SOUND AND HEARING
IDENTIFY and SET UP: Eq.(15.1) gives the wavelength in terms of the frequency. Use Eq.(16.5) to relate the pressure and displacement amplitudes. EXECUTE: (a) = v / f = (344 m/s)/1000 Hz = 0.344 m (b) pmax = BkA and Bk is constan
IDENTIFY: Compare the path difference to the wavelength. SET UP: The separation between sources is 5.00 m, so for points between the sources the largest possible path difference is 5.00 m. EXECUTE: (a) For constructive interfer
IDENTIFY: Use y = x tan to calculate the angular position of the first minimum. The minima are located by m , m = 1, 2,. First minimum means m = 1 and sin 1 = / a and = a sin 1. Use this Eq.(36.2): sin = a equation to calculate . SET
IDENTIFY and SET UP: Consider the distance A to O and B to O as observed by an observer on the ground (Figure 37.1).
(b) d = vt = (0.900) (3.00 108 m s) (5.05 10-6 s) = 1.36 103 m = 1.36 km. 37.3.
1 IDENTIFY and SET UP: The
PHOTONS, ELECTRONS, AND ATOMS
h f - . The e e
IDENTIFY and SET UP: The stopping potential V0 is related to the frequency of the light by V0 = slope of V0 versus f is h/e. The value fth of f when V0 = 0 is related to by = hf th .
EXECUTE: (a) From
THE WAVE NATURE OF PARTICLES
IDENTIFY and SET UP: EXECUTE: (a) =
h h = . For an electron, m = 9.11 10 -31 kg . For a proton, m = 1.67 10 -27 kg . p mv
6.63 10-34 J s = 1.55 10-10 m = 0.155 nm (9.11 10-31 kg)(4.70 106 m/s)
m 9.11 10 -31 kg 1
n2h 2 . 8mL2
IDENTIFY and SET UP: The energy levels for a particle in a box are given by En = EXECUTE: (a) The lowest level is for n = 1, and E1 =
(1)(6.626 10-34 J s) 2 = 1.2 10-67 J. 8(0.20 kg)(1.5 m) 2
1 2E 2(1.2 10-67 J) (b)
L = l (l + 1) . Lz = ml . l = 0, 1, 2,., n - 1. ml = 0, 1, 2,., l . cos = Lz / L .
IDENTIFY and SET UP:
EXECUTE: (a) l = 0 : L = 0 , Lz = 0 . l = 1: L = 2 , Lz = ,0, - . l = 2 : L = 6 , Lz = 2 , ,0, - , -2 . (b) In each case cos
(a) (b) (c)
28 14 85 37
Si has 14 protons and 14 neutrons. Rb has 37 protons and 48 neutrons. Tl has 81 protons and 124 neutrons.
(a) Using R = (1.2 fm)A1 3 , the radii are roughly 3.6 fm, 5.3 fm, and 7.1 fm. (b) Usin
PARTICLE PHYSICS AND COSMOLOGY
(a) IDENTIFY and SET UP: Use Eq.(37.36) to calculate the kinetic energy K. 1 EXECUTE: K = mc 2 - 1 = 0.1547 mc 2 2 2 1- v / c
m = 9.109 10 -31 kg, so K = 1.27 10-14 J (b) IDENTIFY and SET UP: The total energy of th
Grammar Relating to Dating
One way to tell if someones right for you is by analyzing their grammar.
Grammar matters when someone is trying to find a potential date.
Having incorrect grammar might decrease the chance of a relationship being
Creating Graphs to Reach
In this assignment, you will create two graphs and answer questions about Bond's
Gym, the business you are supporting.
1. Gather materials and necessary information.
a) Ask your teacher w
y = 4.85 cm
IDENTIFY and SET UP: Plane mirror: s = - s (Eq.34.1) and m = y / y = - s / s = +1 (Eq.34.2). We are given s and y and are asked to find s and y. EXECUTE: The object and image are shown in Figure 34.1. s =
THE NATURE AND PROPAGATION OF LIGHT
IDENTIFY: For reflection, r = a . SET UP: The desired path of the ray is sketched in Figure 33.1. 14.0 cm EXECUTE: tan = , so = 50.6 . r = 90 - = 39.4 and r = a = 39.4 . 11.5 cm EVALUATE: The angle of incidence
ELECTRIC CHARGE AND ELECTRIC FIELD
(a) IDENTIFY and SET UP: Use the charge of one electron ( -1.602 10 -19 C) to find the number of electrons required to produce the net charge. EXECUTE: The number of excess electrons needed to produce net charge
^ E = E cos dA, where is the angle between the normal to the sheet n and the
(a) IDENTIFY and SET UP:
electric field E . EXECUTE: In this problem E and cos are constant over the surface so
E = E cos dA = E cos A = (14 N/C )( cos 60 )
ra = 0.150 m rb = (0.250 m) 2 + (0.250 m) 2 rb = 0.3536 m
IDENTIFY: Apply Eq.(23.2) to calculate the work. The electric potential energy of a pair of point charges is given by Eq.(23.9). SET UP: Let the initial position of q2 b
CAPACITANCE AND DIELECTRICS
Q Vab SET UP: 1 F = 10 -6 F EXECUTE: Q = CVab = (7.28 10 -6 F)(25.0 V) = 1.82 10 -4 C = 182 C EVALUATE: One plate has charge + Q and the other has charge -Q . Q PA and V = Ed . IDENTIFY and SET UP: C = 0 ,
CURRENT, RESISTANCE, AND ELECTROMOTIVE FORCE
IDENTIFY: I = Q / t . SET UP: 1.0 h = 3600 s EXECUTE: Q = It = (3.6 A)(3.0)(3600 s) = 3.89 104 C. EVALUATE: Compared to typical charges of objects in electrostatics, this is a huge amount of char
IDENTIFY: The newly-formed wire is a combination of series and parallel resistors. SET UP: Each of the three linear segments has resistance R/3. The circle is two R/6 resistors in parallel. EXECUTE: The resista
MAGNETIC FIELD AND MAGNETIC FORCES
! IDENTIFY and SET UP: Apply Eq.(27.2) to calculate F . Use the cross products of unit vectors from Section 1.10. ! ^ j EXECUTE: v = ( +4.19 104 m/s ) i + ( -3.85 104 m/s ) ^ ! ^ (a) B = (1.40 T ) i ! ! ! ^ ^ F
SOURCES OF MAGNETIC FIELD
! ^ EXECUTE: (a) r = ( 0.500 m ) i , r = 0.500 m ! ! ^ v r = vr^ i = -vrk j ^
! IDENTIFY and SET UP: Use Eq.(28.2) to calculate B at each point. ! ! ! ! ! qv r 0 qv r ^ r ^ B= 0 = , since r = . 4 r 2 4 r 3 r ! ! 6 ^ and
IDENTIFY: Altering the orientation of a coil relative to a magnetic field changes the magnetic flux through the coil. This change then induces an emf in the coil. SET UP: The flux through a coil of N turns is = NBA
Apply Eq.(30.4). di (a) E2 = M 1 = (3.25 10-4 H)(830 A/s) = 0.270 V; yes, it is constant. dt
IDENTIFY and SET UP: EXECUTE: (b) E1 = M
di2 ; M is a property of the pair of coils so is the same as in part (a). Thus E1 = 0.270 V. dt EVALU
IDENTIFY: SET UP: EXECUTE:
i = I cos t and I rms = I/ 2.
The specified value is the root-mean-square current; I rms = 0.34 A.
(a) I rms = 0.34 A
(b) I = 2 I rms = 2(0.34 A) = 0.48 A. (c) Since the current is positive hal
IDENTIFY: Since the speed is constant, distance x = ct. SET UP: The speed of light is c = 3.00 108 m/s . 1 yr = 3.156 107 s.
x 3.84 108 m = = 1.28 s c 3.00 108 m/s (b) x = ct = (3.00 108 m/s)(8.61 yr)(3.156 107 s/yr) =
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