P115_Sp23_E2_Ver_A_Solution

Physics 115 Spring 2023 Midterm Exam 2 1. [5 pts] Consider the two circuits at right. Is the brightness of bulb A greater than, less than , or equal to the brightness of bulb B? A) Greater than B) Less than C) Equal to D) Not enough information Bulb A is connected in series to a 5 ࠵? resistor, while bulb B is connected in series to a parallel network of two 10 ࠵? resistors. The equivalent resistance of the two 10 ࠵? resistors is 5 ࠵? as shown below. 1 ࠵? !"! = 1 10 Ω + 1 10 Ω = 2 10 Ω ࠵? !"! = 10 Ω 2 = 5 Ω The total resistance of each circuit is thus the same, and since both batteries are identical, the same current flows through each circuit. Bulb brightness is proportional to current, so both bulbs have equal brightness. 2. [5 pts] In the circuit at right, the current through the battery, I bat , is: A) I 1 /2 B) 2 I 1 /3 C) I 1 D) 3 I 1 /2 E) 2 I 1 For resistors connected in parallel, the voltage across each resistor is the same. The current I 1 can be defined as: ࠵? # = ࠵? $%! ࠵? The current through the 2R resistor is & !"# ’( , which is # ’ ࠵? # . Applying Kirchhoff’s junction law, the current through both resistors combines to flow through the battery. The current through the battery is therefore ) ’ ࠵? # (࠵? # + # ’ ࠵? # ) . A B

Physics 115 Spring 2023 Midterm Exam 2 3. [5 pts] The switch in the RC circuit shown at right is closed at t = 0. The capacitor is initially uncharged. What is the current through the resistor at the time t = 0.56 s? A) 0.13 A B) 0.29 A C) 0.48 A D) 0.62 A E) 0.97 A We can use the following equation to determine the current through the resistor: ࠵?(࠵?) = ࠵? * ࠵? +!/(- = ࠵? $%! ࠵? ࠵? +!/(- ࠵?(0.56 s) = 12 V 16 Ω ࠵? +*./0 2/(#0 4)6’*×#* $% 89 ࠵?(0.56 s) = 0.13 A

Physics 115 Spring 2023 Midterm Exam 2 5. [5 pts] A negative charge moves with a velocity in the positive y -direction through a region of uniform magnetic field. The negative charge experiences a magnetic force that points out of the page. What is the direction of the magnetic field? A) Positive x -direction (to the right) B) Negative x -direction (to the left) C) Negative y -direction (toward bottom of the page) D) Into the page E) Out of the page By the right-hand rule, a force out of the page is consistent with a velocity in the positive y-direction and a magnetic field that points in the negative x-direction. However, since the charge in this question is negative, the magnetic field must point in the positive x-direction. 6. [5 pts] Four charges of equal mass are initially moving to the right at equal speeds and enter a magnetic field. The magnetic field exerts a force on each of these charges resulting in the paths shown in the diagram. Which of the following statements are true? A) All charges are positive, and the magnetic field is into the page. B) All charges are positive, and the magnetic field is out of the page. C) All charges are negative, and the magnetic field is out of the page. D) Charges A and B are positive, C and D are negative, and the magnetic field is into the page. E) More than one of the above is correct. At point P, the velocity of the charges is to the right, and to turn clockwise, the force on the charges must be downward (toward the center of the circular path). By the right-hand rule, a downward force, with a rightward velocity, would be consistent with an out of the page magnetic field for positive charges, or an into the page magnetic field for negative charges. Since there is no choice that has an into the page magnetic field with negative charges, only choice B is correct. P ࠵? ⃗ ࠵? ⃗

Physics 115 Spring 2023 Midterm Exam 2 7. [5 pts] The top wire in the figure at right produces a magnetic field at point P with a magnitude B top . The magnitude of the net magnetic field at point P is 5 B top and it points into the page. What is the magnitude and the direction of current in the bottom wire? A) 50 A to the right B) 50 A to the left C) 80 A to the right D) 80 A to the left E) 120 A to the left Using the right-hand rule, the magnetic field at point P due to the current in the top wire points into the page, and we know it has a magnitude B top . The magnitude of the net magnetic field at point P is 5B top , which means that the bottom wire could produce a magnetic field of 4B top into the page at point P or a 6B top field that points out of the page. Let us first look at the case where the magnetic field needs to be 4B top into the page. For the bottom wire to produce a magnetic field into the page, the current must move to the left through the bottom wire, and the current must be 80 A as shown below. |࠵? $"!!": | = 4=࠵? !"; = ࠵? * ࠵? $"!!": 2࠵?(2 cm) = 4 ࠵? * (10 A) 2࠵?(1 cm) ࠵? $"!!": = 4(10 A) 2 cm 1 cm = 80 A This is consistent with choice D. For the case where the magnetic field needs to be 6B top out of the page, the bottom wire to needs to carry a current to the right, and the current must be 120 A as shown below. |࠵? $"!!": | = 6=࠵? !"; = ࠵? * ࠵? $"!!": 2࠵?(2 cm) = 6 ࠵? * (10 A) 2࠵?(1 cm) ࠵? $"!!": = 6(10 A) 2 cm 1 cm = 120 A Choice E has the correct current magnitude but not the correct current direction.

Physics 115 Spring 2023 Midterm Exam 2 9. [5 pts] A student has added carbon dioxide (CO 2 ) and helium gas to the same container (He). The molecular mass of carbon dioxide is greater than that of helium. The student has added the gases such that the number of carbon dioxide molecules is five times larger than the number of helium molecules ( ࠵? -< & = 5࠵? => ). Consider both gases as ideal. Which of the following statements are correct? Statement I: the carbon dioxide molecules have a slower average speed. Statement II: the helium molecules have a larger average kinetic energy. Statement III: the contribution of all the carbon dioxide molecules to the total internal energy is greater than the contribution from all of the helium molecules. A) Statement I only B) Statement II only C) Statement III only D) Statements II and III E) Statements I and III Exam announcement: Treat carbon dioxide as a monoatomic gas, and that the gases are in the container for a long time. The carbon dioxide and helium molecules are in the same container for a long time and are thus at the same temperature. This means that both molecules have the same average kinetic energy ( statement 2 is incorrect ). Since the mass of the carbon dioxide molecules are larger than that of helium, the average speed of carbon dioxide is less than that of the helium molecules ( statement 1 is correct ). The internal energy is equal to the sum of the kinetic energies of both carbon dioxide and helium molecules. Since they have equal average kinetic energy, but there are more carbon dioxide molecules, the carbon dioxide does contribute more to the total internal energy ( statement III is correct ).

Physics 115 Spring 2023 Midterm Exam 2 10. [5 pts] The rms speed of neon at 20 ° C is v . Is the rms speed of neon at 80 ° C greater than , less than , or equal to 2 v ? A) Greater than 2 v B) Less than 2 v C) Equal to 2 v D) Not possible to determine the answer. The rms speed is given as: ࠵? ?:@ = S 3࠵? A ࠵? ࠵? An increase in temperature from 20 ° C to 80 ° C, is an increase by 1.2 in terms of absolute temperature. The speed thus increases by √1.2 or 1.1, which is less than 2v. 11. [5 pts] A rigid container contains n moles of an ideal gas at room temperature, 20°C , and at pressure P . Suppose you want to use the same container to create a condition where the pressure was 2 P . Which of the following changes could you make? A) Keep the same temperature and number of moles but use an ideal gas with twice the molecular mass as the original gas. B) Keep the same number of moles but use a temperature of 40°C C) Keep the same temperature and increase to the number of moles to 4 n . D) Keep the same temperature and increase to the number of moles to 2 n. E) More than one of the above. The ideal gas law in terms of pressure is given as: ࠵? = ࠵?࠵?࠵? ࠵? Of the choices given, only increasing the number of moles by a factor of two would double the pressure (and keeping the same temperature). For choice A, there is no dependence on pressure with the mass of the molecules. For choice B, an increase in temperature to 40°C is an increase by a factor less than two. Choice C would cause the pressure to increase by a factor of four.

Name: _______________________________________ UW NetID: __________________ Last Name First Name Physics 115 Spring 2023 Midterm Exam 2 As part of an endurance race, a 68.0-kg-athlete cycles a distance of 80.0 km and then climb 170 flights of stairs. 13. [5 pts] The athlete cycles at 15 km/h which corresponds to a total metabolic power of 480 W. How many Calories will the athlete burn during the cycle portion of the endurance race? Show your work. We can solve the Calories burned as follows: ࠵? = ∆࠵? ∆࠵? ∆࠵? = ࠵?∆࠵? ∆࠵? = ࠵? ∆࠵? ࠵? = (480 W) 80.0 km 15 km/hr (3600 s/hr) = 9.2 × 10 0 J ∆࠵? = (9.2 × 10 0 J) i 1 Cal 4190 J k = 2200 Cal 14. [5 pts] After the cycle, the athlete wants to consume a number of energy gels that will provide an energy input equivalent to 1.25 times the energy that they will burn in climbing the 170 flights of stairs (each flight is 2.90 m high). If each energy gel is 130 Calories, how many energy gels does the athlete need to consume? Assume the athlete’s efficiency is 25% and show your work. The energy needed to climb the stairs can be determined as follows: ࠵? = ࠵?ℎ࠵?࠵? ࠵?࠵?࠵? ࠵?࠵?࠵? ࠵?ℎ࠵?࠵? ࠵?࠵?࠵? ࠵?࠵?࠵? = ∆࠵? B ࠵? $C?D>E = ࠵?࠵?ℎ ࠵? $C?D>E ࠵? $C?D>E = ࠵?࠵?ℎ ࠵? = (68 kg)(9.8 m/s ’ )(170)(2.90 m) 0.25 = 1.31 × 10 0 J = 314 Cal The athlete wants to consume 1.25 times the energy needed to climb the stairs, so 392 Cal. The athlete therefore needs 3 gels (392 Cal/130 Cal/gel ~ 3 gels).

Name: _______________________________________ UW NetID: __________________ Last Name First Name Physics 115 Spring 2023 Midterm Exam 2 15. [5 pts] Your instructor carries out a 5-step random walk (fixed to one dimension). Your instructor tosses a coin to determine which direction to move, a head corresponds to one step to the right, a tails refers to one step to the left. If a macrostate is defined as the final location from the start point, how many macrostates are possible for this experiment? Show your work. In a 5-step random walk, the instructor could end up in six different locations (1) 5 steps to the right, (2) 3 steps to the left, (3) 1 step to the right, (4) 1 step to the left, (5) three steps to the left, and (6) five steps to the left. The corresponding configuration of the coins and the locations is shown at right. 16. [5 pts] If the instructor were to carry out 1500 trials of this experiment, how many times would they end up three steps to the right? Show your work. There are 32 total microstates (2 5 ). There are five microstates that corresponds to the macrostate of three steps to the right. The probability of the macrostate of three steps to the right is 5/32 or ~0.16. With a total of 1500 trials, the instructor will end up three steps to the right about 5/32 of the time or ~234 times.

Name: _______________________________________ UW NetID: __________________ Last Name First Name Physics 115 Spring 2023 Midterm Exam 2 The figure at right shows two identical cylinders, A and B that are fitted with identical pistons. Each piston is free to move without friction. Both cylinders have been in the same room for a long time. Cylinder A contains an ideal gas with a smaller molecular mass than the ideal gas in cylinder B. 19. [5 pts] Is the pressure of the gas in cylinder A greater than, less than, or equal to the pressure of the gas in cylinder B? Explain. If not enough information is given, state so explicitly. Since the cylinders are identical and the pistons are identical, we can conclude that the pressure of the gases in each cylinder are equal to each other since we can defined the pressure of the gas as follows: ࠵? B%@ = ࠵? %!: + ࠵?࠵?/࠵? The mass and cross-sectional area of each piston is the same, so the pressure of each gas is the same. Cylinder C has a diameter that is twice as large as that of cylinder A and the mass of piston in cylinder C is four times larger than that of cylinder A. The cylinders have been in the same room for a long time and the height of piston A is twice as high as that in cylinder C. 20. [6 pts] Is the number of particles in container A greater than , less than , or equal to the number of particles in container C? Explain. By using the logic from the previous question, we can conclude that the pressure of the gas in cylinder A is the same as that in cylinder C. (Since the diameter of cylinder C is twice as large as that of cylinder A, the cross-sectional area of the piston in cylinder C is four times larger than the cross-sectional area of the piston in cylinder A.) ࠵? B%@,G = ࠵? %!: + ࠵?࠵?/࠵? ࠵? B%@,- = ࠵? %!: + 4࠵?࠵? 4࠵? = ࠵? B%@,G We can write the number of particles in cylinder A as: ࠵? = ࠵?࠵? ࠵? A ࠵? We know the pressures and the temperatures are the same, so the number of particles is proportional to the volume. Cylinder A has half the volume of cylinder C, so the number of particles is less than that in cylinder C. ࠵? G = ࠵? G ℎ G ࠵? - = 4࠵? G ℎ G 2 = 2࠵? G