125 Pages

# ch10

Course Number: PHYSIC 2, Spring 2008

College/University: Lehigh

Word Count: 15886

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1. (a) The second hand of the smoothly running watch turns through 2 radians during 60 s . Thus, = 2 = 0.105 rad/s. 60 (b) The minute hand of the smoothly running watch turns through 2 radians during 3600 s . Thus, = 2 = 1.75 10 -3 rad / s. 3600 (c) The hour hand of the smoothly running 12-hour watch turns through 2 radians during 43200 s. Thus, = 2 = 145 10 -4 rad / s. . 43200 2. The problem asks us to...

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Lehigh - PHYSIC - 2
1. (a) All the energy in the circuit resides in the capacitor when it has its maximum charge. The current is then zero. If Q is the maximum charge on the capacitor, then the total energy is2.90 10-6 C Q2 U= = = 117 10-6 J. . -6 2C 2 3.60 10 Fc
Lehigh - PHYSIC - 2
1. The vector area A and the electric field E are shown on the diagram below. The angle between them is 180 35 = 145, so the electric flux through the area is = E A = EA cos = (1800 N C ) 3.2 10-3 m cos145 = -1.5 10-2 N m 2 C.2()2. We u
Lehigh - PHYSIC - 2
1. (a) The amplitude is half the range of the displacement, or xm = 1.0 mm. (b) The maximum speed vm is related to the amplitude xm by vm = xm, where is the angular frequency. Since = 2f, where f is the frequency,vm = 2 fxm = 2 (120 Hz ) (1.0 10 -
Lehigh - PHYSIC - 2
1. The time it takes for a soldier in the rear end of the column to switch from the left to the right foot to stride forward is t = 1 min/120 = 1/120 min = 0.50 s. This is also the time for the sound of the music to reach from the musicians (who are
Lehigh - PHYSIC - 2
1. We take p3 to be 80 kPa for both thermometers. According to Fig. 18-6, the nitrogen thermometer gives 373.35 K for the boiling point of water. Use Eq. 18-5 to compute the pressure:pN =T 373.35 K p3 = 273.16 K 273.16 K(80 kPa) = 109.343kPa.
Lehigh - PHYSIC - 2
1. According to Eq. 39-4 En L 2. As a consequence, the new energy level E'n satisfiesEn L = En LFG IJ = FG L IJ H K H L K-22=1 , 2which gives L = 2 L. Thus, the ratio is L / L = 2 = 1.41.2. (a) The ground-state energy is( 6.63 10
Lehigh - PHYSIC - 2
1. Our calculation is similar to that shown in Sample Problem 42-1. We set K = 5.30 MeV=U = (1/ 4 0 )( q qCu / rmin ) and solve for the closest separation, rmin:rmin-19 9 q qCu kq qCu ( 2e )( 29 ) (1.60 10 C )( 8.99 10 V m/C ) = = = 4 0 K 4 0 K
Lehigh - PHYSIC - 2
1. The x and the y components of a vector a lying on the xy plane are given byax = a cos , a y = a sin where a =| a | is the magnitude and is the angle between a and the positive x axis. (a) The x component of a is given by ax = 7.3 cos 250 = 2
A.T. Still University - SDF - asdf
Economics 100b Professor Wood1/31/08 Lecture 4ASUC Lecture Notes Online is the only authorized note-taking service at UC Berkeley. Please do not share, copy or illegally distribute these notes. Our non-profit, student-run program depends on your i
Lehigh - PHYSIC - 2
1. Conservation of momentum requires that the gamma ray particles move in opposite directions with momenta of the same magnitude. Since the magnitude p of the momentum of a gamma ray particle is related to its energy by p = E/c, the particles have th
Lehigh - PHYSIC - 2
1. Each atom has a mass of m = M/NA, where M is the molar mass and NA is the Avogadro constant. The molar mass of arsenic is 74.9 g/mol or 74.9 103 kg/mol. 7.50 1024 arsenic atoms have a total mass of (7.50 1024) (74.9 103 kg/mol)/(6.02 1023 mol
Lehigh - PHYSIC - 2
1. From the time dilation equation t = t0 (where t0 is the proper time interval, = 1 / 1 - 2 , and = v/c), we obtain = 1-FG t IJ . H t K2 0The proper time interval is measured by a clock at rest relative to the muon. Specifically, t0 = 2.2
Lehigh - PHYSIC - 2
1. (a) The flux through the top is +(0.30 T)r2 where r = 0.020 m. The flux through the bottom is +0.70 mWb as given in the problem statement. Since the net flux must be zero then the flux through the sides must be negative and exactly cancel the tota
Lehigh - PHYSIC - 2
1. The air inside pushes outward with a force given by piA, where pi is the pressure inside the room and A is the area of the window. Similarly, the air on the outside pushes inward with a force given by poA, where po is the pressure outside. The mag
Lehigh - PHYSIC - 2
1 1. The potential energy stored by the spring is given by U = 2 kx 2 , where k is the spring constant and x is the displacement of the end of the spring from its position when the spring is in equilibrium. Thusk=2 25 J 2U = 2 x 0.075 mb g b g
Lehigh - PHYSIC - 2
1. If R is the fission rate, then the power output is P = RQ, where Q is the energy released in each fission event. Hence, R = P/Q = (1.0 W)/(200 106 eV)(1.60 10 19 J/eV) = 3.1 1010 fissions/s.2. We note that the sum of superscripts (mass number
Lehigh - PHYSIC - 2
1. (a) An Ampere is a Coulomb per second, so84 A h = 84FG HCh sIJ FG 3600 s IJ = 3.0 10 K H hK5C.(b) The change in potential energy is U = qV = (3.0 105 C)(12 V) = 3.6 106 J.2. The magnitude is U = eV = 1.2 109 eV = 1.2 GeV.3. T
Lehigh - PHYSIC - 2
1. In air, light travels at roughly c = 3.0 108 m/s. Therefore, for t = 1.0 ns, we have a distance of d = ct = (3.0 108 m / s) (1.0 10-9 s) = 0.30 m.2. (a) From Fig. 33-2 we find the smaller wavelength in question to be about 515 nm, (b) and the
Lehigh - PHYSIC - 2
1. We are only concerned with horizontal forces in this problem (gravity plays no direct role). We take East as the +x direction and North as +y. This calculation is efficiently implemented on a vector-capable calculator, using magnitude-angle notati
Lehigh - PHYSIC - 2
1. (a) Let E = 1240 eVnm/min = 0.6 eV to get = 2.1 103 nm = 2.1 m. (b) It is in the infrared region.2. The energy of a photon is given by E = hf, where h is the Planck constant and f is the frequency. The wavelength is related to the frequency b
Lehigh - PHYSIC - 2
1. Comparing the light speeds in sapphire and diamond, we obtainv = vs - vd = cFG 1 - 1 IJ Hn n K F 1 - 1 IJ = 4.55 10 m s. = c2.998 10 m sh G H 177 2.42 K .s d 8 72. (a) The frequency of yellow sodium light is c 2.998 108 m s f = = = 5.09
Lehigh - PHYSIC - 2
1. Charge flows until the potential difference across the capacitor is the same as the potential difference across the battery. The charge on the capacitor is then q = CV, and this is the same as the total charge that has passed through the battery.
Lehigh - PHYSIC - 2
1. (a) For a given value of the principal quantum number n, the orbital quantum number ranges from 0 to n 1. For n = 3, there are three possible values: 0, 1, and 2. (b) For a given value of , the magnetic quantum number m ranges from - to + . For =
Lehigh - PHYSIC - 2
1. With speed v = 11200 m/s, we findK= 1 2 1 mv = (2.9 105 ) (11200) 2 = 18 1013 J. . 2 22. (a) The change in kinetic energy for the meteorite would be1 1 K = K f - K i = - K i = - mi vi2 = - 4 106 kg 15 103 m/s 2 2()()2= -5 1014 J
Lehigh - PHYSIC - 2
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Lehigh - PHYSIC - 2
1. (a) The motion from maximum displacement to zero is one-fourth of a cycle so 0.170 s is one-fourth of a period. The period is T = 4(0.170 s) = 0.680 s. (b) The frequency is the reciprocal of the period:f = 1 1 = = 1.47 Hz. T 0.680 s(c) A sinuso
Lehigh - PHY - 2
1. (a) The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given byB= 0i2r.With r = 20 ft = 6.10 m, we havec4 10 B=hb 2 b6.10 mg-7T m A 100 Ag = 3.3 10-6T = 3.3 T.(b
Lehigh - PHYSIC - 2
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Lehigh - PHYSIC - 2
1. The image is 10 cm behind the mirror and you are 30 cm in front of the mirror. You must focus your eyes for a distance of 10 cm + 30 cm = 40 cm.2. The bird is a distance d2 in front of the mirror; the plane of its image is that same distance d2
Lehigh - PHYSIC - 2
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Lehigh - PHYSIC - 2
1. (a) The charge that passes through any cross section is the product of the current and time. Since 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 103 C. (b) The number of electrons N is given by q = Ne, where e is the magnitu
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Cal Poly Pomona - COM - 413
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Cal Poly Pomona - COM - 413
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