Physics Final Exam - Summer 2014 Version 31 - Answer Key Page 1 of 20

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1 Physics Final Exam - Summer 2014 Version 31 - Answer Key Page 1 of A pendulum on the Earth has a period T. The acceleration due to gravity on Mars is less than that on the Earth, and the acceleration due to gravity on the Moon is even less than on Mars. Identical pendulums have the same mass and length. Where would the period of an identical pendulum be the smallest? (a) The period is smallest on the Earth. (b) The period is the same on the Earth, the Moon, and Mars. (c) The period is smallest on Mars. (d) The period is smallest on the Moon. (e) The period is smallest on both Mars and the Moon. The period of a pendulum is given by: T = 2π The length of the pendulum does not change. The acceleration due to gravity (g) changes on the Earth, the Moon, and Mars. The smallest period is when the acceleration due to gravity (g) is the largest. This occurs on the Earth. L g

2 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 2 of Joule s machine is shown on the right. A block with a mass of 6.1 kg is dropped some distance (z). A rope is attached both to the block and to a pulley which is connected to a rotating paddle. When the block drops, the paddle turns causing the temperature of the water to increase. If the mass of the water is 0.3 kg, what distance (z) must the mass drop to increase the temperature of the water by 0.5 C? (Assume that there is no friction.) (a) m (b) 8.35 m (c) m (d) m (e) 2.46 m Energy is conserved. The potential energy of the block is converted into the heat energy of the water. This gives the following: E initial = E final P E initial = Q final m block gz = m water c water T Solving for z: z = m waterc water T m block g = (0.3)(4186)(0.5) (6.1)(9.8) = m

3 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 3 of Which of the following changes the speed of sound in air? (a) frequency (b) temperature (c) period (d) amplitude (e) wavelength The speed of sound depends in a medium only depends on the properties of the medium. The temperature of the air changes the speed of sound in air.

4 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 4 of A treasure chest filled with gold has a mass of 930 kg and is sitting at the bottom of the ocean in a place that is 2.6 m deep. A chain is attached to the treasure chest so that the treasure chest can be pulled up to the surface. Since the treasure chest is filled with gold, assume that the chest has the density of gold. Assume the ocean water has the density of regular water. What force is required to lift the treasure chest? (a) 4,722 N (b) 3,274 N (c) 15,205 N (d) 8,642 N (e) 9,978 N There are three forces acting on the treasure chest: the force of gravity, tension in the chain, and the force of buoyancy. The sum of the forces for the treasure chest gives the following: Solving for tension: ΣF y : T + F B F g = ma T + F B F g = 0 T = F g F B T = m chest g ρ water V chest g The volume of the chest can be found using the density and mass: ρ = m V V chest = m chest ρ gold Substituting this into the tension equation gives: ( ) mchest T = m chest g ρ water g ρ gold ( ) 930 T = (930)(9.8) (1000) (9.8) 19, 300 T = 8,642 N

5 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 5 of A certain string has a mass of 0.15 kg and a length of 0.64 m. A tension of 39 N is applied to the string. What is the velocity of a wave on the string? (a) 0.69 m/s (b) m/s (c) m/s (d) m/s (e) 3.02 m/s The velocity of a wave on a string is given by: T v = µ The mass density is the mass divided by the length. (µ = M/L) Substituting this into the previous equation gives the following: T L (39)(0.64) v = M = = m/s 0.15

6 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 6 of An air bubble ascends from the bottom of a lake that is 22.3 m deep. The temperature at the bottom of the lake is 263 K, and near the surface the temperature is 289 K. If the volume of the bubble is 83.6 m 3 at top of the lake, what is the volume of the bubble at the bottom of the lake? (a) m 3 (b) m 3 (c) m 3 (d) m 3 (e) m 3 P V = nrt The number of moles is constant. From the Ideal Gas Law, this gives the following ratio: P bot V bot T bot = P topv top T top The pressure at the top of the lake is atmospheric pressure. P top = Pa. The pressure at the bottom of the lake depends on the depth, as shown: P bot = P top + ρgh P bot = (1000)(9.8)(22.3) P bot = Pa Solving for the volume at the bottom of the lake: V bot = P topv top T bot P bot T top V bot = ( )(83.6)(263) ( )(289) V bot = m 3

7 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 7 of Two metal blocks are placed in a room and have the same initial temperature. Both blocks are made of steel, and both blocks are heated by the same amount (Q). Block Two has twice the mass of Block One. How does the change in temperature of Block One compare to the change in temperature of the Block Two? (a) T one = T two /2 (b) T one = T two /4 (c) T one = 4 T two (d) T one = T two (e) T one = 2 T two Both blocks are heated by the same amount, which gives the following: Q one = Q two m one c steel T one = m two c steel T two m one T one = m two T two Block Two has twice the mass of Block One, which means m two = 2 m one. This gives the following: m one T one = (2 m one ) T two T one = 2 T two The temperature change in Block One is twice the temperature change in Block Two.

8 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 8 of A speaker with a power of 34 W is playing music. What is the sound intensity level of the music at a distance of 22 m? Assume that the sound can only be heard in front of the speaker and not behind it. (a) 36.4 db (b) db (c) db (d) 78.5 db (e) 11.2 db In order to find the sound intensity level, first the intensity must be calculated. Since the sound only goes in front of the speaker, the area is half of the surface area of a sphere, which is 2πr 2. The intensity is found by: I = Power Area = Power 2πr 2 = 34 2π(22) 2 = W/m 2 The sound intensity level is given by: β = (10 db) log 10 I I 0 = (10 db) log = (10 db) log 10 ( ) = (10 db)(10.05) = db

9 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 9 of A 2.9 kg mass is attached to a spring on a table. Assume that there is no friction. The mass oscillates with a period of 9.5 s. If the maximum amplitude of the mass is 5.3 m, what is the maximum velocity of the mass on the spring? (a) 1.92 m/s (b) 1.44 m/s (c) 0.62 m/s (d) 6.37 m/s (e) 3.51 m/s The angular velocity can be found from the period as shown: The max velocity is given by: ω = 2π T = 2π 9.5 = 0.66 rad/s v max = A ω = (5.3 m)(0.66 rad/s) = 3.51 m/s

10 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 10 of A bank vault is opened by applying a force of 102 N perpendicular to the plane of the door, 1.5 m from the hinges. Find the torque due to this force about an axis through the hinges. (a) 68.0 Nm (b) Nm (c) 0.0 Nm (d) Nm (e) Nm Torque about an axis is equal to the radius times the force times the sine of the angle between the force and distance. τ = rf sin(θ) = (1.5 m)(102 N)(sin(90)) = Nm

11 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 11 of A suction cup has an air pressure of 20.0 kpa inside the suction cup when it is attached to a ceiling. The suction cup covers an area of 2.04 cm 2, and a mass is hanging from the suction cup. What is the largest mass that the suction cup can hold before falling? (a) 1.18 kg (b) 0.42 kg (c) 2.81 kg (d) 1.69 kg (e) kg The suction cup is held up because of the force due to the difference of pressure inside and outside of the suction cup. The pressure outside is atmospheric pressure. Sum of the forces for the suction cup: Sum of the forces for the hanging mass ΣF y : F out F in T = ma = 0 P out A P in A T = 0 (P out P in )A = T ΣF y : T F g = ma = 0 T = F g = mg Substituting the second equation into the first and solving for mass gives: m = (P out P in )A g = ( )( ) (9.8) = 1.69 kg

12 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 12 of A heat engine has two reservoirs, one at a temperture of 350 K, and the other at 900 K. The heat engine takes 8,400 J of heat from the hot reservoir, expels 7,400 J of heat to the cold reservoir, and does 1,000 J of work. Which of the following is true about this heat engine? (a) This heat engine is possible since the entropy of the system decreases. (b) This heat engine is possible but is an ideal reversible heat engine. (c) This heat engine is possible and is a regular irreversible heat engine. (d) This heat engine is not allowed by the second law of thermodynamics. (e) This heat engine is not allowed by the first law of thermodynamics. The highest possible efficiency for a heat engine is for an ideal or irreversible heat engine and is given by: e ideal = 1 T c T h e ideal = e ideal = e ideal = 0.61 The efficiency for a real heat engine is given by: e = W Q H e = 1,000 8,400 e = 0.12 This heat engine is possible since the calculated efficiency is less than the ideal efficiency.

13 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 13 of A container has a volume of 5.76 m 3 and holds 15.8 moles of an ideal gas. The pressure in the container is 83.5 kpa. Wat is the temperature of the gas? (a) 6, K (b) K (c) 3, K (d) K (e) 1, K P V = nrt From the Ideal Gas Law, solving for temperature gives: T = P V nr = ( Pa)(5.76 m 3 ) (15.8 mol)(8.31 J/(K mol)) = 3, K

14 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 14 of A cart with a mass of 3.3 kg is pushed to the right with a velocity of 8.8 m/s. A second cart with a mass of 7.5 kg is in front of the first cart and is moving to the right with a velocity of 3.2 m/s. The first cart collides with the back of the second cart in an ineslastic collision. After the collision, the first cart moves with a final velocity of 0.5 m/s. What is the final velocity of the second cart? (a) 7.77 m/s (b) 1.85 m/s (c) m/s (d) 6.85 m/s (e) 8.57 m/s Momentum is converserved for all collisions. Conservation of Momentum gives the following: p 1,i + p 2,i = p 1,f + p 2,f m 1 v 1,i + m 2 v 2,i = m 1 v 1,f + m 2 v 2,f This problem only has motion in one direction (the x-direction) so only the x-components of the momentum are needed, and the notation can be simplified. Solving for v 2,f : v 2,f = m 1v 1,i + m 2 v 2,i m 1 v 1,f m 1 v 1,i + m 2 v 2,i = m 1 v 1,f + m 2 v 2,f m 2 (3.3 kg)(8.8 m/s) + (7.5 kg)(3.2 m/s) (3.3 kg)(0.5 m/s) = 7.5 kg = 6.85 m/s

15 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 15 of A figure skater initially has her arms extended. She starts spinning on the ice at 3 rad/s. She then pulls her arms in close to her body. Which of the following happens? (a) She spins at a higher angular velocity. (b) She spins at the same angular velocity. (c) Her angular momentum increases. (d) Her angular momentum decreases. (e) She spins at a lower angular velocity. Angular momentum of the system is convserved. L skater,initial = L skater,final I skater,initial ω skater,initial = I skater,final ω skater,final m skater r 2 skater,initial ω skater,initial = m skater r 2 skater,final ω skater,final As the figure skater pulls in her arms, her radius decreases. In order to conserve angular momentum, her angular speed (ω) increases.

16 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 16 of The equation for a traveling wave is: I) What is the amplitude of the wave? II) What is the frequency of the wave? III) What is the wavelength of the wave? y(x, t) = (3.7 m) sin(83 t 7.7 x) (a) I) 3.7 m II) Hz III) 0.82 m (b) I) 3.7 m II) 6.17 Hz III) 1.94 m (c) I) 7.7 m II) 0.82 Hz III) m (d) I) 7.7 m II) Hz III) 1.94 m (e) I) 3.7 m II) 0.82 Hz III) m The equation for a traveling wave is given by: y(x, t) = A sin(ω t k x) y(x, t) = A sin(2πft 2π λ x) From the equation given in the problem, the amplitude (A) is 3.7 m. The angular frequency (ω) is 83 rad/s. and the wave number (k) is 7.7 m 1. The frequency (f) is related to the angular frequency (ω) and can be found by the following: ω = 2πf f = ω 2π = 83 = Hz 2π The wavelength (λ) is related to the wave number (k) and can be found by the following: k = 2π λ λ = 2π k = 2π 7.7 = 0.82 m

17 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 17 of A Pitot-Static tube is used to measure pressure in order to find an airplane s velocity. The pressure at 1, where the velocity of air is zero, is Pa, and the pressure at 2 is Pa. The airplane is flying at an altitude where the density of air is 0.77 kg/m 3. What is the velocity of the airplane? (a) m/s (b) m/s (c) m/s (d) m/s (e) m/s The height of point 1 and point 2 are similar and aproximately zero. The velocity of point 1 is zero. This gives the following: P ρv2 1 + ρgh 1 = P ρv2 2 + ρgh 2 P ρ(0)2 + ρg(0) = P ρv2 2 + ρg(0) Solving for v 2 : v 2 = 2(P 1 P 2 ) ρ P 1 = P ρv2 2 2( = 4 ) = m/s 0.77

18 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 18 of A ball with a mass of 2.3 kg is thrown up at a 67.6 angle above the horizontal. The ball initially has a velocity of 54.9 m/s and starts approximately on the ground. How long is the ball in the air before it hits the ground? (a) s (b) 5.18 s (c) 2.13 s (d) s (e) 2.36 s The initial velocity in the x-direction and y-direction are given by: v 0x = v 0 cos θ v 0y = v 0 sin θ The ball is being thrown straight up. At the very top, the velocity in the y-direction is zero (v fy = 0). The acceleration in the y-direction is -9.8 m/s 2. Use the kinematic equation with velocity, time, and acceleration. v fy = v 0y + a y t Solving for time gives: t = v f y v 0y a = 0 v 0 sin θ a = (54.9 m/s) sin m/s 2 = 5.18 s The total time the ball in in the air is twice the time it takes for the ball to travel to the top of its path. t total = 2t = 2(5.18 s) = s

19 Version 31 - Answer Key Physics Final Exam - Summer 2014 Page 19 of 20 Refer to the following P-V diagram for Problem 19 and Problem 20. In the figure above, P 1 = 30.2 kpa, P 2 = 62.1 kpa, P 3 = 88.1 kpa, V 1 = 2.24 m 3, and V 2 = 7.01 m A gas undergoes a two part process from A to C as shown in the P-V diagram above. The first part is from A to B and the second from B to C. What is the work done by the gas during this process (all parts from A to C)? (a) kj (b) kj (c) kj (d) kj (e) kj The work for A B is the area under the curve which can be broken up into a rectange and a triangle. This gives the following: W AB = P 2 (V 2 V 1 ) + (1/2)(P 3 P 2 )(V 2 V 1 ) = (62.1 kpa)(7.01 m m 3 ) + (1/2)(88.1 kpa 62.1 kpa)(7.01 m m 3 ) = kj kj = kj The work done on a PV-diagram is the area under the curve. Now calculate the total work: W A C = W AB + W BC W A C = W AB + P BC V BC W A C = W AB + P BC (0) W A C = W AB + 0 = kj

20 Physics 220/Final Exam Page 20 of 20 SOLUTIONS 20. After going through the two part process from A to C as explained in Problem 19, the gas undergoes another two part process. Part one is from C to D, and part two is from D to A. The gas has now undergone one complete cycle and is back at A (where it started). What is the work done by the gas during the entire cycle (all four parts A-B-C-D-A)? (a) kj (b) kj (c) kj (d) kj (e) kj The work done on a PV-diagram is the area under the curve. From Problem 19: W A A = W AB + W BC + W CD + W DA W A A = W AB + P BC V BC + W CD + P DA V DA W A A = W AB + P BC (0) + W CD + P DA (0) W A A = W AB + W CD W AB = P 2 (V 2 V 1 ) + (1/2)(P 3 P 2 )(V 2 V 1 ) = (62.1 kpa)(7.01 m m 3 ) + (1/2)(88.1 kpa 62.1 kpa)(7.01 m m 3 ) = kj kj = kj The work for D A is found by the following: W CD = P 1 (V 1 V 2 ) = (30.2 kpa)(2.24 m m 3 ) = kj Now calculate the total work: W A A = W AB + W CD = kj kj = kj