<p> Phys102-Dr.Aljalal 09-May-06 1</p><p>Quiz-1 Sec-16 A wave in a string, is given by the equation: y(x,t) = 0.24*sin(3.0*x-24*t), where x and y are in meters and t is in seconds. Calculate the magnitude of the transverse speed at x = 2.0 m and t = 1.0 s. (a) 1.8 m/s. (b) 5.5 m/s. (c) 3.8 m/s. (d) 2.1 m/s. (e) 8.0 m/s. u(x,t)= y(x,t) = - 24(0.24)cos(3.0x - 24t) t At x = 2.0 m and t = 1.0 s, u(2.0,1.0)= - 24(0.24)cos(3.0(2.0) - 24(1.0)) = - 3.8 m / s</p><p>Quiz-1 Sec-17 Sinusoidal waves are generated on a string under constant tension by a source vibrating at a constant frequency. If the power delivered by the vibrating source is reduced to one half of the initial value, what is the ratio of the final amplitude to the initial amplitude? (a) 1.0 (b) 2.0 (c) 0.50 (d) 0.71 (e) 1.4</p><p>1 P= m v w2 y 2 avg2 m Constant tension means same speed. 1 P= m v w2 y 2 avg,i2 m,i 1 P= m v w2 y 2 avg,f2 m,f 1 P= P avg,f2 avg,i 1 1 1 mv w2 y 2 = ( m v w 2 y 2 ) 2m,f 2 2 m,i 1 y2= y 2 m,f2 m,i y 1 m,f = = 0.71 ym,i 2</p><p>Quiz-2 Sec-16 A 40 cm string of linear mass density 8.0 g/m is fixed at both ends. The string is driven by a variable frequency audio oscillator ranged from 300 Hz to 800 Hz. It was found that the string is set in oscillation only at the frequencies 440 Hz and 660 Hz. What is the tension in the string? (a) 125 N. (b) 496 N. (c) 500 N. Phys102-Dr.Aljalal 09-May-06 2</p><p>(d) 322 N. (e) 248 N. </p><p>The difference between two consecutive resonant frequencies is v/2L Thus, v 660- 440 = 220 = . 2L t But v = . m t t m -3 . 220 = = 8 10 2L 2(0.4)</p><p> t 176 = . 8 10-3 t = (8� 10-3 )176 2 247.8 N .</p><p>Quiz-2 Sec-17 0.35-66% In figure 4, two small identical speakers are connected (in phase) to the same source. The speakers are 4.10 m apart and at ear level. An observer stands at X, 8.00 m in front of one speaker. In the frequency range 200 Hz-500 Hz, the sound he hears will be most intense if the frequency is: [speed of sound in air is 343 m/s] </p><p>(a) 600 Hz. (b) 346 Hz. (c) 500 Hz. (d) 210 Hz. (e) 422 Hz. </p><p>2 2 L2 = 4.1 + 8.0 = 8.99 m</p><p>L1 = 8.0 m</p><p>DL = L2 - L 1 = 0.99 m</p><p>Most intense sound occurs when we have constructive interference. The condition for constructive interference is Phys102-Dr.Aljalal 09-May-06 3</p><p>DL = n l , where n = 0,1, 2,…</p><p> v But l = , f v v 343 So DL = n or f= n = n = n 346.47 f DL 0.99 for n = 1, f = 346 Hz, which is within the range given in the question.</p><p>Quiz-3 Sec-16 A stationary source emits a sound wave of frequency f. A man travels toward the source at half the speed of sound. The frequency as detected by the man is: (a) 3f (b) 3f/2 (c) 2f/3 (d) f (e) 2f </p><p> v v f= f D v vs</p><p>The source is stationary vs =0. The detector is moving towards the source, so the sign in the numerator should be +.</p><p> v v + v+ v 3 f= fD = f2 = f v v 2 Quiz-3 Sec-17 A driver of a racing car hears a frequency of 1.0*10**4 Hz while moving with a speed of 0.25 v (v is the speed of sound in air) towards a stationary source. Find the frequency of the source. (a) 0.8*10**4 Hz. (b) 0.9*10**4 Hz. (c) 1.2*10**4 Hz. (d) 1.0*10**4 Hz. (e) 0.7*10**4 Hz. </p><p> v v f= f D v vs f'= 1.0x104 Hz is the frequency at the detector. f is the frequency of the source. vD = 0.25 v. We will use + sing since the detector is moving towards the source. vS = 0 since the source is stationary. v+ 0.25v 1.0� 104 = f 1.25f . v 1.0 104 f= = 0.8 104 Hz . 1.25 Quiz-4 Sec-16 One mole of an ideal gas is taken through the cyclic process ABCA as shown in Fig. (2). What is the net heat absorbed, or lost, by the gas? Phys102-Dr.Aljalal 09-May-06 4</p><p>(a) -2.0*10*3 J. (b) -1.0*10*3 J. (c) 1.0*10*3 J. (d) 2.0*10*3 J. (e) 5.0*10*3 J. </p><p>Using the first law of thermodynamics Eint = Q – W, and the fact that for a cyclic process </p><p>Eint = 0, we can say that the heat absorbed by the system is equal to the work done by the system W. Q = W. 1 The work W is the area enclosed by the path, which is -(2 m3 )(1kPa) = - 1kJ . Note that 2 the area is negative since the direction of the process is counterclockwise. You can see this by doing it step by step. The work from A to B is 3 kJ, the work from B to C is zero, and the work from C to A is – 4 kJ. Thus the total work will be – 1 kJ. </p><p>Quiz-4 Sec-17 The wall of a home is 0.2 m thick, 2.0 m high, 10 m wide and has a thermal conductivity of 0.4 watt/m/K. If the inside temperature is 15 degrees Celsius and the outside temperature is -5.0 degrees Celsius, how much energy is lost in 12 hours? (a) 4.5*10**6 J. (b) 4.5*10**5 J. (c) 4.5*10**6 J. (d) 2.7*10**7 J. (e) 3.4*10**7 J. </p><p>QD T P= = kA avg t L DT 15 - ( - 5) Q= kA t = 0.4(10)(2.0) (12创 60 60) = 3.46 107 J. L 0.2 Quiz-5 Sec-16 The volume of an oxygen container is 50.0 L. As oxygen leaks from the container, the pressure inside the container drops from 21.0 to 9.00 atm, and its temperature drops from 303 to 283 K. The number of moles that leaks from the container is: (a) 22.8 mol. (b) 11.1 mol. (c) 19.4 mol. (d) 65.3 mol. (e) 42.2 mol. Phys102-Dr.Aljalal 09-May-06 5</p><p>Use the ideal gas law pV= nRT</p><p> pi V pi V= nRT i i� n i RTi</p><p> pf V pf V= n f RT� n f RTf Number of moles leaked from the container is</p><p> pi V p f VV p i p f ni- n f = - = ( - ) RTi RT f R T i T f 50创 10-3 21 1.01 创 10 5 9 1.01 10 5 =( - ) = 22.8 moles 8.31 303 283 Quiz-5 Sec-17 One mole of a monatomic ideal gas at 410 K is compressed to half its original volume by an isobaric process. How much work is done in the process? (a) 3.3 kJ done on the gas (b) 1.7 kJ done on the gas (c) 3.3 kJ done by the gas (d) 8.3 kJ done on the gas (e) 1.7 kJ done by the gas </p><p>For isobaric process, pressure is kept constant. The work done by the gas is W= p(Vf - V i ). 1 1 Since the final volume is half the initial volume W= p( V - V ) = - pV . 2i i 2 i Using the ideal gas law pV= nRT , we can find the work to be 1 1 W= - nRT = - 1(8.31)(410) = - 1.7 kJ. 2i 2 The work done by the gas is negative. So 1.7 kJ is done on the gas. The answer is (b). Quiz-6 Sec-16 For the arrangement of charges shown in figure (1), the electric field at the point P is: </p><p>(a) Zero. (b) k*q/(d**2) in the positive y-direction. (c) 2*k*q/(d**2) in the negative y-direction. (d) 2*k*q/(d**2) in the positive y-direction. (e) k*q/(d**2) in the negative y-direction. Phys102-Dr.Aljalal 09-May-06 6</p><p>E2,y E1,y E2 E1</p><p>E1,x E2,x 2 d 2 d E3</p><p>1 3 2</p><p>The x-component of E1 and E2 cancel each other. We need only to consider forces along the y-axis. q 1 q E= E cos450 = k cos45 0 = k 1,y 1 ( 2d)2 2 2 d2 q 1 q E= E cos450 = k cos45 0 = k 2,y 2 ( 2d)2 2 2 d2 2q E== k 3 d2 The sum of the forces along the y-axis are 2 q 2q 1 q E+ E - E = k - k = ( - 2)k . 1,y 2,y 3 2 2d2 d 2 2 d 2 It will be along the negative y-axis. Note that none of the choices are correct. There was a problem with this question in that major!</p><p>Quiz-6 Sec-17 In figure 5, four charges are placed on the circumference of a circle of diameter 2 m. If an electron is placed at the center of the circle, then the electron will [Take Q = 60 micro-C, q = 20 micro-C] </p><p>(a) move to the left. (b) move to the right. (c) move downward. (d) stay at the center. (e) move upward. </p><p>The electron will stay in its position. Since the forces cancel each other.</p><p>Quiz-7 Sec-16 A positive point charge q sits at the center of a hollow spherical shell. The shell, with radius R and negligible thickness, has net charge -2q. The electric field strength outside the spherical shell (at r>R) will be: Phys102-Dr.Aljalal 09-May-06 7</p><p>(a) 3*k*q/r**2 radially inwards. (b) k*q/r**2 radially inwards. (c) 3*k*q/r**2 radially outwards. (d) k*q/r**2 radially outwards. (e) zero.</p><p>Since we are calculating the electric field outside the sphere, we can treat the sphere as a point charge located at the center of the sphere. So this is equivalent to a point charge of +q-2q=-q located at the center. The electric field is k*q/r**2 radially inwards.</p><p>Quiz-7 Sec-17 Charge is uniformly distributed over the entire xy plane with a surface charge density of 20 micro- Coulomb/m**2. A sphere has a radius of 1.0 m, and is centered at the origin. What is the net electric flux through the surface of the sphere? (a) 2.8*10**7 N.(m**2)/C (b) 1.4*10**7 N.(m**2)/C (c) zero (d) 7.1*10**6 N.(m**2)/C (e) 2.2*10**6 N.(m**2)/C </p><p>According to the Guss’ law, 0(flux through any closed surface) = the net charge inside the closed surface.</p><p>The net charge enclosed by the sphere is the charge of a circle of radius r on the surface which is r2.</p><p>2 Flux through the sphere = r / 0 = 7.1*10**6 N.(m**2)/C.</p><p>Quiz-8 Sec-16 A long noncoducting cylinder (radius 12.0 cm) has a charge of uniform density 5.0 nano-C/m**3 distributed through its column. Determin the magnitude of the electric field 5.0 cm from the axis of the cylinder. [See figure (3)]. </p><p>(a) 14 N/C. (b) 4 N/C. (c) 22 N/C. (d) 34 N/C. (e) 31 N/C. Phys102-Dr.Aljalal 09-May-06 8</p><p>E</p><p>Gaussian surface</p><p>Imagine a cylindrical Gaussian surface of length L and radius r = 5.0 cm. The flux through this surface is E(2rL). According to Gauss’ law</p><p>0(flux through any closed surface) = the net charge inside the closed surface.</p><p>Thus, </p><p>0 E(2rL)= the net charge inside the closed surface.</p><p>The net charge enclosed by the Gaussian surface = (the charge density)(volume of the Gaussian cylinder).</p><p>2 0 E(2rL)=   r L.</p><p>-9 -12 E =  r/20 = 5 x10 (0.05)/(2 x 8.85x10 ) = 14.1 N/C.</p><p>Quiz-8 Sec-16 An infinite nonconducting sheet has a surface charge density 0.10*10**(-6) C/m**2 on one side. How far apart are equipotential surfaces whose potentials differ by 90 V? (a) 2.0 cm. (b) 1.6 cm. (c) 0.88 cm. (d) 1.8 cm. (e) 2.5 cm. </p><p>V2 V1</p><p> d</p><p> E</p><p>Non-conducting Equipotential Equipotential sheet surface surface</p><p>The electric field due to an infinite non-conducting sheet is uniform and has a magnitude of Phys102-Dr.Aljalal 09-May-06 9</p><p>-6 -12 6 E = /20 = 0.10 x10 /(2 x 8.85x10 ) = 5.6 x10 V/m.</p><p>For a uniform, the electric potential difference between two equipotential surfaces separated by distance d is V = E d.</p><p>Thus d= V/ E = 90/(5.6 x106) = 1.6 cm.</p><p>Quiz-9 Sec-16 A copper wire of cylindrical shape has resistance R. What is the resistance of a second wire, made of the same material, that is twice as long and has twice the diameter ? (a) 4R (b) 2R (c) R/2 (d) R/4 (e) R </p><p>The resistance of the first wire is R= L1/A1, where L1 is the length of the wire and A1 is the cross sectional area. For the second wire R= L1)/(4A1) = R/2.</p><p>Quiz-9 Sec-16</p><p>A 500 W electric heater is designed to operate from a 120-V power supply. The line voltage decreases and the heater takes only 459 W. Find the voltage drop in the line voltage (Assuming the resistance is constant). (a) 10 Volts. (b) 5 Volts. (c) 2 Volts. (d) 15 Volts. (e) 3 Volts. </p><p>2 2 At normal operational voltage P1 = V1 /R = (120) /R = 500 W. Thus R = 1202/500. 2 After the decrease of the voltage, P2 = V2 /R = 459 W. 2 2 Thus, V2 = (459 R) = 459 (120 /500). V2 = 115 V. Voltage drop is V1 – V2 = 120 – 115 = 5 V.</p>