PHYSICS 1307 FALL 2007 EXAM # 3 Thursday, November 15, 2007 SOLUTIONS. P atm = N m 2 v sound =343 m/ s

Transcription

1 1/9 Last Name: First Name: SMU ID: PHYSICS 1307 FALL 2007 EXAM # 3 Thursday, November 15, 2007 SOLUTIONS CQ: NQ: Total: Scientific data air =1.29 kg m 3 steel = C 1 P atm = N m 2 v sound =343 m/ s water = kg m 3 CONCEPTUAL QUESTIONS 1. In the figure 3 table tennis balls are levitated in equilibrium above a vertical column through which air escapes. In which pipe the air velocity is higher and why? A. B. By Bernoulli's equation we know that areas of high velocity are also areas of low pressure. Since the pressure is proportional to the high of the column of air, A has lower pressure than B and therefore the air in A has higher velocity.

2 2/9 2. What is the function a(t) that better describes the motion represented by the graphic below. Tip: This is the acceleration of a SHM. Acceleration 1 m/s t (s) 1 m/s 2 The function represented in the graph is a=a max cos t. We need to determine a max and. Both can be found from information present in the graph. The first one is easy, it is equal to 1m/ s 2. The angular frequency is given by =2 /T where the period is T=1.0 s. Thus we have a=1.0 m/ s 2 cos 2 t. 3. Does a car bounce on its springs faster when it is empty or when it is fully loaded? Explain. The frequency of a SHM is equal to f = 1 k/ m. Thus the car will bounce faster (with higher 2 frequency) with a smaller mass. This is when it is empty. 4. If the amplitude of a sound wave is tripled by what factor will the intensity increase? Explain. The intensity is proportional to the energy emitted and the energy is proportional to the amplitude square. Thus, if the amplitude is tripled, the intensity will be increased by a factor of nine!

3 3/9 5. In an old rerun of the classic Star Trek series it was said that the minimum temperature had dropped below C. Discuss this affirmation using the knowledge acquired in class. The zero of the Kelvin scale is the absolute zero for any temperature. This zero is equal to in the Celsius scale. Thus the affirmation given in the show was false, it is impossible for the temperature of any object to go below this point. NUMERICAL QUESTIONS 6. An solid object has a weight of 5.00 N. When the object is submerged in water the apparent weigh is 3.50 N. The density of the object is x 10 3 kg / m 3 A. 1.4 B. 1.0 C. 3.4 D. 2.5 E. 5 The difference between the real weight and the apparent one is equal to the buoyancy force therefore, W obj W apparent = water gv obj 5.0 N 3.50 N 10 3 kg/m 3 9.8m/s m 3 =V obj to find the density we need also the mass which can be found from the real weight. Finally: obj = m V = 0.5kg m 3=3400kg/m3 7. The Ausable River in upstate New York is about 40 m wide. It's 2.2 m deep and flows at 4.5m/s. Just before it reaches Lake Champlain, the river enters Ausable Chasm, a deep gorge only 3.7 m wide. If the flow rate is 6.0 m/s, the depth of the river at this point is m A.10 B. 14 C.18 D. 22 E. 26

4 4/9 We can use the equation of continuity: A 1 v 1 =A 2 v 2 40m 2.2m 4.5m/ s= 3.7m h 6.0 m/s then h=18 m. _ 8. The air pressure at the center of a category 5 hurricane with a speed at the center of 300 km/h is about kpa A. 97 B. 101 C.110 D. 120 E. 87 We can use Bernoulli's equation in two different points. The first one is at the center of the hurricane and the second point, far away where the atmosphere is quiet. Then: P v 2 =P air 1 atm P 1 = Pa kg/ m3 300/3.6 2 P 1 =96821 Pa 9. The minimum cross section area required of a vertical steel cable to from which is suspended a 320 kg chandelier is cm 2. The ultimate tensile strength of steel is of the order of N /m 2 A. 0.8 B. 0.5 C. 0.1 D E The ultimate tensile strength is the maximum value of tensile stress before having a rupture of the material. So, N /m 2 = F A min 320 kg 9.8m /s2 A min = = m2 2 N /m (Check units!)

5 5/9 An object of mass 5.00 kg moves in simple harmonic motion with amplitude 12 cm on a spring of negligible mass. Its maximum acceleration is 108 cm / s The constant of the spring is N / m A. 50 B. 45 C. 40 D. 35 E. 30 Remember that the maximum acceleration in a SHM which is given by a max =A 2 angular frequency is equal to = k / m. Combining both we obtain: a max = A k m solving for the spring 's constant we get k= 1.08 m/s2 5.00kg =45 N / m m and the 11.The energy of the oscillation is 10 1 J? A. 5.1 B. 4.0 C. 3.2 D. 2.5 E. 2.0 E= 1 2 k A2 E= N / m 0.12 m 2 =0.324 J 12. The angular position of a pendulum is represented by the equation = cos 4.43t with in radians. The length of the pendulum is m. A. 1.5 B. 1.0 C D E Here we need to identify the angular frequency of a simple pendulum. This frequency is given by = g/ L where g is gravity and L is the length of the pendulum. The frequency can also be determined by the coefficient that multiplies the time t in the equation that describes the motion. Therefore: 4.43Hz= 9.8m/s 2 L so L=0.50 m

6 6/9 13. At a distance of 2.0 m from a localized source you measure the intensity level of 75 db. The distance to perceive half the intensity is m. Assume isotropic emission and a homogeneous medium. A. 5.4 B. 4.0 C. 3.2 D. 2.8 E. 1.0 Using the fact that intensity is proportional to the inverse of the distance to the source we have the following relation : 2 I 1 I 2 = r 1 r 2 2 thus if I 2 =2 I 1 we can solve for r 2 to obtain: r 2 = 2r m. 14. A driver travels northbound on a highway at a speed of 25.0 m/s. A police car, traveling southbound at a speed of 40.0 m/s approaches with its siren producing a sound at a frequency of 2500 Hz. The frequency that the driver observes as the police car approaches is Hz A B.2830 C.2682 D.2624 E.2076 This is a typical case of moving source and moving observer in the Doppler effect case. From the picture it is clear that the observer is moving towards the source and the source likewise is moving towards the observer. Thus: 40 m/s 25 m/s N f '= f v v sound observer v sound V source 343 m/s 25 m /s f '=2500 Hz 343m/ s 40 m/s f '=3036Hz

7 7/9 15.The force exerted on your eardrum due to the water when you are swimming at the bottom of a pool that is 5.0 m deep is N. The eardrum has an area of approximately 1cm 2 and do not neglect the internal pressure inside the middle ear. A. 15 B. 10 C. 7.5 D. 5 E. 2.5 The pressure at the bottom of the pool is given by: P=P atm water gh= Pa 10 3 kg/ m m/s 2 5m= Pa Since the internal pressure inside the middle ear is about the atmospheric pressure, the net force is given by the product of the gauge pressure times the area of the eardrum. So F=P gauge A= Pa m 2 =5N _ 16. At 0 0 C, the hole in a steel washer is 9.52 mm in diameter. The temperature that must it be heated in order to fit over a 9.55 mm diameter bolt is 0 C A. 300 B. 283 C. 273 D. 263 E. 253 We just need to use the equation for linear thermal expansion: L=L 0 steel T 9.55mm 9.52mm = 9.52 mm C 0 1 T C= T 17. A transverse wave is propagating on a string. The wave frequency is 44 Hz. The string is under 21 N of tension and has a mass per unit of length of 15 g/m. The wavelength of the wave is m. A B C D.0.05 E Using the equation for the velocity on a string v= F T / m/ L = 21N /0.015 kg/m=37.4 m/ s the relation between velocity, frequency and wavelength : v= f = 37.4m/ s 44 Hz =0.85m and is straightforward to obtain _

8 8/9 18. As you can observe in the figure below a hydraulic lift supports a truck with a mass of 3200 kg. The force F that must be applied to the smaller piston to support the truck is N. Assume circular pistons. 1 2 F 2 A. 490 B. 520 C. 560 D. 600 E cm 15 cm This problem is easily solved using Pascal's principle that states that: points in a fluid that are at the same depth experience the same pressure. Thus right below each piston we have: P 1 =P 2 F 1 A 1 = F 2 A 2 F kg 9.8 m/s 2 = 0.60m m N=F 2 _ 19. Two sound waves have measured intensities of 100 W /m 2 and 200 W /m 2. The intensity of the two sounds combined is db A. 300 B. 250 C. 212 D.145 E. 122 The intensity in decibels is equal to 10 log =144.77dB

9 9/9 20. A figure of a sinusoidal wave whose velocity is 300 m/s is shown below. The wavelength of the wave is m. y (meters) A. 300 B. 275 C. 225 D. 150 E m 1m t(sec) Given that v= /T and the period T is 0.50 s we have =300 m/ s 0.5 s=150 m 21. The normal body temperature is F. The value of this temperature on the Celsius scale is 0 C A. 38 B. 37 C. 36 D. 35 E. 34 T 0 F = C F= 9 5 T 0 C C _ Write your answers here: 6.C 7.C 8.A 9.E 10.B 11.C 12.C 13.D 14.A 15.D 16.D 17.A 18.A 19.D 20.D 21.B