Chapter 14
1. The pressure increase is the applied force divided by the area: p = F/A = F/r2, where r is the radius of the piston. Thus p = (42 N)/(0.011 m)2 = 1.1 105 Pa. This is equivalent to 1.1 at
Chapter 33
1. (a) From Fig. 33-2 we find the smaller wavelength in question to be about 515 nm. (b) Similarly, the larger wavelength is approximately 610 nm. (c) From Fig. 33-2 the wavelength at which
Chapter 32
1. We use
6 n=1
Bn = 0 to obtain
5 n=1
B6 = -
Bn = - ( -1Wb + 2 Wb - 3 Wb + 4 Wb - 5 Wb ) = +3 Wb .
2. (a) The flux through the top is +(0.30 T)r2 where r = 0.020 m. The flux through the
Chapter 31
1. (a) The period is T = 4(1.50 s) = 6.00 s. (b) The frequency is the reciprocal of the period: f = 1 1 = = 167 105 Hz. . T 6.00s
(c) The magnetic energy does not depend on the direction of
Chapter 30
1. (a) The magnitude of the emf is
=
d B d = 6.0t 2 + 7.0t = 12t + 7.0 = 12 2.0 + 7.0 = 31 mV. dt dt
c
h
bg
(b) Appealing to Lenz's law (especially Fig. 30-5(a) we see that the current flo
Chapter 29
1. (a) The field due to the wire, at a point 8.0 cm from the wire, must be 39 T and must be directed due south. Since B = 0i 2 r ,
-6 2 rB 2 0.080 m 39 10 T i= = = 16 A. 0 4 10 -7 T m A
b
g
Chapter 28
1. (a) The force on the electron is r r r FB = qv B = q vx ^ + v y ^ i j
= -1.6 10-19 C 2.0 106 m s ( -0.15 T ) - 3.0 106 m s ( 0.030 T ) -14 ^ = 6.2 10 N k.
( (
(
)
) (
) ( B ^i + B j ) =
Chapter 27
1. (a) The energy transferred is
2t (2.0 V) 2 (2.0 min) (60 s / min) U = Pt = = = 80 J . r+R 1.0 + 5.0
(b) The amount of thermal energy generated is
F I F 2.0 V I (5.0 ) (2.0 min) (60 s /
Chapter 26
1. (a) The charge that passes through any cross section is the product of the current and time. Since t = 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 103 C. (b) The
Chapter 25
1. (a) The capacitance of the system is C= q 70 pC = = 35 pF. . V 20 V
(b) The capacitance is independent of q; it is still 3.5 pF. (c) The potential difference becomes V = q 200 pC = = 57
Chapter 24
1. If the electric potential is zero at infinity then at the surface of a uniformly charged sphere it is V = q/40R, where q is the charge on the sphere and R is the sphere radius. Thus q =
Chapter 23
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 2 = E A = EA cos = ( 1800 N C
Chapter 17
1. (a) When the speed is constant, we have v = d/t where v = 343 m/s is assumed. Therefore, with t = 15/2 s being the time for sound to travel to the far wall we obtain d = (343 m/s) (15/2
Chapter 16
1. (a) The angular wave number is k = 2 2 = = 3.49 m -1. 1.80 m
(b) The speed of the wave is v = f =
( 1.80 m ) ( 110 rad s ) = = 31.5 m s. 2 2
2. The distance d between the beetle and the
Chapter 15
1. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is am = 2xm, where is the angular frequency ( = 2f since there are 2 radians in one cycl
Chapter 13
1. The magnitude of the force of one particle on the other is given by F = Gm1m2/r2, where m1 and m2 are the masses, r is their separation, and G is the universal gravitational constant. We
Chapter 12
1. (a) The center of mass is given by xcom = 0 + 0 + 0 + (m)(2.00 m) + ( m)(2.00 m) + ( m)(2.00 m) = 1.00 m. 6m
(b) Similarly, we have ycom = 0 + (m)(2.00 m) + (m)(4.00 m) + ( m)(4.00 m) +
Chapter 11
1. The velocity of the car is a constant r ^ v = + ( 80 km/h ) (1000 m/km)(1 h/3600 s) ^ = (+22 m s)i, i and the radius of the wheel is r = 0.66/2 = 0.33 m. (a) In the car's reference frame
Chapter 10
1. The problem asks us to assume vcom and are constant. For consistency of units, we write 5280 ft mi vcom = 85 mi h = 7480 ft min . 60 min h
b
F g G H
I J K
Thus, with x = 60 ft , the time
Chapter 9
1. We use Eq. 9-5 to solve for ( x3 , y3 ). (a) The x coordinates of the system's center of mass is: xcom = m1 x1 + m2 x2 + m3 x3 (2.00 kg)(-1.20 m) + ( 4.00 kg ) ( 0.600 m ) + ( 3.00 kg ) x
Chapter 8
1. (a) Noting the vertical displacement is 10.0 m 1.50 m = 8.50 m downward (same that direction as Fg ), Eq. 7-12 yields Wg = mgd cos = (2.00 kg)(9.80 m/s 2 )(8.50 m) cos 0 = 167 J. (b) One
Chapter 7
1. (a) The change in kinetic energy for the meteorite would be 1 1 K = K f - K i = - K i = - mi vi2 = - 4 106 kg 15 103 m/s 2 2
(
)(
)
2
= -5 1014 J ,
or | K |= 5 1014 J . The negative sign
Chapter 6
1. We do not consider the possibility that the bureau might tip, and treat this as a purely horizontal motion problem (with the person's push F in the +x direction). Applying Newton's second
Chapter 5
1. We apply Newton's second law (specifically, Eq. 5-2). (a) We find the x component of the force is Fx = max = ma cos 20.0 = ( 1.00kg ) ( 2.00m/s 2 ) cos 20.0 = 1.88N. (b) The y component o
Chapter 4
1. The initial position vector ro satisfies r - ro = r , which results in
^ ^ ^ ^ ^ ro = r - r = (3.0^ - 4.0k)m - (2.0i - 3.0^ + 6.0 k)m = (-2.0 m) i + (6.0 m) ^ + (- 10 m) k . j j j
^ 2.
Chapter 3
r 1. A vector a can be represented in the magnitude-angle notation (a, ), where
2 a = ax + a 2 y
is the magnitude and a = tan - 1 y a x r is the angle a makes with the positive x axis. (a) G
Chapter 1
1. The metric prefixes (micro, pico, nano, .) are given for ready reference on the inside front cover of the textbook (see also Table 12). (a) Since 1 km = 1 103 m and 1 m = 1 106 m, 1km = 1
Chapter 44
1. The total rest energy of the electron-positron pair is E = me c 2 + me c 2 = 2me c 2 = 2(0.511 MeV) = 1.022 MeV . With two gamma-ray photons produced in the annihilation process, the wav
Chapter 43
1. If MCr is the mass of a 52Cr nucleus and MMg is the mass of a 26Mg nucleus, then the disintegration energy is Q = (MCr 2MMg)c2 = [51.94051 u 2(25.98259 u)](931.5 MeV/u) = 23.0 MeV. 2. Ad
Chapter 42
1. Kinetic energy (we use the classical formula since v is much less than c) is converted into potential energy (see Eq. 24-43). From Appendix F or G, we find Z = 3 for Lithium and Z = 90 f