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# Practice Exam 5 for General Physics 1 PHYS113

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NAME_________________________________________ Date__________ Quiz instructions You should have both an exam and a machine-gradeable answer card (Scantron). On the other side of this page is an official “physics fact sheet”. Detach it if you wish. Not all of the information on the “fact sheet” will be useful. Before you start: ● Fill in the bubbles for your Wildcat ID number. Scantron cards without ID number bubbles filled out won't be counted. ● Fill in the the test version number (written in big letters before the first question), and the answer sheet number (1). ● Also print your name, the course, and the date in the box at the top of the card. ● Also write your name on the first and last pages of the exam. (If you detach this page, the next page is the first page, so write your name on that page, and also the last page. If you leave this page attached, then write your name on this page, and the last page.) ● Take a deep breath. Relax and pay attention. It's really not that bad. Please note: ● Your score on the multiple choice questions will be determined by your answers on the answer card. However, it is a good idea to mark your answers on the exam sheet too, for your own reference. ● Do not fold or roll the card. Unreadable cards won't be counted. ● Mark your responses on the answer card by filling in the bubbles completely with a #2 pencil. ● Do not fill in more than one response for each question. ● If you erase an answer, make sure you erase it completely. ● If you need a new card, please ask. When you are done: ● Recheck your filled-in circles against your exam sheet. ● Be sure that you have answered all questions, that you have all of the pages of the exam, and that both the question number and your response match agree with your exam sheet. You should see the words “The End” after the last question. ● Turn in both the Scantron card and your exam at the front of the room. ● Your exam will be returned to you after all exams are scanned and checked and the written part is graded. ● Please write your name on your exam. Constants SI prefixes Gravity at Earth's surface g = 9.80 m/s , down pico 10-12 p tera 1012 T -11 3 -1 -2 -9 9 Gravitational constant G = 6.674 × 10 5 kg 2 nano 10-6 n giga 106 G Atmospheric pressure 1 atm = 1.013×10 N/m micro 10 μ mega 10 M Boltzmann's constant k = 1.38×10 J/K milli 10-3 m kilo 103 k Ideal gas constant R = 8.31 J/(mole·K) centi 10-2 c hecto 102 h 23 Avagadro's number N A 6.02×10 8 Speed of light c = 2.998 × 10 m/s Trigonometry sin θ = “opposite” / “hypotenuse” Kinematics, dynamics, energy, momentum cos θ = “adjacent” / “hypotenuse” Δ x Δθ hypotenuse = opposite + adjacent ⃗= ⃗ ω= Δt Δt t h Δ⃗v α= Δω s y pote a= Δt o n use Δt ∑ τ=I α p θ ∑ F=ma ⃗ o adjacent 1 2 θ(t)=θ +ω t+ 1 α t ⃗(t)=x0+v⃗0+ a ⃗const 0 0 2 const 2 ω =ω +2αΔθ Impulses and collisions v =v 02aΔ x 0 Impulse: F Δ t=Δ p⃗ 1 2 KE = mv 2 KE rotati2n Collision: ∑ ⃗initial ⃗ final linear2 W=F dcosθ W=τΔθ Simple harmonic motion and waves Fd L=Iω ⃗=mv ⃗ inertia Period: T=2π √restoring Some kinds of potential energy Spring: F=k x Uniform gravity: PE Gmgy Pendulum: F=mg x/L 1 2 Frequency: f =1/T Spring: PE S k2 Oscillator: v0/ A=2π f Waves: v= f λ Geometry of circles and spheres Standing: f =n(v/2L),any n 360 degrees = 2π radians = 1 full circle n Circumference of a circle: 2πr fn=n(v/4L),odd n Rolling distance = (radius)*(angle in radians) Small angles: (linear size of object) ≈ (arc length) Other definitions and relations Area of a sphere: 4πr Density: ρ=m/V Pressure: P=F/A Circular or rotational motion 2 In a fluid: Δ P=ρ gΔ y v=rω, a = v =rω 2 Buoyancy: FB=ρ fluidg centrir m m 2 Gravity (sphere): F =G 1 2 τ=r Fsinθ, I= ∑ mr G r2 Friction: F =μF Static equilibrium F N Ideal gas: PV=nRT=NkT ax=a y0 → ∑ F x ∑ F y0 3 α=0 → ∑ τ=0 Avg transl. KE: KE= kT 2 Specific heat: Q=cmΔT Latent heat: Q=mL Name: Date: General Physics exam 5 Practice, version 1 1. An amusement park ride swings two people of mass 60 kg in a circle of radius 2.6 m. If the total angular momentum of the two people is 1160 kg*m /s, what is their angular velocity? A. 1.4 rad/s, calculated from (1160 kg*m /s)/(2*(60 kg)*(2.6 m) ) +02 B. 1.6×10 rad/s, calculated from (60 kg)*(2.6 m) C. 0.7 rad/s, calculated from 2*(60 kg)*(2.6 m) /(1160 kg*m /s) +07 2 2 D. 3.1×10 rad/s, calculated from (60 kg)*(1160 kg*m /s) /(2.6 m) 2. If air resistance and other sources of friction exert a torque 26 m*N on the ride, what torque must be provided to the ride to keep it spinning at a constant angular velocity? A. 26 m*N B. 0.038 m*N C. 52 m*N D. 13 m*N E. 0 m*N 3. If the people on the ride pull themselves in closer to the axis of rotation by pullin on the chain connected to their swing, their angular momentum _____ and their angular speed _____. A. decreases; stays the same B. increases; stays the same C. decreases; increases D. stays the same; stays the same E. stays the same; increases 3 4. Water has a density of 1.00 g/cm What is the mass of 1 cubic meter of water? A. 1 kg, calculated from (1.00 g/cm ) * (1.00 m ) +06 3 3 3 B. 1×10 kg, calculated from (1.00 g/cm ) * (100 cm/m) * (1.00 m ) C. 1×10 +03 kg, calculated from (1.00 g/cm ) * (0.001 kg/g) * (100 cm/m) * 3 (1.00 m ) D. 1×10 -09kg, calculated from ((1.00 g/cm ) * (0.001 kg/g) / (100 cm/m) ) * 3 (1.00 m ) page 1 of 6 5. How much higher is the absolute pressure 1.2 m under the surface of a pool of fresh 3 3 water (density = 1.00×10 kg/m ) than at the surface of the pool? A. 1.2×10 +03 N/m , calculated from (1.00×10 kg/m )*(1.2 m) B. 1.2×10 +04 N/m , calculated from (1.00×10 kg/m )*(9.8 m/s )*(1.2 m) 2 2 C. 12 N/m , calculated from (9.8 m/s )*(1.2 m) D. 0.012 N/m , calculated from (9.8 m/s )*(1.2 m)/(1.00×10 kg/m ) 3 6. If the pool in the previous problem were filled with ocean water (density = 1.025×10 kg/m ), what then would be the difference between absolute pressure at 1.2 m* depth and pressure at the surface of the water? A. 12 N/m , calculated from (9.8 m/s )*(1.2 m) B. 0.011 N/m , calculated from (9.8 m/s )*(1.2 m)/(1.025×10 kg/m ) 3 +04 2 3 3 2 C. 1.2×10 N/m , calculated from (1.025×10 kg/m )*(9.8 m/s )*(1.2 m) D. 1.2×10 +03 N/m , calculated from (1.025×10 kg/m )*(1.2 m) 2 7. A hydraulic brake cylinder with area 0.00061 m applies 1300 N of force to the brakes. What is the pressure in the hydraulic line? 2 2 A. 0.79 N*m , calculated from (1300 N)*(0.00061 m ) B. 1.3×10 +03 N, calculated from (1300 N) C. 5.3×10 +04 N/m, calculated from (1300 N)/sqrt((0.00061 m )) +06 2 2 D. 2.1×10 N/m , calculated from (1300 N)/(0.00061 m ) 8. The brake pedal of a car is attached to a "master cylinder" which transits pressure into 2 the hydaulic system. If the master cylinder has area 0.00023 m what force has to be applied in order to cause the 1300 N* force in the brake cylinder of the previous problem? A. 3.4×10 +03 N, calculated from (1300 N)*(0.00061 m )/(0.00023 m ) 2 +06 2 B. 2.1×10 N*m, calculated from (1300 N)/(0.00061 m ) C. 0.79 N, calculated from (1300 N)*(0.00061 m ) 2 D. 4.9×10 +02 N, calculated from (1300 N)*(0.00023 m )/(0.00061 m ) 2 +06 2 E. 5.7×10 N, calculated from (1300 N)/(0.00023 m ) page 2 of 6 9. A distillery in Russia pumps vodka (density = 0.789×10 kg/m ) from one level to a higher level in their factory, a height difference of 3.4 m. How much higher is the pressure at the lower floor compare to the higher floor? +02 2 3 3 A. 2.3×10 N/m , calculated from (0.789×10 kg/m )/(3.4 m) B. 2.6×10 +04N/m , calculated from (0.789×10 kg/m )*(9.8 m/s )*(3.4 m) C. 2.7×10 +03N/m , calculated from (0.789×10 kg/m )*(3.4 m) +04 2 3 3 2 D. 1.3×10 N/m , calculated from (0.789×10 kg/m )*(9.8 m/s )*(3.4 m)/2 3 10. A spherical balloon is filled with helium (density = 0.176 kg/m ) The balloon itself has a mass of 0.15 kg unfilled. When filled, it has a volume of 0.76 m Air has a density of 1.23 kg/m What is the buoyant force of air on the balloon when filled? 3 3 2 A. 1.31 N, calculated from (0.176 kg/m )*(0.76 m )*(9.8 m/s ) B. 11 N/m , calculated from (1.23 kg/m )*(9.8 m/s )*(0.76 m ) 3 1/3 C. 12.1 N, calculated from (1.23 kg/m )*(9.8 m/s ) 2 3 3 2 D. 9.16 N, calculated from (1.23 kg/m )*(0.76 m )*(9.8 m/s ) 11. What is the force of gravity on the full balloon including the helium inside it, when the balloon is filled? A. 2.8 N, calculated from ((0.176 kg/m )*(0.76 m ) + (0.15 kg))*(9.8 m/s ) 2 3 3 2 B. 9.2 N, calculated from (1.23 kg/m )*(0.76 m )*(9.8 m/s ) 2 C. 1.5 N, calculated from (0.15 kg)*(9.8 m/s ) D. 11 N, calculated from ((1.23 kg/m )*(0.76 m ) + (0.15 kg))*(9.8 m/s ) 2 12. If the full balloon is tied to the ground by a string, what is the tension in the string? A. 1.5 N, calculated from (0.15 kg)*(9.8 m/s ) 2 B. 6.4 N, calculated from (1.23 kg/m )*(0.76 m )*(9.8 m/s ) - ((0.176 kg/m )* 3 3 2 (0.76 m ) + (0.15 kg))*(9.8 m/s ) C. 1.3 N, calculated from (0.176 kg/m )*(0.76 m )*(9.8 m/s ) 2 D. 9.2 N, calculated from (1.23 kg/m )*(0.76 m )*(9.8 m/s ) 2 page 3 of 6 13. A 0.75 kg piece of solid styrofoam floats on water. What is the volume of styrofoam in 3 3 3. the water? (Water density = 1.00×10 kg/m , styrofoam density = 23 kg/m ) A. 9.8×10 +03 m , calculated from (9.8 m/s )*(1.00×10 kg/m ) 3 B. 0.033 m , calculated from (0.75 kg)/(23 kg/m )3 +12 3 2 3 3 3 C. 2.2×10 m , calculated from ((9.8 m/s )*(1.00×10 kg/m )/(0.75 kg)) D. 0.0074 m , calculated from (0.75 kg)*(9.8 m/s )/(1.00×10 kg/m ) 3 3 3 3 E. 0.00075 m , calculated from (0.75 kg)/(1.00×10 kg/m ) 14. What fraction of the styrofoam's volume is in the water, as a percentage? A. 1×10 +02 %, calculated from 100 3 3 3 B. 2.3 %, calculated from 100*((0.75 kg)/(1.00×10 kg/m ))/((0.75 kg)/(23 kg/m )) C. 4.3×10 +03 %, calculated from 100*(1.00×10 kg/m )/(23 kg/m ) 3 3 3 3 D. 43 %, calculated from (1.00×10 kg/m )/(23 kg/m ) 15. A concrete slab 6.8 m long expands 7.9 m when the temperature increase by 5.5 ℃. What is the coefficient of expansion of this concrete? A. 0.21 /℃, calculated from ((7.9 m)/(6.8 m))/(5.5 ℃) B. 1.2, calculated from (7.9 m)/(6.8 m) C. 1.4 m/℃, calculated from (7.9 m)/(5.5 ℃) D. 1.2 m/℃, calculated from (6.8 m)/(5.5 ℃) 3 16. A helium-filled balloon has a volume of 6.3 m at atmospheric pressure and temperature 16.8 ℃. When cooled to 6.7 ℃, what is its volume at the same pressure? 3 3 A. 2.5 m , calculated from (6.3 m )*(6.7 ℃)/(16.8 ℃) B. 6.1 m , calculated from (6.3 m )*((6.7 ℃)+273.15)/((16.8 ℃)+273.15) 3 3 C. 6.5 m , calculated from (6.3 m )*((16.8 ℃)+273.15)/((6.7 ℃)+273.15) D. 16 m , calculated from (6.3 m )*(16.8 ℃)/(6.7 ℃) 17. One cubic meter of air at 0℃ and standard atmospheric pressure (1.000 atm) has a mass of 1.28 kg. How much volume does the same mass of air occupy at the same temperature if the pressure increases to 1.025 atm? A. 1 m , calculated from (1.000 m ) 3 B. 1.02 m , calculated from (1.000 m )*(1.025 atm)/(1.000 atm) 3 3 C. 0.976 m , calculated from (1.000 m )*(1.000 atm)/(1.025 atm) D. 0.976 m /atm, calculated from (1.000 m )/(1.025 atm) page 4 of 6 18. If the root-mean-square (RMS) speed of molecules in a gas increases by a factor 5, by what factor does the temperature increase? A. 5, calculated from (5) B. 25, calculated from (5) 2 C. 0.2, calculated from 1/(5) D. 10, calculated from 2*(5) 19. The sun keeps the earth warm. What is the primary way heat reaches the earth from the sun through the vacuum of space? A. radiation B. conduction C. convection 20. Koreans use steel chopsticks. When eating in a restaurant in Korea, an American tourist foolishly (and impolitely) leaves the ends of the steel chopsticks sticking in a hot bowl of rice while talking on his cell phone. When he tries to pick up the chopsticks, the chopsticks are hot everywhere. How did heat travel through the steel? A. conduction B. convection C. radiation 21. If the heat capacity of water is 4184 J/(kg·℃), how much energy does it take to raise the temperature of 7.5 kg of water by 5.6 ℃ using an electric heater? Assume perfect thermal insulation around the water. A. 1×10 +02 J, calculated from (4184 J/(kg·℃))/((5.6 ℃)*(7.5 kg)) +04 B. 2.3×10 J, calculated from (4184 J/(kg·℃))*(5.6 ℃) C. 4.2×10 +03 J, calculated from (4184 J/(kg·℃)) +05 D. 1.8×10 J, calculated from (4184 J/(kg·℃))*(7.5 kg)*(5.6 ℃) E. 3.1×10 +04 J/kg, calculated from (4184 J/(kg·℃))*(7.5 kg) 22. How much time should it take to defrost a 8.6 kg frozen dinner in a 1150 W microwave oven if all the power of the microwave goes into melting the dinner? Assume the frozen dinner has the same latent heat of fusion as ice, 334×10 J/kg. +07 3 A. 4.5×10 s, calculated from (334×10 J/kg)*(1150 W)/(8.6 kg) B. 2.5×10 +03 s, calculated from (334×10 J/kg)*(8.6 kg)/(1150 W) -08 3 C. 2.2×10 s, calculated from (8.6 kg)/((1150 W)*(334×10 J/kg)) D. 2.9×10 +06 s, calculated from (334×10 J/kg)*(8.6 kg) +02 3 E. 2.9×10 s, calculated from (334×10 J/kg)/(1150 W) Don't stop here, you're not done yet.... page 5 of 6 Short written answer questions A circus stunt diver whose mass is 60.0 kg belly-flops into a tub containing 8000 kg of water from 5.00 m above the surface of the water. (i) How much does the temperature of the water increase due to the diver stopping quickly in the water? Assume all the water stays in the tub, as does all the energy transferred from the belly flop. Use 4.186 J/(g K) for the specific heat of water. (ii) If the diver stays in the tub of for 60 seconds and loses 100 J per second of heat to the water, how much further does the water temperature increase? Assume no heat transfer through the walls and floor of the tub and surface of the water. Don't turn in your exam until you've checked your WID bubbles, your version number bubble, your name.... The End page 6 of 6 NAME_________________________________________ Date__________ Quiz instructions You should have both an exam and a machine-gradeable answer card (Scantron). On the other side of this page is an official “physics fact sheet”. Detach it if you wish. Not all of the information on the “fact sheet” will be useful. Before you start: ● Fill in the bubbles for your Wildcat ID number. Scantron cards without ID number bubbles filled out won't be counted. ● Fill in the the test version number (written in big letters before the first question), and the answer sheet number (1). ● Also print your name, the course, and the date in the box at the top of the card. ● Also write your name on the first and last pages of the exam. (If you detach this page, the next page is the first page, so write your name on that page, and also the last page. If you leave this page attached, then write your name on this page, and the last page.) ● Take a deep breath. Relax and pay attention. It's really not that bad. Please note: ● Your score on the multiple choice questions will be determined by your answers on the answer card. However, it is a good idea to mark your answers on the exam sheet too, for your own reference. ● Do not fold or roll the card. Unreadable cards won't be counted. ● Mark your responses on the answer card by filling in the bubbles completely with a #2 pencil. ● Do not fill in more than one response for each question. ● If you erase an answer, make sure you erase it completely. ● If you need a new card, please ask. When you are done: ● Recheck your filled-in circles against your exam sheet. ● Be sure that you have answered all questions, that you have all of the pages of the exam, and that both the question number and your response match agree with your exam sheet. You should see the words “The End” after the last question. ● Turn in both the Scantron card and your exam at the front of the room. ● Your exam will be returned to you after all exams are scanned and checked and the written part is graded. ● Please write your name on your exam. Constants SI prefixes Gravity at Earth's surface g = 9.80 m/s , down pico 10-12 p tera 1012 T -11 3 -1 -2 -9 9 Gravitational constant G = 6.674 × 10 5 kg 2 nano 10-6 n giga 106 G Atmospheric pressure 1 atm = 1.013×10 N/m micro 10 μ mega 10 M Boltzmann's constant k = 1.38×10 J/K milli 10-3 m kilo 103 k Ideal gas constant R = 8.31 J/(mole·K) centi 10-2 c hecto 102 h 23 Avagadro's number N A 6.02×10 8 Speed of light c = 2.998 × 10 m/s Trigonometry sin θ = “opposite” / “hypotenuse” Kinematics, dynamics, energy, momentum cos θ = “adjacent” / “hypotenuse” Δ x Δθ hypotenuse = opposite + adjacent ⃗= ⃗ ω= Δt Δt t h Δ⃗v α= Δω s y pote a= Δt o n use Δt ∑ τ=I α p θ ∑ F=ma ⃗ o adjacent 1 2 θ(t)=θ +ω t+ 1 α t ⃗(t)=x0+v⃗0+ a ⃗const 0 0 2 const 2 ω =ω +2αΔθ Impulses and collisions v =v 02aΔ x 0 Impulse: F Δ t=Δ p⃗ 1 2 KE = mv 2 KE rotati2n Collision: ∑ ⃗initial ⃗ final linear2 W=F dcosθ W=τΔθ Simple harmonic motion and waves Fd L=Iω ⃗=mv ⃗ inertia Period: T=2π √restoring Some kinds of potential energy Spring: F=k x Uniform gravity: PE Gmgy Pendulum: F=mg x/L 1 2 Frequency: f =1/T Spring: PE S k2 Oscillator: v0/ A=2π f Waves: v= f λ Geometry of circles and spheres Standing: f =n(v/2L),any n 360 degrees = 2π radians = 1 full circle n Circumference of a circle: 2πr fn=n(v/4L),odd n Rolling distance = (radius)*(angle in radians) Small angles: (linear size of object) ≈ (arc length) Other definitions and relations Area of a sphere: 4πr Density: ρ=m/V Pressure: P=F/A Circular or rotational motion 2 In a fluid: Δ P=ρ gΔ y v=rω, a = v =rω 2 Buoyancy: FB=ρ fluidg centrir m m 2 Gravity (sphere): F =G 1 2 τ=r Fsinθ, I= ∑ mr G r2 Friction: F =μF Static equilibrium F N Ideal gas: PV=nRT=NkT ax=a y0 → ∑ F x ∑ F y0 3 α=0 → ∑ τ=0 Avg transl. KE: KE= kT 2 Specific heat: Q=cmΔT Latent heat: Q=mL Name: Date: General Physics exam 5 Practice, version 2 1. An amusement park ride swings two people of mass 65 kg in a circle of radius 3.3 m. If the total angular momentum of the two people is 1250 kg*m /s, what is their angular velocity? A. 1.1 rad/s, calculated from 2*(65 kg)*(3.3 m) /(1250 kg*m /s) 2 2 B. 0.88 rad/s, calculated from (1250 kg*m /s)/(2*(65 kg)*(3.3 m) ) C. 3.1×10 +07 rad/s, calculated from (65 kg)*(1250 kg*m /s) /(3.3 m) +02 D. 2.1×10 rad/s, calculated from (65 kg)*(3.3 m) 2. If air resistance and other sources of friction exert a torque 27 m*N on the ride, what torque must be provided to the ride to keep it spinning at a constant angular velocity? A. 0.037 m*N B. 54 m*N C. 27 m*N D. 0 m*N E. 14 m*N 3. If the people on the ride pull themselves in closer to the axis of rotation by pullin on the chain connected to their swing, their angular momentum _____ and their angular speed _____. A. decreases; stays the same B. increases; stays the same C. stays the same; increases D. stays the same; stays the same E. decreases; increases 3 4. Water has a density of 1.00 g/cm What is the mass of 1 cubic meter of water? A. 1 kg, calculated from (1.00 g/cm ) * (1.00 m ) +03 3 3 B. 1×10 kg, calculated from (1.00 g/cm ) * (0.001 kg/g) * (100 cm/m) * (1.00 m ) -09 3 3 C. 1×10 kg, calculated from ((1.00 g/cm ) * (0.001 kg/g) / (100 cm/m) ) * (1.00 m ) +06 3 3 3 D. 1×10 kg, calculated from (1.00 g/cm ) * (100 cm/m) * (1.00 m ) page 1 of 6 5. How much higher is the absolute pressure 1.2 m under the surface of a pool of fresh 3 3 water (density = 1.00×10 kg/m ) than at the surface of the pool? A. 1.2×10 +04 N/m , calculated from (1.00×10 kg/m )*(9.8 m/s )*(1.2 m) B. 0.012 N/m , calculated from (9.8 m/s )*(1.2 m)/(1.00×10 kg/m ) 3 2 2 C. 12 N/m , calculated from (9.8 m/s )*(1.2 m) D. 1.2×10 +03 N/m , calculated from (1.00×10 kg/m )*(1.2 m) 6. If the pool in the previous problem were filled with ocean water (density = 1.025×10 kg/m ), what then would be the difference between absolute pressure at 1.2 m* depth and pressure at the surface of the water? A. 0.011 N/m , calculated from (9.8 m/s )*(1.2 m)/(1.025×10 kg/m ) 3 B. 12 N/m , calculated from (9.8 m/s )*(1.2 m) +04 2 3 3 2 C. 1.2×10 N/m , calculated from (1.025×10 kg/m )*(9.8 m/s )*(1.2 m) D. 1.2×10 +03 N/m , calculated from (1.025×10 kg/m )*(1.2 m) 2 7. A hydraulic brake cylinder with area 0.00065 m applies 1200 N of force to the brakes. What is the pressure in the hydraulic line? +06 2 2 A. 1.8×10 N/m , calculated from (1200 N)/(0.00065 m ) B. 0.78 N*m , calculated from (1200 N)*(0.00065 m ) 2 C. 1.2×10 +03 N, calculated from (1200 N) +04 2 D. 4.7×10 N/m, calculated from (1200 N)/sqrt((0.00065 m )) 8. The brake pedal of a car is attached to a "master cylinder" which transits pressure into 2 the hydaulic system. If the master cylinder has area 0.00023 m what force has to be applied in order to cause the 1200 N* force in the brake cylinder of the previous problem? A. 3.4×10 +03 N, calculated from (1200 N)*(0.00065 m )/(0.00023 m ) 2 +06 2 B. 5.2×10 N, calculated from (1200 N)/(0.00023 m ) C. 0.78 N, calculated from (1200 N)*(0.00065 m ) 2 D. 4.2×10 +02 N, calculated from (1200 N)*(0.00023 m )/(0.00065 m ) 2 +06 2 E. 1.8×10 N*m, calculated from (1200 N)/(0.00065 m ) page 2 of 6 9. A distillery in Russia pumps vodka (density = 0.789×10 kg/m ) from one level to a higher level in their factory, a height difference of 3.2 m. How much higher is the pressure at the lower floor compare to the higher floor? +04 2 3 3 2 A. 2.5×10 N/m , calculated from (0.789×10 kg/m )*(9.8 m/s )*(3.2 m) B. 2.5×10 +03N/m , calculated from (0.789×10 kg/m )*(3.2 m) C. 2.5×10 +02N/m , calculated from (0.789×10 kg/m )/(3.2 m) +04 2 3 3 2 D. 1.2×10 N/m , calculated from (0.789×10 kg/m )*(9.8 m/s )*(3.2 m)/2 3 10. A spherical balloon is filled with helium (density = 0.176 kg/m ) The balloon itself has a mass of 0.15 kg unfilled. When filled, it has a volume of 0.81 m Air has a density of 1.23 kg/m What is the buoyant force of air on the balloon when filled? 3 3 2 A. 9.76 N, calculated from (1.23 kg/m )*(0.81 m )*(9.8 m/s ) B. 1.4 N, calculated from (0.176 kg/m )*(0.81 m )*(9.8 m/s ) 2 C. 12.1 N, calculated from (1.23 kg/m )*(9.8 m/s ) 2 2 3 2 3 1/3 D. 11.2 N/m , calculated from (1.23 kg/m )*(9.8 m/s )*(0.81 m ) 11. What is the force of gravity on the full balloon including the helium inside it, when the balloon is filled? A. 2.9 N, calculated from ((0.176 kg/m )*(0.81 m ) + (0.15 kg))*(9.8 m/s ) 2 3 3 2 B. 9.8 N, calculated from (1.23 kg/m )*(0.81 m )*(9.8 m/s ) 2 C. 1.5 N, calculated from (0.15 kg)*(9.8 m/s ) D. 11 N, calculated from ((1.23 kg/m )*(0.81 m ) + (0.15 kg))*(9.8 m/s ) 2 12. If the full balloon is tied to the ground by a string, what is the tension in the string? A. 9.8 N, calculated from (1.23 kg/m )*(0.81 m )*(9.8 m/s ) 2 B. 1.5 N, calculated from (0.15 kg)*(9.8 m/s )2 3 3 2 C. 1.4 N, calculated from (0.176 kg/m )*(0.81 m )*(9.8 m/s ) D. 6.9 N, calculated from (1.23 kg/m )*(0.81 m )*(9.8 m/s ) - ((0.176 kg/m )* 3 (0.81 m ) + (0.15 kg))*(9.8 m/s )2 page 3 of 6 13. A 0.57 kg piece of solid styrofoam floats on water. What is the volume of styrofoam in 3 3 3. the water? (Water density = 1.00×10 kg/m , styrofoam density = 23 kg/m ) A. 0.0056 m , calculated from (0.57 kg)*(9.8 m/s )/(1.00×10 kg/m ) 3 B. 0.00057 m , calculated from (0.57 kg)/(1.00×10 kg/m ) 3 3 3 C. 0.025 m , calculated from (0.57 kg)/(23 kg/m ) D. 9.8×10 +03 m , calculated from (9.8 m/s )*(1.00×10 kg/m ) 3 +12 3 2 3 3 3 E. 5.1×10 m , calculated from ((9.8 m/s )*(1.00×10 kg/m )/(0.57 kg)) 14. What fraction of the styrofoam's volume is in the water, as a percentage? A. 43 %, calculated from (1.00×10 kg/m )/(23 kg/m ) 3 +02 B. 1×10 %, calculated from 100 C. 4.3×10 +03 %, calculated from 100*(1.00×10 kg/m )/(23 kg/m ) 3 3 3 3 D. 2.3 %, calculated from 100*((0.57 kg)/(1.00×10 kg/m ))/((0.57 kg)/(23 kg/m )) 15. A concrete slab 6.9 m long expands 8.6 m when the temperature increase by 7.9 ℃. What is the coefficient of expansion of this concrete? A. 0.16 /℃, calculated from ((8.6 m)/(6.9 m))/(7.9 ℃) B. 0.87 m/℃, calculated from (6.9 m)/(7.9 ℃) C. 1.2, calculated from (8.6 m)/(6.9 m) D. 1.1 m/℃, calculated from (8.6 m)/(7.9 ℃) 3 16. A helium-filled balloon has a volume of 6.9 m at atmospheric pressure and temperature 18.1 ℃. When cooled to 6.5 ℃, what is its volume at the same pressure? 3 3 A. 7.2 m , calculated from (6.9 m )*((18.1 ℃)+273.15)/((6.5 ℃)+273.15) B. 2.5 m , calculated from (6.9 m )*(6.5 ℃)/(18.1 ℃) 3 3 C. 19 m , calculated from (6.9 m )*(18.1 ℃)/(6.5 ℃) D. 6.6 m , calculated from (6.9 m )*((6.5 ℃)+273.15)/((18.1 ℃)+273.15) 17. One cubic meter of air at 0℃ and standard atmospheric pressure (1.000 atm) has a mass of 1.28 kg. How much volume does the same mass of air occupy at the same temperature if the pressure increases to 1.033 atm? A. 0.968 m /atm, calculated from (1.000 m )/(1.033 atm) B. 0.968 m , calculated from (1.000 m )*(1.000 atm)/(1.033 atm) 3 3 C. 1 m , calculated from (1.000 m ) D. 1.03 m , calculated from (1.000 m )*(1.033 atm)/(1.000 atm) page 4 of 6 18. If the root-mean-square (RMS) speed of molecules in a gas increases by a factor 4, by what factor does the temperature increase? A. 16, calculated from (4) 2 B. 0.25, calculated from 1/(4) C. 8, calculated from 2*(4) D. 4, calculated from (4) 19. The sun keeps the earth warm. What is the primary way heat reaches the earth from the sun through the vacuum of space? A. conduction B. convection C. radiation 20. Koreans use steel chopsticks. When eating in a restaurant in Korea, an American tourist foolishly (and impolitely) leaves the ends of the steel chopsticks sticking in a hot bowl of rice while talking on his cell phone. When he tries to pick up the chopsticks, the chopsticks are hot everywhere. How did heat travel through the steel? A. radiation B. convection C. conduction 21. If the heat capacity of water is 4184 J/(kg·℃), how much energy does it take to raise the temperature of 7.3 kg of water by 7.8 ℃ using an electric heater? Assume perfect thermal insulation around the water. A. 4.2×10 +03 J, calculated from (4184 J/(kg·℃)) +05 B. 2.4×10 J, calculated from (4184 J/(kg·℃))*(7.3 kg)*(7.8 ℃) C. 73 J, calculated from (4184 J/(kg·℃))/((7.8 ℃)*(7.3 kg)) D. 3.1×10 +04 J/kg, calculated from (4184 J/(kg·℃))*(7.3 kg) E. 3.3×10 +04 J, calculated from (4184 J/(kg·℃))*(7.8 ℃) 22. How much time should it take to defrost a 7.6 kg frozen dinner in a 850 W microwave oven if all the power of the microwave goes into melting the dinner? Assume the frozen 3 dinner has the same latent heat of fusion as ice, 334×10 J/kg. A. 3.7×10 +07 s, calculated from (334×10 J/kg)*(850 W)/(7.6 kg) -08 3 B. 2.7×10 s, calculated from (7.6 kg)/((850 W)*(334×10 J/kg)) C. 3.9×10 +02 s, calculated from (334×10 J/kg)/(850 W) +03 3 D. 3×10 s, calculated from (334×10 J/kg)*(7.6 kg)/(850 W) E. 2.5×10 +06 s, calculated from (334×10 J/kg)*(7.6 kg) Don't stop here, you're not done yet.... page 5 of 6 Short written answer questions A circus stunt diver whose mass is 60.0 kg belly-flops into a tub containing 8000 kg of water from 5.00 m above the surface of the water. (i) How much does the temperature of the water increase due to the diver stopping quickly in the water? Assume all the water stays in the tub, as does all the energy transferred from the belly flop. Use 4.186 J/(g K) for the specific heat of water. (ii) If the diver stays in the tub of for 60 seconds and loses 100 J per second of heat to the water, how much further does the water temperature increase? Assume no heat transfer through the walls and floor of the tub and surface of the water. Don't turn in your exam until you've checked your WID bubbles, your version number bubble, your name.... The End page 6 of 6

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