Physics 202 Quiz 1. Apr 8, 2013

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1 Name: Physics 202 Quiz 1 Apr 8, 2013 Word Problems Show all your work and circle your final answer. (Ten points each.) 1. One end of a piano wire is wrapped around a cylindrical tuning peg and the other end is fixed in place. The tuning peg is turned so as to stretch the wire. The piano wire is made from steel (Y = N/m 2. It has a radius of 0.80 mm, and an unstrained length of 0.76 meters. The radius of the tuning peg is 1.8 mm. Initially, there is no tension in the wire. Find the tension in the wire when the tuning peg is turned through two revolutions. Answer: 12,000 newtons Since the wire is being stretched, we are dealing with longitudinal stress. Another clue is the fact that we are given Young s modulus in the statement of the problem. The main equation to use is F A = Y L L 0 We are asked for the tension in the wire which is F. The cross-sectional area can be obtained from the radius of the wire using A = πr 2. Thus, A = π( ) 2 = We are given Y explicitly and the initial length L 0 is 0.76 meters. The change in length is due to the turning of the peg. Since the wire is made to twist twice around peg, the change in length is twice the circumference of the peg (C = 2πr). Thus, L = 2C = (2)(2π)( ) = We now have all the numbers we need. Plugging them it, we get: (F ) ( ) = ( ) 0.76 = F = 11968

2 2 2. An 86.0-kilogram climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of 1200 N/m. He accidentally slips and falls freely for meters before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest? Answer: 1.95 meters As the climber is falling his gravitational potential energy is being converted into the kinetic energy of his fall. The total amount of gravitational potential energy converted is P E grav = mgh = (86.0)(9.80)(0.750) = This is the climber s kinetic energy once the rope begins to stretch. At this point his kinetic energy is then converted into the elastic potential energy (P E elas = 1 2 kx2 ) in the rope. However, he is still falling, so there is more gravitational potential energy to take into account (P E grav = mgh). In other words, at the moment the rope begins to stretch, the types of energy are: KE = P E grav = mgh = (86.0)(9.80)(x) = (842.80)(x) P E elas = 1 2 kx2 = 1 2 (1200)(0)2 = 0 At the bottom of the stretch, the climber is momentarily at rest (no kinetic energy), and he has exhausted his gravitational potential energy, but now the spring is stretched. Now the types of energy are: KE = 0 P E grav = mgh = (86.0)(9.80)(0) = 0 P E elas = 1 2 kx2 = 1 2 (1200)(x)2 = (600)(x) 2 Conservation of energy says that these two totals are the same. Thus, (632.10) + (842.80)(x) = (600)(x) 2 This is a quadratic equation in x. Rewriting this in standard form yields: 600x x = 0 And applying the quadratic formula yields: x = b ± b 2 4ac 2a = (842.80) ± (842.80) 2 (4)( 600)(632.10) (2)( 600) ± = 1200 = or

3 The negative solution corresponds to an upward stretch we want the positive solution. 3

4 Name: Physics 202 Quiz 2 Apr 15, 2013 Word Problems Show all your work and circle your final answer. (Ten points each.) 1. A submersible pump is put under the water at the bottom of a well and is used to push water up through a pipe. What minimum output gauge pressure must the pump generate to make the water reach the nozzle at ground level, 71 meters above the pump? Answer: 700 kilopascals The pressure differential in any water column is given by P = ρgh. In this case we have P = (1000)(9.80)(71) = Now, to get the absolute pressure at the bottom we need to take into account the atmospheric pressure at the top. But we are asked for the gauge pressure instead, so we need to back this correction back out. So the final calculation still yields 695,800 Pa.

5 2 2. A blood transfusion is being set up in an emergency room for an accident victim. Blood has a density of 1060 kg/m 3 and a viscosity of η is Pa-s. The needle being used has a length of 3.0 cm and an inner radius of 0.25 mm. The doctor wishes to use a volume flow rate through the needle of m 3 /s. What is the distance h above the victim s arm where the level of the blood in the transfusion bottle should be located? As an approximation, assume that the level of the blood in the transfusion bottle and the point where the needle enters the vein in the arm have the same pressure of one atmosphere. (In reality, the pressure in the vein is slightly above atmospheric pressure.) Answer: 0.34 meters This is an application of Poiseuille s law: Plugging in what we have Q = πr4 P 8ηL ( ) = (π)( ) 4 ( P ) (8)( )(0.030) = P = This is the pressure differential that will create the required volume flow rate out of the needle. It remains to calculate the height that will create this pressure differential. Since the volume flow rate is so small, this can be done using the hydrostatic equation, P = ρgh. Thus, = (1060)(9.80)(h) = h =

6 Name: Physics 202 Quiz 3 Apr 22, 2013 Word Problems Show all your work and circle your final answer. (Ten points each.) 1. A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.16 m 2 and whose thickness is 2.0 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 18 m 2 and 0.10 meters, respectively. Heat is lost via conduction through the window and the wall. The temperature difference between the inside and outside is the same for the window and the wall. Of the total heat lost by the wall and the window, what is the percentage lost by the window? Answer: 97% The formula for the rate of the conduction of heat is P = Q t = ka L T For the glass, we have an overall thermal conductivity of And for the wall, we have ka L = (0.80)(0.16) = (0.0020) ka L = (0.010)(18) = (0.10) The total thermal conductivity is the sum of these two, W/ C. Therefore, the percent of the total heat lost by the window is =

7 2 2. 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 =

8 3 This is the temperature differential from the aluminum plate: C. So, the temperature on the other side of the steel plate is T = =

9 Name: Physics 202 Quiz 4 May 6, 2013 Word Problems Show all your work and circle your final answer. (Ten points each.) 1. The pressure and volume of an ideal monatomic gas change from A to B to C, as Figure 1 shows. The curved line between A and C is an isotherm. (a) Determine the total heat for the process and (b) state whether the flow of heat is into or out of the gas. Figure 1: Problem Answer: (a) -80 kj (b) Out We cannot tell the heat flow from the pressure-volume graph, but we can determine the work done. The process from A to B is under constant volume (isochoric), so no work is done in this step, but the process from B to C is under constant pressure (isobaric), so the work done in this phase is W = P V = (400000)( 0.200) = In addition, we are told that the temperature of the gas returns to the same value from A to C (this is what it means when it says the are on an isotherm). The internal energy of a gas is directly related to its temperature. This implies that no net change to internal energy occurs in the process. By the first law

10 2 of thermodynamics U = Q W, if internal energy does not change, then the work and heat must balance. Thus, Q = (b) Doing work on a gas would normally increase its temperature. Since the temperature remains constant, the excess heat must be taken out of the gas. You can also see this in the equation U = Q W.

11 3 2. A Carnot engine operates between temperatures of 650 and 350 K. To improve the efficiency of the engine, it is decided either to raise the temperature of the hot reservoir by 40 K or to lower the temperature of the cold reservoir by 40 K. Which change gives the greatest improvement? Justify your answer by calculating the efficiency in each case. Answer: Lower the cold The efficiency is given by e = 1 T c /T h. Thus, e = = If we raise the temperature of the hot reservior we get: e = = If we lower the temperature of the cold reservior we get: e = =

12 Name: Physics 202 Quiz 5 May 13, 2013 Word Problems Show all your work and circle your final answer. (Ten points each.) 1. The middle C string on a piano is under a tension of 944 newtons. The period and wavelength of a wave on this string are 3.82 milliseconds and 1.26 meters, respectively. Find the linear density of the string. Answer: kg/m We will need to use the equation v = F µ We are given F is 944 newtons, but we do not know the velocity. We need to calculate that using v = fλ We are given the wavelength λ = 1.26 meters, but not the frequency. We are given the period, however. So, we need to recall that Thus the velocity of the wave is: Therefore, f = 1 T = 1 = v = ( )(1.26) = ( ) = 944 µ = µ = Remember, this is in kilograms per meter. So this is something like nine grams per meter of string, which seems reasonable.

13 2 2. The mass of a string is 5.0 grams, and it is stretched so that the tension in it is 180 newtons. A transverse wave traveling on this string has a frequency of 260 Hz and a wavelength of 0.60 meters. What is the length of the string? Answer: 0.68 meters The speed of the wave on the string is given by the formula v = F/µ. In addition, since we know the frequency and wavelength, we can determine this speed in this case: v = fλ = (260)(0.60) = 156 Thus, (156) = 180 µ = µ = By definition, the linear density of the string is mass per unit length. In symbols, µ = M L We can use this last formula to figure out the length of this string: ( ) = L = L =

14 Name: Physics 202 Quiz 6 May 20, 2013 Word Problems Show all your work and circle your final answer. (Ten points each.) 1. One method for measuring the speed of sound uses standing waves. A cylindrical tube is open at both ends, and one end admits sound from a tuning fork. A movable plunger is inserted into the other end at a distance L from the end of the tube where the tuning fork is. For a fixed frequency, the plunger is moved until the smallest value of L is measured that allows a standing wave to be formed. Suppose that the tuning fork produces a 485-Hz tone, and that the smallest value observed for L is m. What is the speed of the sound in the gas in the tube? Answer: 512 m/s Since we are dealing with a tube closed at one end, the frequency of the standing waves is given by the formula f = (2n 1) v 4L n = 1, 2, 3,... We are looking for the fundamental frequency so n = 1. Plugging in our data yields v (485) = (1) = v = (4)(0.264)

15 2 2. The range of human hearing is roughly from 20 Hz to 20 khz. Based on these limits and a value of 343 m/s for the speed of sound, what are the lengths of the longest and shortest pipes (open at both ends and producing sound at their fundamental frequencies) that you expect to find in a pipe organ? Answer: 8.6 meters and 8.6 millimeters Since the tube is open at both ends, we want to use the equation f = (2n) v 4L n = 1, 2, 3,... with the fundamental frequency associated with n = 1. Thus, The other extreme is (20) = (2) 343 4L (20000) = (2) 343 4L = L = = L =

16 Name: Physics 202 Quiz 7 May 27, 2013 Word Problems Show all your work and circle your final answer. (Ten points each.) 1. The outside mirror on the passenger side of a car is convex and has a focal length of -7.0 meters. Relative to this mirror, a truck traveling in the rear has an object distance of 11 meters. Find (a) the image distance of the truck and (b) the magnification of the mirror. Answer: (a) -4.3 meters (b) 0.39 (a) The lens equation is Thus, 1 d o + 1 d i = 1 f d i = 1 7 = d i = (b) The magnification can be derived from m = d i d o thus, m = =

17 2 2. A spotlight on a boat is 2.5 meters above the water, and the light strikes the water at a point that is 8.0 meters horizontally displaced from the spotlight (see Figure 2). The depth of the water is 4.0 meters. Determine the distance d, which locates the point where the light strikes the bottom. Figure 2: Problem Answer: 12 meters The key here is to apply Snell s law at the point of refraction. The incident angle is the complement to the acute angle formed by the light ray and the surface of the water. The tangent of this acute angle is related to the distances in the diagram: tan θ = 2.5 = θ = So the angle of incidence is θ 1 = = Snell s law, n 1 sin θ 1 = n 2 sin θ 2, will tell us the refracted angle: (1) sin( ) = (1.33) sin θ 2 = θ 2 = This angle is also related to the distances in the drawing. In this case, the adjacent side is the 4.00 meter height and the opposite side is what we want to know. Thus, tan( ) = x = x = Of course, we must add the 8.00 meters to get the full length d: d = =

18 Name: Physics 202 Quiz 8 Jun 5, 2013 Word Problems Show all your work and circle your final answer. (Ten points each.) 1. A flat screen is located 0.60 meters away from a single slit. Light with a wavelength of 510 nm (in vacuum) shines through the slit and produces a diffraction pattern. The width of the central bright fringe on the screen is meters. What is the width of the slit? Answer: meters The width of the central bright fringe is defined by the location of the dark fringes on either side. Thus, the distance to the first dark fringe is half the width of the central bright fringe: meters. The formula for the location of the dark fringes is sin θ = m λ W The θ is in a right triangle. The opposite side is on the screen and the adjacent side is the from the slit to the screen. Thus, Plugging this in, we have tan θ = = θ = sin = (1)( ) W = W =

19 2 2. 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) =