Physics 202 Exam 1. May 1, 2013

Transcription

1 Name: Physics 202 Exam 1 May 1, 2013 Word Problems Show all your work and circle your final answer. (Ten points each.) 1. If 2.4 m 3 of a gas initially at STP is compressed to 1.6 m 3 and its temperature raised to 30 C, what is the final pressure? Answer: 1.7 atm This is an application of the ideal gas law, P V = nrt. It s pretty typical to be asked this before-and-after type question. When that happens I think about dividing this equation by itself: P 1 V 1 P 2 V 2 = n 1R 1 T 1 n 2 R 2 T 2 This one equation contains all the other gas laws. The contents of the gas remain the same, we know that n 1 = n 2. So the n cancels. The R cancels because its a constant. We are left with: P 1 V 1 = T 1 P 2 V 2 T 2 STP stands for standard temperature and pressure, which correspond to T 1 = K and P 1 = kpa. We can plug in all our data then solve for P 2. (101325)(2.4) (P 2 )(1.6) = (273.15) (303.15) P 2 = Or, 170 kpa. We want to quote this in atmospheres. The conversion is Pa 1 atm Pa =

2 2 2. What is the average speed of the molecules in low-density oxygen gas at 0 C? (The mass of an oxygen molecule, O 2, is kg.) Answer: 461 m/s The average kinetic energy of a gas is given by the formula KE avg = 3 2kT. From the kinetic energy we can derive the average speed. Thus, KE avg = ( 3 2 )( )(273.15) = The kinetic energy is related to the speed according to KE = 1 2 mv2, so ( ) = 1 2 ( )(v avg ) 2 v avg =

3 3 3. A popgun uses an ideal spring for which k = 2000 N/m. When cocked, the spring is compressed 3.0 centimeters. How high can the gun shoot a 5.0-gram projectile? Answer: 18 meters This looks like an energy problem. At the top of the trajectory, all the energy is potential: P E = mgh. Initially, the energy is also all potential, but this time because of the spring: P E = 1 2 kx2. The initial energy is P E = 1 2 (2000)(0.030)2 = Using this for the final energy, we have: = mgh = (0.0050)(9.8)(h) = h =

4 4 4. A platform is suspended by four wires at its corners. The wires are 3.0 meters long and have a diameter of 2.0 millimeters. Young s modulus for the material of the wires is N/m 2. How far will the platform drop (due to elongation of the wires) if a 50-kilogram load is placed at the center of the platform? Answer: 0.65 millimeters We are going to use the normal stress formula ( ) F L A = Y In order to use it, we need to calculate the force being applied (this is the weight of the load) and the cross-sectional area providing support. The weight is L 0 W = mg = (50)(9.8) = 490 The area of support from one wire is πr 2. Since the radius of each wire is 1.0 millimeter (half of the diameter), we have A = (4)(π)(0.001) 2 = I have multiplied by four because there are four wires providing support. Now we can use the stress formula: ( ) 490 L = ( ) = L =

5 5 5. A molten plastic flows out of a tube that is 8.0 centimeters long at a rate of 13 cm 3 /min when the pressure differential between the two ends of the tube is 0.24 atm. Find the viscosity of the plastic. The inner diameter of the tube 1.30 millimeters. Answer: kg/m-s This is a viscosity problem and we will use Poiseuille s equation: P = 8ηLQ πr 4 But our numbers are not in SI units. Let s convert: and and and Therefore, Q = L = 0.08 m ( ) 3 13 cm3 1 m min 1 min 100 cm 60 sec = m 3 /s P = 0.24 mmhg Pa 1 atm = Pa r = 1 2 (1.30 mm) 1 m = mm = (8)(η)(0.08)( ) (π)( ) 4 = η =

6 6 6. Determine the temperature that results when 1.0 kilograms of ice at exactly 0 C is mixed with 9.0 kilograms of water at 50 C and no heat is lost. Answer: 37 C We expect the hot water to reduce to a final temperature call it T. The heat lost by the hot water is Q hot = cm T = (4186)(9.0)(T 50) Notice that this is a negative number because the heat is being lost. This heat is gained by the ice which does two things: melts the ice then brings the ice-water up to the temperature T. In symbols: Q cold = ml + cm T = (1.0)(334000) + (4186)(1.0)(T 0) The heat gained by the cold is lost by the hot, i.e., Q cold = Q hot. Thus (1.0)(334000) + (4186)(1.0)(T 0) = (4186)(9.0)(T 50) = T = 37674T = T = /41860 =

7 Name: Physics 202 Exam 2 Jun 10, 2013 Word Problems Show all your work and circle your final answer. (Ten points each.) 1. A water tower is a familiar sight in many towns. The purpose of such a tower is to provide storage capacity and to provide sufficient pressure in the pipes that deliver the water to customers. Figure 1 shows a spherical reservoir that contains kilograms of water when full. The reservoir is vented to the atmosphere at the top. For a full reservoir, find the gauge pressure that the water has at the faucet in (a) house A and (b) house B. Ignore the diameter of the delivery pipes. Figure 1: Problem Answer: (a) 245 kpa (b) 174 kpa (a) We need to know the height of the spherical reservoir to use the basic hydrostatic equation P = ρgh. Fortunately we are told that the reservoir is a sphere.footnote{remember that the pressure differential only depends on the height and not the shape of the container. Whether a pipe or a sphere, only the height matters.} If we knew its volume, we could determine its radius. But the volume is connected to the weight of water (which we know) through the density of water. By definition, density is ρ = M/V, so (1000) = ( ) V = V = 525

8 2 The radius is can be derived from V = 4 3 πr3 : So, the height of the reservoir is (525) = 4 3 π(r)3 = r = r = So, the total height from the top (exposed to air) and the bottom is 2r = This height introduces a pressure differential of P = ρgh = (1000)(9.80)(25.009) = This is the gauge pressure because it is the differential relative to atmospheric pressure (because the top of the reservior is exposed to air). (b) The extra height of 7.30 meters reduces the water pressure based on the same formula, P = ρgh. P = (1000)(9.80)(7.30) = So the gauge pressure at the elevated house is P = =

9 3 2. Figure 2 shows a hydraulic system used with disc brakes. The force F is applied perpendicularly to the brake pedal. The pedal rotates about the axis shown in the drawing and causes a force to be applied perpendicularly to the input piston (radius = 9.50 mm) in the master cylinder. The resulting pressure is transmitted by the brake fluid to the output plungers (radii = 19.0 mm), which are covered with the brake linings. The linings are pressed against both sides of a disc attached to the rotating wheel. Suppose that the magnitude of F is 9.00 newtons. Assume that the input piston and the output plungers are at the same vertical level, and find the force applied to each side of the rotating disc. Figure 2: Problem Answer: 108 newtons This is an application of Pascal s principle: the pressure in a continuous fluid is the same at a particular height. This means that the pressure applied to the input piston is transmitted to the output plungers. So we need to determine this input pressure which is coming from the brake pedal. Since the pedal is pinned at the top, it acts like a lever. The initial force is converted to a torque about the axis in the diagram. Since the line of action is perpendicular to the pedal, the torque applied is given by τ = F x, where x is the distance to the axis, which in this case is meters. τ = F x = τ = (9.00)(0.150) = This torque is transmitted to the piston from a distance of meters, so the force actually applied to the input piston is τ = F x = (1.3500) = (F )(0.0500) = F = Now, the pressure input is this force over the cross-sectional area of the piston. We have: A = πr 2 = (π)( ) 2 =

10 4 So the pressure at this point is P = F A = = Since this input pressure equals the output pressure, we can calculate the output force once we figure out its cross-sectional area: Finally, using P = F/A, we have: A = πr 2 = (π)(0.0190) 2 = (95228) = (F ) ( ) = F = We get a three-fold magnification of force from the lever and a four-fold magnification from the hydraulics.

11 5 3. In an aluminum pot, 0.15 kilograms of water at 100 C boils away in four minutes. The bottom of the pot is 3.1 mm thick and has a surface area of m 2. To prevent the water from boiling too rapidly, a stainless steel plate has been placed between the pot and the heating element. The plate is 1.4 mm thick, and its area matches that of the pot. Assuming that heat is conducted into the water only through the bottom of the pot, find the temperature at (a) the aluminum-steel interface and (b) the steel surface in contact with the heating element. Answer: (a) 101 C (b) 111 C (a) The rate at which the heat is flowing into the water is P = Q t We know the time frame is four minutes (or 240 seconds). The amount of heat that flows is given by Thus, Q = ml v = (0.15)( ) = P = ( ) (240) = This power flow is driven by a temperature differential between the water (100 C) and the other side of the aluminum. This conduction is governed by the equation P = Q t = ka L T For the aluminum plate, the total conductance is ka L = (240)(0.015) ( ) = Plugging this into the conduction equation gives (1412.5) = (1161.3)( T ) = T = Since the water is at 100 C, the other side of the aluminum is T = = (b) The power that flows into the water through the aluminum plate also flows through the steel plate. The total conductance of the steel plate is ka L = (14)(0.015) ( ) = So, the conductance equation for the steel plate is (1412.5) = (150.00)( T ) = T =

12 6 This is the temperature differential from the aluminum plate: C. So, the temperature on the other side of the steel plate is T = =

13 7 4. When one person shouts at a football game, the sound intensity level at the center of the field is 60.0 db. When all the people shout together, the intensity level increases to 109 db. Assuming that each person generates the same sound intensity at the center of the field, how many people are at the game? Answer: 79,400 people The intensity level (β) is related to the intensity (I) according to β = 10 log(i/10 12 ) Therefore, the sound intensity from one person is (60) = (10) log(i/10 12 ) = I = And the sound intensity from all the people is (109) = (10) log(i/10 12 ) = I = Since intensity is additive the number of people must be: N = =

14 8 5. The transmitting antenna for a radio station is 7.00 km from your house. The frequency of the electromagnetic wave broad cast by this station is 536 khz. The station builds a second transmitting antenna that broadcasts an identical electromagnetic wave in phase with the original one. The new antenna is 8.12 km from your house. Does constructive or destructive interference occur at the receiving antenna of your radio? Show your calculations. Answer: Constructive In these interference calculations, the important quantity is the wavelength of the wave. Since the speed of light is c is m/s, we can figure out the wavelength based on the frequency using v = fλ: = ( )(λ) = λ = The question of interference centers on the question of how many wavelengths fit in the path-length difference from the two sources. In this case, that difference is l = = 1120 The number of wavelengths is N = = Since this is nearly a whole integer, there will be constructive interference.

15 9 6. A spotlight sends red light (wavelength of nm) to the moon. At the surface of the moon, which is meters away, the light strikes a reflector left there by astronauts. The reflected light returns to the earth, where it is detected. When it leaves the spotlight, the circular beam of light has a diameter of about 0.20 meters, and diffraction causes the beam to spread as the light travels to the moon. In effect, the first circular dark fringe in the diffraction pattern defines the size of the central bright spot on the moon. Determine the diameter (not the radius) of the central bright spot on the moon. Answer: 3.2 kilometers The diffraction angle for the red light is given by sin θ = (1.22)(λ/D) since the aperature is circular. Thus, sin θ = (1.22) 0.20 = θ = degrees The edge of the beam effectively makes a right triangle with the opposite side being the radius of the spot on the moon. The adjacent side is the distance to the moon. So, tan( r degrees) = = r = Since we are asked for the diameter we need to double this length: D = 2r = (2)(1596.7) =