/constant, but K_rot = (1/2)*I*omega^2 becomes 3x larger)
30. D 1.7 rad/s^2 (alpha = delta(omega) / t)
31. B 13 rev (use rotational kinematics to find: delta(theta) = 84.0 rad, then convert rad -> revolutions)
32. A 190 kg/m^3 (density = m/V, where m = 14*75kg, and V = 5500 L = 5.5 m^3 )
33. A yes (object floats if: density(object) < density(fluid); here: 190 kg/m^3 < 1000 kg/m^3)
34. D 97 atm (P = F/A, where F = m*g, and convert final answer from [Pa] to [atm])
35. A roof will bulge upward/outward (Bernoulli's Principle: as v increases, P decreases)
36. E 20. kcal (Q = m*c*delta(T), where delta(T) = 37degC - 10degC = 27degC = 27K)
37. B heat of vaporization ("vaporization"=boiling -- NOT needed here; "fusion"=melting)
38. A entropy increases (delta(S) = Q/T, where Q = heat of fusion and T=273K, so delta(S) is positive)
39. D 30. m (wavelength = 2*L for fundamental mode)
40. A wire A has higher-frequency fundamental (wire A has lower (m/L), so wire A has faster v, and both wires have same wavelength)
41. B transverse waves (displacement of wire is perpendicular to direction of wave propagation)
42. E 60. m/s (v = wavelength * frequency)
43. D 10. Hz (frequency of nth overtone = n*f_1 = 2 Hz, 4 Hz, 6 Hz, 8 Hz, 10 Hz, ...)
44. E 7.0 m (period of ideal pendulum: T = 2*pi*sqrt(L/g) )
45. B period is unchanged (formula for T does NOT depend on mass of pendulum bob)
46. C 1 (period: T = 2*pi*sqrt(m/k) -- does NOT depend on amplitude of oscillations)
47. B 1/4 (E_tot = (1/2)*k*A^2, so when A=d, energy is 1/4 as large as when A=2d )
48. A 1/2 (v_max = A*sqrt(k/m), so when A=d, v_max is 1/2 as large as when A=2d )
49. D 7.2 cm (Hooke's Law: F_spring = k*x)
50. A 0.92 Hz (resonance occurs when f_driving = f_natural = sqrt(k/m) / (2*pi) )
51. C amplitude grows larger (description of resonant growth)
52. D isothermal (constant temperature)
53. A adiabatic (no heat added/removed)
54. B isobaric (constant pressure)
55. C Newton's 3rd Law of Motion
56. A Newton's 1st Law of Motion
57. E Kepler's 3rd Law
58. D Newton's Law of Universal Gravitation
59. C Doppler Effect
60. D Pascal's Principle
61. A Archimedes' Principle
62. E Principle of Superposition (of waves)
63. E Energy, Momentum, and Angular Momentum (three major conserved quantities & Conservation Laws)