Winter 2017 PHYSICS 115 MIDTERM EXAM 1 Section X PRACTICE EXAM SOLUTION Seat No Name (Print): Name (Print): Honor Pledge: All work presented here is my own. Signature: Student ID: READ THIS ENTIRE PAGE NOW Do not open the exam until told to do so. You will have 50 minutes to complete the examination. NO CELL PHONES, TEXT MSG, etc. ALLOWED AT ANY TIME. Before the exam begins: Print and sign your name, and write your student ID number in the spaces above. During the exam When the exam begins, print your name and student ID number on the top of each page. Do this first when you are told to open your exam. If you are confused about a question, raise your hand and ask for an explanation. If you cannot do one part of a problem, move on to the next part. This is a closed book examination. All equations and constants are provided. You may use a calculator, but not a computer, or other internet connected devices (smart-phones, ipads, etc.). For multiple-choice questions: Clearly circle your answer choice. Make no stray marks. If you must erase, erase completely. For free-response questions: Please write neatly and legibly. Do not use scratch paper; it will be ignored. Show your work in enough detail so that the grader can follow your reasoning and your method of solution. Circle your answers, and state units if appropriate. End of exam: Out of respect to other students, please remain seated for the last 20 minutes of the exam. At the end of the exam, please remain seated until all exams have been collected.
1. [5 pts] A 1.0 m tall cylindrical container that is open to the atmosphere contains mercury to a certain depth, d. The rest of the cylinder is filled with water. If the pressure at the bottom of the cylinder is two atmospheres, what is the depth d? (r water = 1000 kg/m 3, r mercury = 13,600 kg/m 3, P atm = 101.3 kpa) A) 0.74 m B) 0.83 m C) 0.66 m D) 0.79 m E) 0.88 m p = p 0 + ρgh p p 0 = ρ Hg gd + ρ H2 g(1 d) =101300Pa 0 ρ Hg gd + ρ g ρ H2 gd =101300Pa 0 H 2 0 d(ρ Hg g ρ H2 0 g) =101300Pa ρ H 2 0 g d = 101300Pa ρ H 2 0 g g(ρ Hg ρ H2 0 ) d = 0.74m 2. [5 pts] A solid block is suspended from a spring scale. When the block is in air, the scale reads 35.0 N, when immersed in water the scale reads 31.1 N, and when immersed in oil the scale reads 31.8 N. What is the density of the oil? (r water = 1000 kg/m 3 ) A) 800 kg/m 3 B) 870 kg/m 3 C) 890 kg/m 3 D) 821 kg/m 3 E) 910 kg/m 3 W EB = ρ B gv B = 35.0N T SB =W EB ρ W gv B = 31.1N V B = W EB T SWB ρ W g T SOB =W EB ρ o gv B ρ o = W T! EB SOB = W T $! ρ EB SOB # & W g $! W # & = ρ EB T $ W # SOB & gv B " g %" W EB T SWB % " W EB T SWB % ρ o = 820 kg/m 3
3. Water enters a house at 1.5 m/s through a pipe with an inside radius of 1.0 cm and at a pressure of 400,000 Pa. The water then travels up a vertical distance of 5.0 m to a second floor bathroom through a pipe of radius 0.5 cm. Calculate the pressure of the water when it reaches the bathroom. A) 3.87 x 10 5 Pa B) 3.34 x 10 5 Pa C) 4.66 x 10 5 Pa D) 2.85 x 10 5 Pa E) 1.49x 10 5 Pa p 1 + ρgy 1 + 1 2 ρv 2 = p 1 2 + ρgy 2 + 1 2 ρv 2 2 p 2 = p 1 + ρg(y 1 y 2 ) + 1 2 ρ v 2 2 ( v 1 2 ) v 1 A 1 = v 2 A 2 v 2 = v A 1 1 A 2 p 2 = p 1 + ρg(y 1 y 2 ) + 1 " 2 ρ v 2 v A 2 " % % $ 1 1 ' $ 1 $ # A ' # 2 & ' & p 2 = 334000Pa 4. [5 pts] During a marathon race, a runner s blood flow increases to 10.0 times her resting rate. Her blood s viscosity has dropped to 95.0% of its normal value due to an increase in core temperature, and the blood pressure difference across the circulatory system has increased by 50.0%. By what factor has the average radii of her blood vessels increased? A) r new = 1.2r B) r new = 1.6r C) r new = 2.4r D) r new = 3.1r E) r new = 4.4r Q' =10Q η ' = 0.95η Δp' =1.5Δp r' 4 = Q'8lη ' Δp'π r' 4 r 4 r 4 = Q8lη Δpπ ( ) 0.95η ( )Δp = Q'8lη ' Δp'π. Δpπ Q'η 'Δp = Q8lη Δp'Qη = 10Q 1.5ΔpQη r' 4 r 4 = 6.3 r' =1.6r
5. [5 pts] In a continuous-caster process, steel sheets 25.4 cm thick, 2.03 m wide, and 10.0 m long are produced at a temperature of 872 C. When the sheets are cooled, their volume changes by 0.158 m 3. What is the final temperature of the sheets? The coefficient of volume expansion for steel is 36 x 10-6 K -1. A) 18 C B) 21 C C) 24 C D) 25 C E) 28 C V = βv % T ' T ) V βv % + T ) = T ' T ' = 0.158m 1 36 10 5 K 78 + 1145K = 293.8K 5.156m1 T ' = 20.8 6. [5 pts] Two neighboring houses are built of the same materials, but one is twice as long, high, and wide as the other and also has walls twice as thick. To keep the houses equally warm on a cold day, at what rate must the heater in the big house supply heat, compared with the heater in the small house? Assume conductive heat losses only. A) 2 B) 4 C) 8 D) 16 E) same rate Area dimension 2 Area large = 4Area small Δx large = 2Δx small " Q % $ ' # Δt & " Q % $ ' # Δt & small large = ka ΔT Δx = k4a ΔT 2Δx " ΔT % = 2$ ka # Δx & ' = 2 " Q % $ # Δt ' & small
7. [5 pts] The glowing filament in a lamp is radiating energy at a rate of 60 W. At the filament s temperature of 1500 C, the emissivity is 0.23. What is the surface area of the filament? A) 2.5 cm 2 B) 3.8 cm 2 C) 4.7 cm 2 D) 7.4 cm 2 E) 9.1 cm 2 P = eσat A A = P eσt A = 60W 0.23 5.67 10 7D W. m 7F K 7A 1773K A = 4.7cmF 8. [5 pts] The rms speed of an diatomic oxygen at 0 C is 460 m/s. The molar mass of oxygen is 32 g/mol and of helium is 4 g/mol. Determine the rms speed of a helium molecule at 0 C. A) 230 m/s B) 326 m/s C) 650 m/s D) 920 m/s E) 1300 m/s v rms = v rms =1300m / s 3RT M = 3(8.31J/mol-K)(273K) 0.004kg/mol v rms,helium = M oxygen M helium v rms,oxygen =1300m / s 9. [5 pts] One mole of an ideal gas is at a pressure of 5.4 atm and a temperature of 22 C. What is the average kinetic energy of a molecule of this gas? A) 4.6 x 10-22 J B) 5.4.1 x 10-21 J C) 6.1 x 10-21 J D) 7.4 x 10-21 J E) 7.9 x 10-21 J K IJK = 3 2 kt = 3 2 (1.38 107F1 J/K)(295K) = 6.1 10 7F8 J
10. [5 pts] An ideal gas undergoes the process at right. Is the magnitude of the heat exchanged with the gas greater than, less than, or equal to the magnitude of the work done on the gas? A) Greater than B) Less than C) Equal to D) Not enough information. The temperature of the gas is proportional to PV. Since the gas changes from a value of 3PV to 2PV, the temperature of the gas decreases. We can therefore conclude that the thermal energy of the gas decreases. The work done on the gas is positive, since the gas is compressed. (The area under the curve is negative, and the work is thus positive.) The sum of the work done on the gas and the heat exchanged with the gas is equal to the change in thermal energy. Since the change in thermal energy is negative, and the work done is positive, the heat exchanged with the gas must be negative and greater in magnitude than the work done on the gas.
Marianne really likes coffee, but on summer days she doesn t want to drink a hot beverage. She would prefer the temperature of her coffee to be 30 C. 11. [5 pts] If she is served 200 ml of coffee at 80 C in a well-insulated container, how much energy would need to be removed from the coffee to lower its temperature to 30 C? (c coffee = 4190 J/(kg-K)) Show your work. Q = m R c R T ' T ) Q = 0.200kg 4190 J kg. K 50K = 41900J 41900 J are removed from the coffee. 12. [5 pts] How much ice at 0 C would Marianne need to add to her coffee to accomplish this change in temperature? Show your work. Q R + Q )RV = 0 m )RV L ' + m )RV c XIYVZ 30K = 41900J m )RV = 41900J L ' + c XIYVZ 30K = 0.091kg A 5.0 m-diameter garden pond holds 5.9 x 10 3 kg of water. Solar energy is incident on the pond at an average rate of 400 W/m 2. 13. [5 pts] How much energy must the water absorb to change temperature from 15 C to 25 C? Show your work. Q = m X c X T = 5.9 10 1 kg 4186 J kg. K 10K = 2.5 10D J 14. [5 pts] If we assume that the energy absorbed is from solar energy only and the water does not exchange energy with the surroundings, how long will it take to change its temperature from 15 C to 25 C? Show your work. P A = 400W/mF P = 400W/m F (π) 2.5m F = 7854W P = Q t t = Q P = 2.5 10D J 7854J/s = 31445s = 8.7hr
Consider the two U-tubes at right. A small volume of oil with density r 1 is poured into the left column of the left U-tube, and a small volume of oil with density r 2 is poured into the right column of the right U-tube. The densities of both oils are less than that of water. The heights of the oil columns are also identical. Points A and D are at the same height above the table, as are points B and C. ρ 1 B F C ρ 2 A G E D 15. [5 pts] Is the pressure at point A greater than, less than or equal to the pressure at point D? Explain. Consider a point E. Points A and E are at the same depth below the surface. However, point A sits below a volume of oil, which is less dense than water. Thus the pressure at point A is less than that at point E. The pressure at point E is the same as that at point D, since both points are at the same depth and are in the same fluid. I can therefore conclude that the pressure at point A is less than that at point D. 16. [5 pts] Is the pressure at point A greater than, less than or equal to the pressure at point C? Explain. The pressure at point C is equal to that at point F or P IY` + ρ X g(h d7efz'irv ). The pressure at poin G is the same as that at point A or P IY` + ρ X g(h g7h ). Since (h g7h )> (h d7efz'irv ), the pressure at point A is greater than that at point C. 17. [5 pts] Two objects of the same mass and volume but different shape are suspended from strings in a tank of water as shown. A student is asked to compare the tension in both strings, and claims: The bottom of object A is deeper in the water where the pressure is higher. Therefore the buoyant force on object A must be greater and the tension in the string is less. Do you agree or disagree with this student? Explain. Object B By Newton s second law, the magnitude of the tension in the Object A string is equal to the magnitude of the weight of the bock minus the magnitude of the buoyant force exerted on the block. Since both blocks have the same mass, the weight of each block is the same. The buoyant force exerted on each block is dependent on the density of the fluid and the volume of fluid displaced. Since both blocks are in the same fluid, and displace an equal volume of fluid, the buoyant force on object A is equal to that on object B. As above, the tension in each string must be equal, and I thus disagree with the student statement.
18. [5 pts] Rank in order, from largest to smallest, the densities of objects A, B, and C in the figure at right. Explain. From Archimedes principle we can conclude: F h = W ρ ' gv lfm = ρ nmo gv nmo ρ nmo = V lfm V nmo ρ ' So the density of the object is related to the ratio of the submerged volume to the volume of the object. This ratio is largest for block A, then block C and then block B. A > C > B. 19. [5 pts] Two cylinders, A and B, of the same cross-sectional area contain the same number of moles of gas, and both cylinders are at room temperature. The pistons in each cylinder are at rest. Is the mass of the piston in cylinder A greater than, less than, or equal to the mass of the piston in cylinder B? Explain. Each cylinder contains the same number of moles at the same temperature. From the ideal gas law, the product of pressure and volume for each gas must be the same. The volume in cylinder B is larger than that in cylinder A, so the pressure of gas B is less than that of gas A. The pressure scales positively with the mass of the piston, so the mass of the piston in cylinder A must be greater than that in cylinder B. 20. [5 pts] Sketch a PV diagram for a gas that goes through a cycle consisting of (a) an isobaric expansion, (b) a constant-volume reduction and (c) an isothermal process that returns the gas to its initial state. Label your graph appropriately and briefly explain your reasoning. An isobaric expansion is represented by a horizontal line on a PV diagram with the process proceeding to the right. An constantvolume reduction is represented by a vertical line on a PV diagram proceeding downward. Isotherms are represented by hyperbola on a PV diagram. A B